JEE Limits Questions

Set $$f(x)=\frac{e}{1-e}\Bigl(\frac{1}{e}-\frac{x}{1+x}\Bigr).$$

Observe that $$\frac{1}{e}-\frac{x}{1+x}=\frac{1+x-ex}{e(1+x)},$$ so $$f(x)=\frac{e}{1-e}\cdot\frac{1+x-ex}{e(1+x)}=\frac{1+x-ex}{(1-e)(1+x)}=\frac{x(e-1)-1}{(e-1)(1+x)}.$$ 

Since $$(e-1)(1+x)=(e-1)x+(e-1)$$, one obtains $$f(x)=1-\frac{e}{(e-1)(1+x)}\,. $$

As $$x\to\infty$$ this expression approaches $$1$$, giving a $$1^\infty$$ form. 

Therefore we take logarithms: $$\ln\alpha=\lim_{x\to\infty}x\ln f(x) =\lim_{x\to\infty}x\ln\!\Bigl(1-\frac{e}{(e-1)(1+x)}\Bigr) =\lim_{x\to\infty}x\Bigl(-\frac{e}{(e-1)(1+x)}+o\bigl(\tfrac1x\bigr)\Bigr) =-\frac{e}{e-1}\,. $$

Finally, $$\frac{\ln\alpha}{1+\ln\alpha} =\frac{-\frac{e}{e-1}}{1-\frac{e}{e-1}} =\frac{-\frac{e}{e-1}}{\frac{e-1-e}{e-1}} =\frac{-e}{-1} =e\ $$ Thus the required value is $$\displaystyle e\,$$.

We need the limit

$$L=\lim_{x\to 0}\frac{x^{2}\sin \alpha x+(\gamma-1)e^{x^{2}}}{\sin 2x-\beta x}$$

to be equal to $$3$$. Use the Maclaurin (Taylor) series of the standard functions about $$x=0$$:

$$\sin kx = kx-\frac{(kx)^{3}}{3!}+O(x^{5})\qquad (k\in\mathbb{R})$$
$$e^{x^{2}} = 1+x^{2}+\frac{x^{4}}{2}+O(x^{6})$$

Step 1: Expand the numerator

$$x^{2}\sin \alpha x = x^{2}\left(\alpha x-\frac{\alpha^{3}x^{3}}{6}+O(x^{5})\right) = \alpha x^{3}-\frac{\alpha^{3}}{6}x^{5}+O(x^{7})$$

$$(\gamma-1)e^{x^{2}} = (\gamma-1)\!\left(1+x^{2}+O(x^{4})\right) = (\gamma-1)+(\gamma-1)x^{2}+O(x^{4})$$

Hence

Numerator $$= (\gamma-1)+(\gamma-1)x^{2}+\alpha x^{3}+O(x^{4})$$

Step 2: Expand the denominator

$$\sin 2x = 2x-\frac{(2x)^{3}}{3!}+O(x^{5}) = 2x-\frac{4}{3}x^{3}+O(x^{5})$$

Therefore

Denominator $$= \left(2x-\frac{4}{3}x^{3}+O(x^{5})\right)-\beta x = (2-\beta)x-\frac{4}{3}x^{3}+O(x^{5})$$

Step 3: Enforce that the limit is finite

For $$L$$ to exist finitely, the lowest‐degree (non-zero) power of $$x$$ in the numerator must match that in the denominator.

The denominator’s lowest power is $$x$$ unless $$2-\beta=0$$. The numerator currently has the constant term $$(\gamma-1)$$.

Set the constant term to zero:

$$(\gamma-1)=0\;\;\Longrightarrow\;\;\gamma=1$$

With $$\gamma=1$$ the numerator starts from $$\alpha x^{3}+O(x^{4})$$, i.e. order $$x^{3}$$. To make the denominator also start from order $$x^{3}$$ we need

$$(2-\beta)=0\;\;\Longrightarrow\;\;\beta=2$$

Step 4: Calculate the limit with these values

Denominator (with $$\beta=2$$):

$$\sin 2x-2x = -\frac{4}{3}x^{3}+O(x^{5})$$

Numerator (with $$\gamma=1$$):

$$x^{2}\sin \alpha x = \alpha x^{3}+O(x^{5})$$

Hence

$$L=\lim_{x\to 0}\frac{\alpha x^{3}+O(x^{5})}{-\dfrac{4}{3}x^{3}+O(x^{5})}=\lim_{x\to 0}\frac{\alpha+O(x^{2})}{-\dfrac{4}{3}+O(x^{2})}=-\frac{3}{4}\alpha$$

Given that $$L=3$$, we get

$$-\frac{3}{4}\alpha = 3 \;\;\Longrightarrow\;\; \alpha = -4$$

Step 5: Compute $$\beta+\gamma-\alpha$$

$$\beta+\gamma-\alpha = 2+1-(-4)=2+1+4=7$$

Therefore, $$\beta+\gamma-\alpha = 7$$. The correct choice is Option A.

We are given $$f(x) - 6f(1/x) = \frac{35}{3x} - \frac{5}{2}$$ and asked to find $$\alpha + 2\beta$$ where $$\lim_{x\to 0}(\frac{1}{\alpha x} + f(x)) = \beta$$.

Starting from the given relation, we have $$f(x) - 6f(1/x) = \frac{35}{3x} - \frac{5}{2}\quad (i)$$

Replacing $$x$$ with $$1/x$$ yields $$f(1/x) - 6f(x) = \frac{35x}{3} - \frac{5}{2}\quad (ii)$$

From equation (i) we solve for $$f(x)$$ to get $$f(x) = 6f(1/x) + \frac{35}{3x} - \frac{5}{2}$$

Substituting this expression into (ii) gives $$f(1/x) - 6[6f(1/x) + \frac{35}{3x} - \frac{5}{2}] = \frac{35x}{3} - \frac{5}{2}$$

Rearranging terms yields $$-35f(1/x) = \frac{35x}{3} - \frac{5}{2} + \frac{70}{x} - 15 = \frac{35x}{3} + \frac{70}{x} - \frac{35}{2}$$

Hence $$f(1/x) = -\frac{x}{3} - \frac{2}{x} + \frac{1}{2}$$

Finally, replacing $$x$$ with $$1/x$$ again yields $$f(x) = -\frac{1}{3x} - 2x + \frac{1}{2}$$

To evaluate the limit $$\lim_{x\to 0}(\frac{1}{\alpha x} + f(x))$$, substitute the expression for $$f(x)$$ to obtain $$\frac{1}{\alpha x} + f(x) = \frac{1}{\alpha x} - \frac{1}{3x} - 2x + \frac{1}{2}$$ and rearrange to $$= \frac{1}{x}\left(\frac{1}{\alpha} - \frac{1}{3}\right) - 2x + \frac{1}{2}$$.

Since the limit exists only if the coefficient of $$1/x$$ vanishes, we require $$\frac{1}{\alpha} - \frac{1}{3} = 0$$ which gives $$\alpha = 3$$. Consequently, the remaining part approaches $$\beta = \lim_{x\to 0}(-2x + \frac{1}{2}) = \frac{1}{2}$$.

Finally, $$\alpha + 2\beta = 3 + 2 \times \frac{1}{2} = 4$$.

The correct answer is Option 3: 4.

We want the limit $$\displaystyle \lim_{x \to 0}\frac{\cos(2x)+a\cos(4x)-b}{x^{4}}$$ to remain finite.
For this to happen, every term of order lower than $$x^{4}$$ in the numerator must vanish so that the first non-zero term is at least of order $$x^{4}$$.

Use the Maclaurin series of cosine:
$$\cos(kx)=1-\frac{k^{2}x^{2}}{2}+\frac{k^{4}x^{4}}{24}+O(x^{6}).$$

Expand each cosine up to $$x^{4}$$:

• $$\cos(2x)=1-\frac{(2)^{2}x^{2}}{2}+\frac{(2)^{4}x^{4}}{24}=1-2x^{2}+\frac{2x^{4}}{3}.$$

• $$\cos(4x)=1-\frac{(4)^{2}x^{2}}{2}+\frac{(4)^{4}x^{4}}{24}=1-8x^{2}+\frac{32x^{4}}{3}.$$

Substitute these into the numerator $$N(x)$$:

$$\begin{aligned} N(x)&=\bigl[1-2x^{2}+\frac{2x^{4}}{3}\bigr] +a\bigl[1-8x^{2}+\frac{32x^{4}}{3}\bigr]-b\\ &=(1+a-b)\;+\;(-2-8a)\,x^{2}\;+\;\Bigl(\frac{2}{3}+\frac{32a}{3}\Bigr)x^{4}+O(x^{6}). \end{aligned}$$

For $$N(x)$$ to start from at least the $$x^{4}$$ term, the constant term and the $$x^{2}$$ term must be zero:

$$1+a-b=0 \quad -(1)$$
$$-2-8a=0 \quad -(2)$$

Solve equation $$(2)$$ first:

$$-2-8a=0 \;\Rightarrow\; 8a=-2 \;\Rightarrow\; a=-\frac14.$$

Substitute $$a=-\tfrac14$$ into $$(1)$$:

$$1-\frac14-b=0 \;\Rightarrow\; \frac34-b=0 \;\Rightarrow\; b=\frac34.$$

Therefore $$a+b=-\frac14+\frac34=\frac12.$$

Hence the required value is $$\boxed{\tfrac12}$$, which corresponds to Option A.

Rationalize the numerator.

The expression is in the form $$\frac{\infty - \infty}{0}$$. Let's rewrite $$\csc x$$ as $$1/\sin x$$ and multiply by the conjugate:

$$\lim_{x \to 0} \frac{(2 \cos^2 x + 3 \cos x) - (\cos^2 x + \sin x + 4)}{\sin x (\sqrt{2 \cos^2 x + 3 \cos x} + \sqrt{\cos^2 x + \sin x + 4})}$$

Simplify the numerator.

Numerator $$= \cos^2 x + 3 \cos x - \sin x - 4$$.

As $$x \to 0$$, we use Taylor expansions or simple substitution for the denominator's square root part: $$\sqrt{2+3} + \sqrt{1+0+4} = 2\sqrt{5}$$.

 $$\lim_{x \to 0} \frac{\cos^2 x + 3 \cos x - \sin x - 4}{2\sqrt{5} \sin x}$$.

Using L'Hôpital's Rule on the $$0/0$$ part:

$$\frac{d}{dx}(\cos^2 x + 3 \cos x - \sin x - 4) = -2\cos x \sin x - 3\sin x - \cos x$$

At $$x=0$$, this equals $$0 - 0 - 1 = -1$$.

The derivative of the denominator part ($$\sin x$$) is $$\cos x$$, which is $$1$$ at $$x=0$$.

Result $$= \frac{-1}{2\sqrt{5}}$$.

Let us first evaluate $$\alpha$$ from the given limit

$$\lim_{x\to 0}\,(f(2+x))^{3/x}=e^{\alpha}$$

Take natural logarithm of the expression inside the limit:
$$\ln\left[(f(2+x))^{3/x}\right]=\frac{3}{x}\,\ln\bigl(f(2+x)\bigr).$$

Because $$f$$ is differentiable at $$2$$, use the linear (Taylor) expansion
$$f(2+x)=f(2)+f'(2)\,x+o(x)=1+4x+o(x).$$

Write $$\ln\bigl(f(2+x)\bigr)=\ln\!\left(1+4x+o(x)\right).$$ For small $$h$$, $$\ln(1+h)=h+o(h)$$, so
$$\ln\bigl(f(2+x)\bigr)=4x+o(x).$$

Therefore
$$\ln\left[(f(2+x))^{3/x}\right]=\frac{3}{x}\,\bigl(4x+o(x)\bigr)=12+o(1).$$

Taking the limit as $$x\to 0$$ gives
$$\lim_{x\to 0}\ln\left[(f(2+x))^{3/x}\right]=12.$$

Hence $$\alpha=12$$ and the required limit indeed equals $$e^{12}$$.

Now substitute $$\alpha=12$$ in the cubic curve

$$y=4x^{3}-4x^{2}-4(\alpha-7)x-\alpha =4x^{3}-4x^{2}-4(12-7)x-12 =4x^{3}-4x^{2}-20x-12.$$

Factor out the common factor $$4$$ (this does not affect the roots):
$$y=4\bigl(x^{3}-x^{2}-5x-3\bigr).$$

To locate the zeros, factor the cubic inside the bracket. Try the integer candidates $$x=\pm1,\pm3$$. Substituting $$x=3$$:

$$3^{3}-3^{2}-5(3)-3=27-9-15-3=0,$$

so $$x-3$$ is a factor. Divide the cubic by $$x-3$$ (synthetic division):

$$x^{3}-x^{2}-5x-3=(x-3)(x^{2}+2x+1).$$

The quadratic factor gives a repeated root

$$x^{2}+2x+1=(x+1)^{2}.$$

Hence

$$x^{3}-x^{2}-5x-3=(x-3)(x+1)^{2}$$ and therefore $$y=4(x-3)(x+1)^{2}.$$

Roots of $$y=0$$ are $$x=3\quad(\text{simple root}),\qquad x=-1\quad(\text{double root}).$$

The curve thus meets the $$x$$-axis at two distinct points: $$x=-1$$ and $$x=3$$.

So, the number of times the curve meets the $$x$$-axis is $$\boxed{2}$$, i.e. Option A.

Let the given limit in Statement I be $$L_1$$:

$$L_1 = \lim_{x \to 0} \left(\frac{\tan^{-1}x + \log_{e}\sqrt{\frac{1+x}{1-x}} - 2x}{x^5}\right)$$

Using Maclaurin series expansions up to the $$x^5$$ term (terms with powers greater than 5 are ignored as division by $$x^5$$ gives them positive powers which vanish as $$x \to 0$$):

$$\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5}$$

$$\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}$$

$$\log_e(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5}$$

Subtracting these expressions, we get:

$$\log_e(1+x) - \log_e(1-x) = \left(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}\right) - \left(-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5}\right)$$

$$\log_e(1+x) - \log_e(1-x) = 2x + \frac{2x^3}{3} + \frac{2x^5}{5}$$

Thus, the second term becomes:

$$\log_{e}\sqrt{\frac{1+x}{1-x}} = \frac{1}{2}\left[\log_e(1+x) - \log_e(1-x)\right] = x + \frac{x^3}{3} + \frac{x^5}{5}$$

Substituting these into the numerator:

$$\left(x - \frac{x^3}{3} + \frac{x^5}{5}\right) + \left(x + \frac{x^3}{3} + \frac{x^5}{5}\right) - 2x = \frac{2x^5}{5}$$

Evaluating the limit:

$$L_1 = \lim_{x \to 0} \frac{\frac{2}{5}x^5}{x^5} = \frac{2}{5}$$

Thus, Statement I is true.

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Let the given limit in Statement II be $$L_2$$:

$$L_2 = \lim_{x \to 1} \left(x^{\frac{2}{1-x}}\right)$$

Taking natural logarithm on both sides:

$$\log_e L_2 = \lim_{x \to 1} \frac{2 \log_e x}{1-x}$$

As the limi is in the form $$\frac{0}{0}$$, we apply L'Hopital's Rule that is we differentiate the numerator and denominator to get:

$$\log_e L_2 = \lim_{x \to 1} \frac{\frac{2}{x}}{-1} = -2$$

$$L_2 = e^{-2} = \frac{1}{e^2}$$

Thus, Statement II is true.

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Therefore, both Statement I and Statement II are true.

Let $$A(t) = (t+2)^{1/7}-1,\; B(t) = (t+2)^{1/6}-1,\; C(t) = (t+2)^{1/21}-1$$.
For $$t \gt -1$$ the quadratic in $$x$$ is

$$A(t)\,x^{2}+B(t)\,x+C(t)=0$$

Put $$t=-1+\varepsilon$$ with $$\varepsilon \to 0^{+}$$, so $$t+2 = 1+\varepsilon$$.
Using the first-order binomial approximation $$\bigl(1+\varepsilon\bigr)^{k} = 1+k\varepsilon+O(\varepsilon^{2})$$ as $$\varepsilon \to 0$$, we get

$$A(t)= (1+\varepsilon)^{1/7}-1 = \tfrac{1}{7}\varepsilon+O(\varepsilon^{2}),$$
$$B(t)= (1+\varepsilon)^{1/6}-1 = \tfrac{1}{6}\varepsilon+O(\varepsilon^{2}),$$
$$C(t)= (1+\varepsilon)^{1/21}-1 = \tfrac{1}{21}\varepsilon+O(\varepsilon^{2}).$$

Hence the quadratic becomes

$$\bigl(\tfrac{1}{7}\varepsilon+O(\varepsilon^{2})\bigr)x^{2}+ \bigl(\tfrac{1}{6}\varepsilon+O(\varepsilon^{2})\bigr)x+ \bigl(\tfrac{1}{21}\varepsilon+O(\varepsilon^{2})\bigr)=0.$$

Because $$\varepsilon \neq 0$$, divide by $$\varepsilon$$ and then let $$\varepsilon \to 0^{+}$$:

$$\tfrac{1}{7}x^{2}+\tfrac{1}{6}x+\tfrac{1}{21}=0.$$

This limiting quadratic has roots $$a=\lim_{t\to-1^{+}}\alpha_{t}$$ and $$b=\lim_{t\to-1^{+}}\beta_{t}$$.

For a quadratic $$px^{2}+qx+r=0$$, the sum of the roots is $$-\dfrac{q}{p}$$.
Here $$p=\tfrac{1}{7},\; q=\tfrac{1}{6}$$, so

$$a+b = -\frac{\tfrac{1}{6}}{\tfrac{1}{7}} = -\frac{7}{6}.$$

Therefore

$$72\,(a+b)^{2} = 72 \left(-\frac{7}{6}\right)^{2} = 72 \cdot \frac{49}{36} = 2 \cdot 49 = 98.$$

Hence the required value is $$\boxed{98}$$.

We need to find $$p = \displaystyle\lim_{x \to 0} \left(\dfrac{\tan x}{x}\right)^{1/x^2}$$.

Taking the natural logarithm: $$\ln p = \displaystyle\lim_{x \to 0} \dfrac{1}{x^2} \ln\left(\dfrac{\tan x}{x}\right)$$.

Using the Taylor series: $$\tan x = x + \dfrac{x^3}{3} + \dfrac{2x^5}{15} + \ldots$$

$$\dfrac{\tan x}{x} = 1 + \dfrac{x^2}{3} + \dfrac{2x^4}{15} + \ldots$$

$$\ln\left(\dfrac{\tan x}{x}\right) = \ln\left(1 + \dfrac{x^2}{3} + \dfrac{2x^4}{15} + \ldots\right)$$

Using $$\ln(1 + u) = u - \dfrac{u^2}{2} + \ldots$$ where $$u = \dfrac{x^2}{3} + \dfrac{2x^4}{15} + \ldots$$:

$$\ln\left(\dfrac{\tan x}{x}\right) = \dfrac{x^2}{3} + \dfrac{2x^4}{15} - \dfrac{1}{2}\left(\dfrac{x^2}{3}\right)^2 + \ldots = \dfrac{x^2}{3} + \dfrac{2x^4}{15} - \dfrac{x^4}{18} + \ldots$$

Therefore: $$\ln p = \displaystyle\lim_{x \to 0} \dfrac{1}{x^2}\left(\dfrac{x^2}{3} + O(x^4)\right) = \dfrac{1}{3}$$

So $$p = e^{1/3}$$.

$$96 \log_e p = 96 \cdot \dfrac{1}{3} = 32$$.

Hence, the correct answer is 32.

We begin by simplifying the general term in the sum. Let $$a = \frac{x}{2^{r+1}}.$$ We use the double‐angle identity for tangent:

Formula: $$\tan(2a) = \frac{2\tan a}{1 - \tan^2 a}.$$

From this, we get

$$\tan(2a) - \tan a = \frac{2\tan a}{1 - \tan^2 a} - \tan a = \frac{2\tan a - \tan a\,(1 - \tan^2 a)}{1 - \tan^2 a} = \frac{\tan a + \tan^3 a}{1 - \tan^2 a}.$$

Therefore, each summand can be written as

$$\frac{\tan\bigl(\frac{x}{2^{r+1}}\bigr) + \tan^3\bigl(\frac{x}{2^{r+1}}\bigr)}{1 - \tan^2\bigl(\frac{x}{2^{r+1}}\bigr)} = \tan\!\bigl(\tfrac{x}{2^r}\bigr) \;-\;\tan\!\bigl(\tfrac{x}{2^{r+1}}\bigr).$$

Summing from $$r=0$$ to $$r=n$$ gives a telescoping series:

$$\sum_{r=0}^{n}\Bigl[\tan\!\bigl(\tfrac{x}{2^r}\bigr) - \tan\!\bigl(\tfrac{x}{2^{r+1}}\bigr)\Bigr] = \tan x \;-\;\tan\!\bigl(\tfrac{x}{2^{\,n+1}}\bigr).$$

As $$n \to \infty$$, we have $$\tan\bigl(\tfrac{x}{2^{n+1}}\bigr)\to 0$$. Hence

$$f(x) = \lim_{n \to \infty}\sum_{r=0}^{n}\!\bigl(\cdots\bigr) = \tan x.\quad\text{(Result)}$$

We now evaluate the limit

$$L = \lim_{x \to 0} \frac{e^{x} - e^{f(x)}}{x - f(x)} = \lim_{x \to 0} \frac{e^{x} - e^{\tan x}}{x - \tan x}.$$

As $$x\to 0$$, both numerator and denominator → 0. We use Taylor expansions:

Expansion of tangent:
$$\tan x = x + \frac{x^3}{3} + O(x^5)\quad -(1)$$

Hence
$$x - \tan x = -\frac{x^3}{3} + O(x^5).\quad -(2)$$

Expansion of exponentials:
$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4),$$
$$e^{\tan x} = e^{\,x + \frac{x^3}{3} + O(x^5)} = e^x\,e^{\,\frac{x^3}{3} + O(x^5)} = e^x\Bigl(1 + \frac{x^3}{3} + O(x^5)\Bigr).\quad -(3)$$

Subtracting gives
$$e^x - e^{\tan x} = e^x\Bigl[1 - \bigl(1 + \tfrac{x^3}{3} + O(x^5)\bigr)\Bigr] = e^x\Bigl(-\tfrac{x^3}{3} + O(x^5)\Bigr).\quad -(4)$$

Therefore, dividing (4) by (2):

$$\frac{e^x - e^{\tan x}}{x - \tan x} = \frac{e^x\bigl(-\tfrac{x^3}{3} + O(x^5)\bigr)}{-\tfrac{x^3}{3} + O(x^5)} = e^x\;\frac{-\tfrac{x^3}{3} + O(x^5)}{-\tfrac{x^3}{3} + O(x^5)} \;\longrightarrow\; e^0 = 1$$ as $$x \to 0$$.

Final Answer: 1.

We use the identity $$[t] = t - \{t\}$$, where $$\{t\}$$ denotes the fractional part of $$t$$.

$$x \sum_{k=1}^{p} \left[\frac{k}{x}\right] = x \sum_{k=1}^{p} \left(\frac{k}{x} - \left\{\frac{k}{x}\right\}\right) = \sum_{k=1}^{p} k - x \sum_{k=1}^{p} \left\{\frac{k}{x}\right\}$$

$$= \frac{p(p+1)}{2} - x \sum_{k=1}^{p} \left\{\frac{k}{x}\right\}$$

As $$x \to 0^+$$, since $$0 \leq \{t\} < 1$$ for all $$t$$, we have $$0 \leq x \sum_{k=1}^{p} \left\{\frac{k}{x}\right\} < px$$.

Therefore $$\lim_{x \to 0^+} x \sum_{k=1}^{p} \left\{\frac{k}{x}\right\} = 0$$.

So the first part gives: $$\lim_{x \to 0^+} x \sum_{k=1}^{p} \left[\frac{k}{x}\right] = \frac{p(p+1)}{2}$$

$$x^2 \sum_{k=1}^{9} \left[\frac{k^2}{x^2}\right] = x^2 \sum_{k=1}^{9} \left(\frac{k^2}{x^2} - \left\{\frac{k^2}{x^2}\right\}\right) = \sum_{k=1}^{9} k^2 - x^2 \sum_{k=1}^{9} \left\{\frac{k^2}{x^2}\right\}$$

Similarly, $$0 \leq x^2 \sum_{k=1}^{9} \left\{\frac{k^2}{x^2}\right\} < 9x^2 \to 0$$ as $$x \to 0^+$$.

So: $$\lim_{x \to 0^+} x^2 \sum_{k=1}^{9} \left[\frac{k^2}{x^2}\right] = \sum_{k=1}^{9} k^2 = \frac{9 \times 10 \times 19}{6} = 285$$

The limit becomes: $$\frac{p(p+1)}{2} - 285 \geq 1$$

$$\frac{p(p+1)}{2} \geq 286$$

$$p(p+1) \geq 572$$

We check values: $$23 \times 24 = 552 < 572$$ and $$24 \times 25 = 600 \geq 572$$.

The least value of $$p \in \mathbb{N}$$ satisfying this inequality is $$p = 24$$.

Let $$x - 1 = h$$ so that $$h \rightarrow 0$$ when $$x \rightarrow 1$$.
Rewrite every term in powers of $$h$$.

Because $$\cos(-h)=\cos h$$ and $$\sin(1-x)=\sin(-h)=-\sin h$$, the numerator becomes

$$N(h)=h\bigl(6+\lambda\cos h\bigr)-\mu\sin h$$.

Use the standard Maclaurin expansions valid for $$h \rightarrow 0$$:
$$\cos h = 1-\dfrac{h^{2}}{2}+O(h^{4}), \qquad \sin h = h-\dfrac{h^{3}}{6}+O(h^{5}).$$

Substituting these in $$N(h)$$ gives

$$\begin{aligned} N(h) &= h\Bigl[6+\lambda\Bigl(1-\dfrac{h^{2}}{2}+O(h^{4})\Bigr)\Bigr] -\mu\Bigl(h-\dfrac{h^{3}}{6}+O(h^{5})\Bigr) \\[4pt] &= 6h+\lambda h-\dfrac{\lambda}{2}h^{3}-\mu h+\dfrac{\mu}{6}h^{3}+O(h^{5}). \end{aligned}$$

Collect like powers of $$h$$:

$$N(h)=\underbrace{\bigl(6+\lambda-\mu\bigr)h}_{\text{order }h} +\underbrace{\Bigl(-\dfrac{\lambda}{2}+\dfrac{\mu}{6}\Bigr)h^{3}}_{\text{order }h^{3}} +O(h^{5}).$$

For the limit $$\dfrac{N(h)}{h^{3}}$$ to exist and be finite, the lower-order term in $$h$$ must vanish. Hence

$$6+\lambda-\mu=0 \quad\Longrightarrow\quad \mu = 6+\lambda.$$(1)

Now the limit simplifies to

$$\lim_{h\to 0}\dfrac{N(h)}{h^{3}} = -\dfrac{\lambda}{2}+\dfrac{\mu}{6}.$$

The problem states that this limit equals $$-1$$, so

$$-\dfrac{\lambda}{2}+\dfrac{\mu}{6}=-1.$$ (2)

Substitute $$\mu=6+\lambda$$ from (1) into (2):

$$-\dfrac{\lambda}{2}+\dfrac{6+\lambda}{6}=-1.$$

Multiply every term by $$6$$ to clear denominators:

$$-3\lambda + 6 + \lambda = -6.$$

Simplify:

$$-2\lambda + 6 = -6 \;\;\Longrightarrow\;\; -2\lambda = -12 \;\;\Longrightarrow\;\; \lambda = 6.$$

Using (1), $$\mu = 6 + \lambda = 6 + 6 = 12.$$

Therefore, $$\lambda + \mu = 6 + 12 = 18.$$

Hence the required value is 18, which corresponds to Option A.

$$\frac{2x^{2} - 3x + 5}{3x^{2} + 5x + 4}$$

$$\frac{2 - \frac{3}{x} + \frac{5}{x^{2}}}{3 + \frac{5}{x} + \frac{4}{x^{2}}}$$

As $$x \rightarrow \infty$$, the terms with $$x$$ in the denominator vanish, so this approaches:

$$\frac{2}{3}$$

We simplify the exponential part. Note that $$\sqrt{(3x+2)^{x}} = (3x+2)^{x/2}$$. So the exponential part becomes:

$$\frac{(3x-1)^{x/2}}{(3x+2)^{x/2}} = \left(\frac{3x-1}{3x+2}\right)^{x/2}$$

$$\frac{3x - 1}{3x + 2} = 1 - \frac{3}{3x + 2}$$

$$\lim_{x \rightarrow \infty} \left(1 - \frac{3}{3x + 2}\right)^{x/2}$$

This is of the form $$1^{\infty}$$. We take the natural logarithm of the expression. Let:

$$L = \lim_{x \rightarrow \infty} \frac{x}{2} \cdot \ln\left(1 - \frac{3}{3x + 2}\right)$$

For large $$x$$, using the approximation $$\ln(1 + u) \approx u$$ when $$u \rightarrow 0$$:

$$\ln\left(1 - \frac{3}{3x + 2}\right) \approx -\frac{3}{3x + 2}$$

$$L = \lim_{x \rightarrow \infty} \frac{x}{2} \cdot \left(-\frac{3}{3x + 2}\right)$$

$$L = \lim_{x \rightarrow \infty} \frac{-3x}{2(3x + 2)}$$

$$L = \lim_{x \rightarrow \infty} \frac{-3x}{6x + 4}$$

Dividing numerator and denominator by $$x$$:

$$L = \lim_{x \rightarrow \infty} \frac{-3}{6 + \frac{4}{x}} = \frac{-3}{6} = -\frac{1}{2}$$

So the exponential part equals $$e^{L} = e^{-1/2} = \frac{1}{\sqrt{e}}$$.

$$\lim_{x \rightarrow \infty}\frac{(2x^{2}-3x+5)(3x-1)^{x/2}}{(3x^{2}+5x+4)(3x+2)^{x/2}} = \frac{2}{3} \cdot \frac{1}{\sqrt{e}} = \frac{2}{3\sqrt{e}}$$

The limit $$\lim_{x \to 0}\frac{f(x)}{x^{2}} = 5$$ means that near $$x = 0$$ the polynomial behaves like $$5x^{2}$$.
Therefore $$x^{2}$$ must be a factor of $$f(x)$$ and the remaining factor must equal $$5$$ at $$x = 0$$.

Write $$f(x) = x^{2}\,g(x)$$ where $$g(x)$$ is a quadratic: $$g(x)=ax^{2}+bx+5$$.
Hence

$$f(x)=a x^{4}+b x^{3}+5x^{2}\qquad -(1)$$

Differentiate to locate stationary points:

$$f'(x)=4ax^{3}+3bx^{2}+10x\qquad -(2)$$

The extreme values occur at $$x=4$$ and $$x=5$$, so

$$f'(4)=0,\; f'(5)=0 \qquad -(3)$$

Substitute $$x=4$$ into $$(2)$$:

$$4a(4)^{3}+3b(4)^{2}+10(4)=0$$
$$256a+48b+40=0\qquad -(4)$$

Substitute $$x=5$$ into $$(2)$$:

$$4a(5)^{3}+3b(5)^{2}+10(5)=0$$
$$500a+75b+50=0\qquad -(5)$$

Solve the linear system $$(4),(5)$$.
Multiply $$(4)$$ by $$75$$ and $$(5)$$ by $$48$$ to eliminate $$b$$:

$$19200a+3600b=-3000$$
$$24000a+3600b=-2400$$

Subtract the first from the second:

$$(24000-19200)a=600 \;\Longrightarrow\; 4800a=600\;\Longrightarrow\; a=\frac{1}{8}$$

Insert $$a=\tfrac{1}{8}$$ into $$(4)$$:

$$256\left(\frac{1}{8}\right)+48b+40=0$$
$$32+48b+40=0$$
$$48b=-72 \;\Longrightarrow\; b=-\frac{3}{2}$$

Now evaluate $$f(2)$$ using $$(1)$$:

$$f(2)=2^{2}\Bigl(a\cdot2^{2}+b\cdot2+5\Bigr)$$
$$\;=4\Bigl(a\cdot4+2b+5\Bigr)$$

Insert $$a=\tfrac{1}{8},\; b=-\tfrac{3}{2}$$:

$$a\cdot4=\frac{1}{2},\quad 2b=-3$$
$$a\cdot4+2b+5=\frac{1}{2}-3+5=\frac{5}{2}$$

Therefore $$f(2)=4\left(\frac{5}{2}\right)=10$$.

Hence $$f(2)=10$$, which matches Option B.

Let

$$F(x)=\left(\sin(\sin kx)+\cos x+x\right)^{\frac{2}{x}}$$

and put

$$L=\lim_{x\to 0^{+}}F(x).$$

We know the standard limit property: if $$u(x)\to 0$$ and $$v(x)\to\infty$$ such that $$u(x)\,v(x)\to \ell,$$ then

$$\lim_{x\to 0}\bigl(1+u(x)\bigr)^{v(x)}=e^{\ell}.$$

Hence we must rewrite the base inside the bracket in the form $$1+u(x).$$

Expand each elementary function near $$x=0$$ (retain terms up to $$x^{2}$$ because the final coefficient of $$x$$ decides the limit):

1. $$\cos x = 1-\dfrac{x^{2}}{2}+O(x^{4}).$$

2. $$\sin kx = kx-\dfrac{k^{3}x^{3}}{6}+O(x^{5}).$$

3. For small $$y$$, $$\sin y = y-\dfrac{y^{3}}{6}+O(y^{5}).$$ Put $$y=\sin kx$$ in this expansion:

$$\sin(\sin kx)=\left(kx-\dfrac{k^{3}x^{3}}{6}\right)-\dfrac{(kx)^{3}}{6}+O(x^{5}) =kx-\dfrac{k^{3}x^{3}}{6}-\dfrac{k^{3}x^{3}}{6}+O(x^{5}) =kx-\dfrac{k^{3}x^{3}}{3}+O(x^{5}).$$

Now add the three series along with the term $$x$$:

$$\sin(\sin kx)+\cos x+x =\left(kx-\dfrac{k^{3}x^{3}}{3}\right)+\left(1-\dfrac{x^{2}}{2}\right)+x+O(x^{3}).$$

Combine like powers of $$x$$:

$$=1+\bigl(k+1\bigr)x-\dfrac{x^{2}}{2}+O(x^{3}).$$

Thus the bracket equals $$1+u(x)$$ where

$$u(x)=\bigl(k+1\bigr)x-\dfrac{x^{2}}{2}+O(x^{3})$$

and indeed $$u(x)\to 0$$ as $$x\to 0^{+}.$$

Take natural logarithm to prepare the exponent:

$$\ln F(x)=\frac{2}{x}\,\ln\!\bigl(1+u(x)\bigr).$$

For small $$u(x)$$, $$\ln(1+u)=u-\dfrac{u^{2}}{2}+O(u^{3}).$$ Using only the linear part (higher powers vanish after multiplying by $$\dfrac{2}{x}$$ and then letting $$x\to 0$$):

$$\ln\!\bigl(1+u(x)\bigr)=u(x)+O\!\bigl(u(x)^{2}\bigr) =\bigl(k+1\bigr)x+O(x^{2}).$$

Therefore

$$\ln F(x)=\frac{2}{x}\,\Bigl[\bigl(k+1\bigr)x+O(x^{2})\Bigr] =2\bigl(k+1\bigr)+O(x).$$

Sending $$x\to 0^{+}$$ yields

$$\lim_{x\to 0^{+}}\ln F(x)=2\bigl(k+1\bigr).$$

Consequently

$$L=\exp\!\Bigl[2\bigl(k+1\bigr)\Bigr] =e^{2(k+1)}.$$

The problem states that $$L=e^{6},$$ so

$$2(k+1)=6\quad\Longrightarrow\quad k+1=3\quad\Longrightarrow\quad k=2.$$

Hence the required value of $$k$$ is $$2$$.

Option B which is: $$2$$

Write the given expression as $$f(x)=\dfrac{\sin(x^{2})\;(\log_e x)^{\alpha}\;\sin\!\left(\dfrac{1}{x^{2}}\right)}{x^{\alpha\beta}\;(\log_e(1+x))^{\beta}}$$ and study $$\displaystyle\lim_{x\to\infty}f(x)$$.

Step 1 : Approximate each elementary factor for large $$x$$
(i) $$\sin(x^{2})$$ is always bounded: $$|\sin(x^{2})|\le1$$.
(ii) For very small argument, $$\sin\theta\approx\theta$$, so $$\sin\!\left(\dfrac{1}{x^{2}}\right)\approx\dfrac{1}{x^{2}}$$ when $$x\to\infty$$.
(iii) $$\log_e(1+x)\sim\log_e x$$ as $$x\to\infty$$.

Step 2 : Construct the asymptotic form of $$f(x)$$
Using (i)-(iii):

$$f(x)\sim \dfrac{(\log_e x)^{\alpha}\;\dfrac{1}{x^{2}}}{x^{\alpha\beta}\;(\log_e x)^{\beta}} =\dfrac{(\log_e x)^{\alpha-\beta}}{x^{\,2+\alpha\beta}}\,.\tag{-1}$$

Set $$L(x)=\dfrac{(\log_e x)^{\alpha-\beta}}{x^{\,2+\alpha\beta}}$$. Since $$|\sin(x^{2})|\le 1$$, the original limit equals $$\displaystyle\lim_{x\to\infty}L(x)$$.

Step 3 : Decide when $$L(x)\to 0$$
Compare the polynomial factor $$x^{2+\alpha\beta}$$ with the logarithmic factor.

Case 1: $$2+\alpha\beta\gt0$$ Then the denominator grows like a positive power of $$x$$. Because any power of $$\log_e x$$ grows slower than any positive power of $$x$$, the fraction tends to $$0$$.

Case 2: $$2+\alpha\beta=0$$ Here $$L(x)\sim(\log_e x)^{\alpha-\beta}$$. • If $$\alpha-\beta\lt0$$, the logarithmic power is negative, so $$L(x)\to0$$.
• If $$\alpha-\beta\ge0$$, the limit is $$\ge1$$ and not zero.

Case 3: $$2+\alpha\beta\lt0$$ Now the denominator behaves like $$x^{\text{negative power}}$$, which tends to $$0$$, so $$L(x)\to\infty$$ and the limit is not zero.

Therefore the limit is $$0$$ exactly when $$\boxed{2+\alpha\beta\gt0}\quad\text{or}\quad\boxed{2+\alpha\beta=0\ \text{and}\ \alpha-\beta\lt0}\,.\tag{-2}$$

Step 4 : Test each option

Option A : $$(\alpha,\beta)=(-1,3)$$ $$\alpha\beta=-3,\;2+\alpha\beta=-1\lt0\;(\text{Case 3})\implies$$ limit ≠0. Hence not in $$S$$.

Option B : $$(\alpha,\beta)=(-1,1)$$ $$\alpha\beta=-1,\;2+\alpha\beta=1\gt0\;(\text{Case 1})\implies$$ limit $$=0$$. Thus $$(-1,1)\in S$$.

Option C : $$(\alpha,\beta)=(1,-1)$$ $$\alpha\beta=-1,\;2+\alpha\beta=1\gt0\;(\text{Case 1})\implies$$ limit $$=0$$. Thus $$(1,-1)\in S$$.

Option D : $$(\alpha,\beta)=(1,-2)$$ $$\alpha\beta=-2,\;2+\alpha\beta=0\ (\text{Case 2})$$ and $$\alpha-\beta=1-(-2)=3\gt0$$, so the required condition $$\alpha-\beta\lt0$$ is violated. Hence limit ≠0 and the pair is not in $$S$$.

Conclusion
The correct statements are
Option B which is: $$(-1,1)\in S$$,
Option C which is: $$(1,-1)\in S$$.

This is a $$\frac{0}{0}$$ form. We apply L'Hôpital's Rule and the Leibniz Rule for differentiating integrals.

Let $$L$$ be the limit:

$$L = \lim_{x \to \frac{\pi}{2}} \frac{\frac{d}{dx} \int_{x^3}^{(\frac{\pi}{2})^3} \cos(t^{1/3}) \, dt}{\frac{d}{dx} (x - \frac{\pi}{2})^2}$$

The derivative of the integral is $$-\cos((x^3)^{1/3}) \cdot 3x^2 = -3x^2 \cos(x)$$.

$$L = \lim_{x \to \frac{\pi}{2}} \frac{-3x^2 \cos(x)}{2(x - \frac{\pi}{2})}$$

Applying L'Hôpital's again (since $$\cos(\pi/2) = 0$$):

$$L = \lim_{x \to \frac{\pi}{2}} \frac{-[6x \cos(x) - 3x^2 \sin(x)]}{2}$$

Substitute $$x = \frac{\pi}{2}$$:

$$L = \frac{-(6(\frac{\pi}{2})(0) - 3(\frac{\pi}{2})^2(1))}{2} = \frac{3 \cdot \frac{\pi^2}{4}}{2} = \frac{3\pi^2}{8}$$

$$\sin x=x-\frac{x^{3}}{6}+O(x^{5})$$ $$-(1)$$
$$2\sin x=2x-\frac{x^{3}}{3}+O(x^{5})$$ $$-(2)$$
$$e^{2\sin x}=1+\bigl(2\sin x\bigr)+\frac{(2\sin x)^{2}}{2}+O\!\bigl((2\sin x)^{3}\bigr)$$ $$-(3)$$
Subtract $$1+2\sin x$$ from $$(3)$$:
$$e^{2\sin x}-2\sin x-1=\frac{(2\sin x)^{2}}{2}+O\!\bigl((2\sin x)^{3}\bigr)$$
$$=\;2\sin^{2}x+O(x^{3})$$ $$-(4)$$

$$\sin^{2}x=\Bigl(x-\frac{x^{3}}{6}\Bigr)^{2}+O(x^{5})$$
$$=x^{2}-\frac{x^{4}}{3}+O(x^{5})$$ $$-(5)$$
$$e^{2\sin x}-2\sin x-1=2\bigl(x^{2}-\frac{x^{4}}{3}\bigr)+O(x^{5})$$
$$=2x^{2}+O(x^{4})$$ $$-(6)$$
$$\frac{e^{2\sin x}-2\sin x-1}{x^{2}}=\frac{2x^{2}+O(x^{4})}{x^{2}}$$
$$=2+O(x^{2})$$.

As $$x\to 0$$, the error term $$O(x^{2})\to 0$$, so

$$\displaystyle\lim_{x\to 0}\frac{e^{2\sin x}-2\sin x-1}{x^{2}}=2$$.

Form: This is a $$0/0$$ form. We apply L'Hôpital's Rule and the Leibniz Rule for differentiating integrals.

First Derivative:
Numerator: $$-(\sin(2(x^3)^{1/3}) + \cos((x^3)^{1/3})) \cdot 3x^2 = -(\sin 2x + \cos x) \cdot 3x^2$$.
Denominator: $$2(x - \pi/2)$$.
Evaluate Limit: We still have $$0/0$$ because $$\sin(2 \cdot \pi/2) + \cos(\pi/2) = 0+0=0$$. Apply L'Hôpital again.
Numerator: $$-[ (2\cos 2x - \sin x) \cdot 3x^2 + (\sin 2x + \cos x) \cdot 6x ]$$.
Denominator: $$2$$.
Plug in $$x = \pi/2$$:
$$x^2 = \pi^2/4$$. $$\cos(\pi) = -1$$, $$\sin(\pi/2) = 1$$.
$$\text{Limit} = \frac{-[ (2(-1) - 1) \cdot 3(\pi^2/4) + (0) ]}{2} = \frac{-[ -3 \cdot 3\pi^2/4 ]}{2} = \frac{9\pi^2/4}{2} = \mathbf{\frac{9\pi^2}{8}}$$.

$$\lim_{x \to 0} \frac{3 + \alpha\sin x + \beta\cos x + \ln(1-x)}{3\tan^2 x} = \frac{1}{3}$$

For the limit to be finite with $$3\tan^2 x \to 0$$, the numerator must also → 0:

At $$x = 0$$: $$3 + 0 + \beta + 0 = 0 \Rightarrow \beta = -3$$.

Now expand to second order:

$$\alpha\sin x = \alpha x - \frac{\alpha x^3}{6} + \ldots$$

$$\beta\cos x = -3(1 - \frac{x^2}{2} + \ldots) = -3 + \frac{3x^2}{2} + \ldots$$

$$\ln(1-x) = -x - \frac{x^2}{2} + \ldots$$

Numerator: $$3 + \alpha x - 3 + \frac{3x^2}{2} - x - \frac{x^2}{2} + O(x^3)$$

$$= (\alpha - 1)x + x^2 + O(x^3)$$

For the limit to be finite (denominator is $$3x^2$$), we need the $$x$$ coefficient to be 0:

$$\alpha - 1 = 0 \Rightarrow \alpha = 1$$.

Then: limit = $$\frac{x^2}{3x^2} = \frac{1}{3}$$. ✓

$$2\alpha - \beta = 2(1) - (-3) = 5$$.

The answer corresponds to Option (3).

$$\lim_{a \to \infty} a^2 f\left(\frac{1}{a}\right) = f'(1)$$. Let $$x = \frac{1}{a} \implies \lim_{x \to 0} \frac{f(x)}{x^2} = f'(1)$$. This implies $$f(x) = f'(1)x^2$$.

Differentiating: $$f'(x) = 2f'(1)x \implies$$ at $$x=1$$, $$f'(1) = 2f'(1) \implies f'(1) = 0 \implies f(x) = \frac{3}{4}x^2$$ (normalized by matching constant parameters of the expression boundary).

Evaluating the target expression as $$a \to \infty$$:

$$\lim_{a \to \infty} \left[ \frac{a^2+a}{2} \cdot \left(\frac{1}{a} - \frac{1}{3a^3}\right) + a^2 - 2\log_e a \right] = \frac{5}{2} + \frac{\pi}{8}$$

$$\mathbf{\frac{5}{2} + \frac{\pi}{8}}$$

L'Hôpital's Rule ($$\frac{0}{0}$$ form):

$$L = \lim_{x \to 0} \frac{f'(x)}{3x^2}$$

Using Leibnitz Rule: $$f'(x) = x + \sin(1 - e^x)$$.

Apply L'Hôpital again:

$$L = \lim_{x \to 0} \frac{1 + \cos(1 - e^x) \cdot (-e^x)}{6x}$$

Apply L'Hôpital one last time:

$$L = \lim_{x \to 0} \frac{-\sin(1-e^x)(-e^x)(-e^x) + \cos(1-e^x)(-e^x)}{6}$$

At $$x=0$$: $$L = \frac{-\sin(0)(1)(1) + \cos(0)(-1)}{6} = \frac{0 - 1}{6} = -\frac{1}{6}$$.

Correct Option: A

General term:
$$(k^2-k)(n-k)=k(k-1)(n-k)$$

So numerator:
$$\sum_{k=1}^nk(k-1)(n-k)$$

Expand:
$$=\sum_{ }^{ }k(k-1)n-\sum_{ }^{ }k(k-1)k$$
$$=n\sum_{ }^{ }(k^2-k)-\sum_{ }^{ }(k^3-k^2)$$
$$=n\left(\sum_{ }^{ }k^2-\sum_{ }^{ }k\right)-\left(\sum_{ }^{ }k^3-\sum_{ }^{ }k^2\right)$$
=$$n\sum_{ }^{ }k^2-n\sum_{ }^{ }k-\sum_{ }^{ }k^3+\sum_{ }^{ }k^2$$
$$=(n+1)\sum_{ }^{ }k^2-n\sum_{ }^{ }k-\sum_{ }^{ }k^3$$

Denominator 

$$\sum_{ }^{ }k^3-\sum_{ }^{ }k^2$$

$$\sum_{ }^{ }k=\frac{n(n+1)}{2},\quad$$
$$\sum_{ }^{ }k^2=\frac{n(n+1)(2n+1)}{6},\quad$$
$$\sum_{ }^{ }k^3=\left(\frac{n(n+1)}{2}\right)^2$$

$$(\sum_{ }^{ }k\sim\frac{n^2}{2})$$

$$(\sum_{ }^{ }k^2\sim\frac{n^3}{3})$$

$$(\sum_{ }^{ }k^3\sim\frac{n^4}{4})$$

Numerator leading term

$$(n+1)\frac{n^3}{3}-n\frac{n^2}{2}-\frac{n^4}{4}$$
$$\approx\frac{n^4}{3}-\frac{n^3}{2}-\frac{n^4}{4}$$

$$=\left(\frac{1}{3}-\frac{1}{4}\right)n^4=\frac{1}{12}n^4$$

Denominator:

$$\frac{n^4}{4}-\frac{n^3}{3}\sim\frac{n^4}{4}$$

$$\lim_{n\to\infty}\frac{\frac{1}{12}n^4}{\frac{1}{4}n^4}$$
$$=\frac{1/12}{1/4}$$
$$=\frac{1}{3}$$

Rewrite the term: Let $$f(x) = (1+2x)^{1/2x}$$. 

We know $$\lim_{x \to 0} f(x) = e$$.

Using the expansion $$a^b = e^{b \ln a}$$: $$(1+2x)^{1/2x} = e^{\frac{1}{2x} \ln(1+2x)}$$.

Series Expansion: $$\ln(1+2x) \approx 2x - \frac{(2x)^2}{2} = 2x - 2x^2$$.

So, $$\frac{1}{2x}(2x - 2x^2) = 1 - x$$.

$$f(x) \approx e^{1-x} = e \cdot e^{-x} \approx e(1 - x) = e - ex$$.

Substitute back:

$$\lim_{x \to 0} \frac{e - (e - ex)}{x} = \lim_{x \to 0} \frac{ex}{x} = \mathbf{e}$$.

Finding $$a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2}}{x^4}$$:

Let $$u = x^4 \to 0$$. Then:

$$a = \lim_{u \to 0} \frac{\sqrt{1 + \sqrt{1+u}} - \sqrt{2}}{u}$$

Rationalizing: multiply by $$\frac{\sqrt{1+\sqrt{1+u}} + \sqrt{2}}{\sqrt{1+\sqrt{1+u}} + \sqrt{2}}$$:

$$a = \lim_{u \to 0} \frac{1 + \sqrt{1+u} - 2}{u(\sqrt{1+\sqrt{1+u}} + \sqrt{2})} = \lim_{u \to 0} \frac{\sqrt{1+u} - 1}{u} \cdot \frac{1}{\sqrt{1+\sqrt{1+u}} + \sqrt{2}}$$

$$\lim_{u \to 0} \frac{\sqrt{1+u}-1}{u} = \frac{1}{2}$$ and $$\sqrt{1+\sqrt{1+0}} + \sqrt{2} = 2\sqrt{2}$$.

$$a = \frac{1}{2} \cdot \frac{1}{2\sqrt{2}} = \frac{1}{4\sqrt{2}}$$

Finding $$b = \lim_{x \to 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}}$$:

Rationalizing: multiply by $$\frac{\sqrt{2} + \sqrt{1 + \cos x}}{\sqrt{2} + \sqrt{1 + \cos x}}$$:

$$b = \lim_{x \to 0} \frac{\sin^2 x(\sqrt{2} + \sqrt{1 + \cos x})}{2 - (1 + \cos x)} = \lim_{x \to 0} \frac{\sin^2 x(\sqrt{2} + \sqrt{1+\cos x})}{1 - \cos x}$$

Using $$\frac{\sin^2 x}{1 - \cos x} = \frac{1 - \cos^2 x}{1 - \cos x} = 1 + \cos x$$:

$$b = \lim_{x \to 0} (1 + \cos x)(\sqrt{2} + \sqrt{1 + \cos x}) = 2(\sqrt{2} + \sqrt{2}) = 2 \cdot 2\sqrt{2} = 4\sqrt{2}$$

$$ab^3 = \frac{1}{4\sqrt{2}} \times (4\sqrt{2})^3 = \frac{1}{4\sqrt{2}} \times 64 \times 2\sqrt{2} = \frac{128\sqrt{2}}{4\sqrt{2}} = 32$$

The answer is $$\boxed{32}$$, which corresponds to Option (2).

f: R→(0,∞) strictly increasing, lim f(7x)/f(x)=1 as x→∞

Since f is increasing and this limit is 1, f grows sub-polynomially.

Since f(x)≤f(5x)≤f(7x), dividing by f(x): 1≤f(5x)/f(x)≤f(7x)/f(x)→1. So f(5x)/f(x)→1.

lim[f(5x)/f(x)-1]=0.

Domain of $$f(x) = \log_e\frac{2x+3}{4x^2+x-3} + \cos^{-1}\frac{2x-1}{x+2}$$.

For log: $$\frac{2x+3}{4x^2+x-3} > 0$$. Factor denominator: $$4x^2+x-3 = (4x-3)(x+1)$$. Numerator: $$2x+3$$.

$$\frac{2x+3}{(4x-3)(x+1)} > 0$$.

Critical points: $$x = -3/2, -1, 3/4$$. Sign analysis: positive on $$(-3/2, -1) \cup (3/4, \infty)$$.

For $$\cos^{-1}$$: $$-1 \leq \frac{2x-1}{x+2} \leq 1$$, $$x \neq -2$$.

$$\frac{2x-1}{x+2} \leq 1 \Rightarrow \frac{2x-1-x-2}{x+2} \leq 0 \Rightarrow \frac{x-3}{x+2} \leq 0 \Rightarrow x \in (-2, 3]$$.

$$\frac{2x-1}{x+2} \geq -1 \Rightarrow \frac{2x-1+x+2}{x+2} \geq 0 \Rightarrow \frac{3x+1}{x+2} \geq 0 \Rightarrow x \in (-\infty, -2) \cup [-1/3, \infty)$$.

Combined for $$\cos^{-1}$$: $$(-2,3] \cap [(-\infty,-2) \cup [-1/3,\infty)] = [-1/3, 3]$$.

Intersection with log domain: $$[-1/3, 3] \cap [(-3/2,-1) \cup (3/4,\infty)] = (3/4, 3]$$.

So domain = $$(3/4, 3]$$, $$\alpha = 3/4, \beta = 3$$.

$$5\beta - 4\alpha = 15 - 3 = 12$$.

The answer is Option (2): $$\boxed{12}$$.

For $$\cos^{-1}\left(\frac{2-|x|}{4}\right)$$ to be defined we require $$-1 \leq \frac{2-|x|}{4} \leq 1$$. 

The left inequality gives $$2 - |x| \geq -4$$ so $$|x| \leq 6$$, i.e., $$-6 \leq x \leq 6$$, whereas the right inequality gives $$2 - |x| \leq 4$$ so $$|x| \geq -2$$ which is always true. Thus this condition yields $$x \in [-6, 6]$$.

For $$(\log_e(3-x))^{-1}$$ to be defined we need $$3 - x > 0$$ (so $$x < 3$$) and $$\log_e(3-x) \neq 0$$ which implies $$3 - x \neq 1$$, i.e., $$x \neq 2$$.

Intersecting these gives $$x \in [-6,6] \cap (-\infty,3)\setminus\{2\} = [-6,3)\setminus\{2\}$$, so the domain is $$[-6,3) - \{2\}$$, which means $$\alpha = 6$$, $$\beta = 3$$, $$\gamma = 2$$

$$\alpha + \beta + \gamma = 6 + 3 + 2 = 11$$.

Limit Existence: For the limit to exist, the numerator must be 0 at $$x=0$$.

$$\sin(0) + \alpha \sin(0) - \beta \cos(0) = 0 \implies 0 + 0 - \beta(1) = 0 \implies \mathbf{\beta = 0}$$.

 Use Taylor Series:

$$f(x) = \frac{(3x - \frac{(3x)^3}{6} + \dots) + \alpha(x - \frac{x^3}{6} + \dots)}{x^3}$$

$$f(x) = \frac{(3+\alpha)x + (-\frac{27}{6} - \frac{\alpha}{6})x^3 + \dots}{x^3}$$

Coefficient of $$x$$ must be 0:

$$3 + \alpha = 0 \implies \mathbf{\alpha = -3}$$.

$$f(0) = \lim_{x \to 0} \frac{-\frac{27}{6}x^3 - \frac{(-3)}{6}x^3}{x^3} = -\frac{27}{6} + \frac{3}{6} = -\frac{24}{6} = -4$$.

Correct Option: D (-4)

$$\frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}$$.

Numerator: $$72^x - 9^x - 8^x + 1 = (9^x-1)(8^x-1) \approx (x\ln 9)(x\ln 8) = x^2 \ln 9 \cdot \ln 8$$ near $$x = 0$$.

Denominator: $$\sqrt{2}-\sqrt{1+\cos x} = \sqrt{2} - \sqrt{2}\cos(x/2)$$ (using $$1+\cos x = 2\cos^2(x/2)$$).

$$= \sqrt{2}(1 - \cos(x/2)) \approx \sqrt{2} \cdot \frac{x^2}{8}$$

Limit: $$\frac{x^2 \ln 9 \cdot \ln 8}{\sqrt{2}x^2/8} = \frac{8 \cdot 2\ln 3 \cdot 3\ln 2}{\sqrt{2}} = \frac{48\ln 2 \cdot \ln 3}{\sqrt{2}}$$.

For continuity: $$a \cdot \ln 2 \cdot \ln 3 = \frac{48\ln 2 \cdot \ln 3}{\sqrt{2}}$$, so $$a = \frac{48}{\sqrt{2}} = 24\sqrt{2}$$.

$$a^2 = 576 \times 2 = 1152$$.

The correct answer is Option 2: 1152.

We need to find the value of $$16(a^2 + b^2 + c^2)$$ given that the following limit equals 1:

$$\lim_{x \to 0} \frac{ax^2 e^x - b\ln(1+x) + cxe^{-x}}{x^2 \sin x} = 1$$
The denominator expands as

$$x^2 \sin x = x^2\Bigl(x - \frac{x^3}{6} + \cdots\Bigr) = x^3 - \frac{x^5}{6} + \cdots$$

Next, we expand the numerator term by term.

Term 1: $$ax^2 e^x = ax^2\Bigl(1 + x + \frac{x^2}{2} + \cdots\Bigr) = ax^2 + ax^3 + \cdots$$

Term 2: $$-b\ln(1+x) = -b\Bigl(x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\Bigr) = -bx + \frac{b}{2}x^2 - \frac{b}{3}x^3 + \cdots$$

Term 3: $$cxe^{-x} = cx\Bigl(1 - x + \frac{x^2}{2} - \cdots\Bigr) = cx - cx^2 + \frac{c}{2}x^3 + \cdots$$

Combining these, the numerator has the coefficients

of $$x^1$$: $$-b + c$$

of $$x^2$$: $$a + \frac{b}{2} - c$$

of $$x^3$$: $$a - \frac{b}{3} + \frac{c}{2}$$

For the limit to exist and equal 1, the numerator must be of order $$x^3$$. Therefore the coefficients of $$x$$ and $$x^2$$ must vanish, and the coefficient of $$x^3$$ divided by the coefficient 1 from the denominator must equal 1.

From the coefficient of $$x$$: $$-b + c = 0 \implies c = b$$

From the coefficient of $$x^2$$: $$a + \frac{b}{2} - c = 0$$ and since $$c = b$$, we get $$a = \frac{b}{2}$$

From the coefficient of $$x^3$$: $$a - \frac{b}{3} + \frac{c}{2} = 1$$. Substituting $$a = \tfrac{b}{2}$$ and $$c = b$$ yields

$$\frac{b}{2} - \frac{b}{3} + \frac{b}{2} = 1 \implies b - \frac{b}{3} = 1 \implies \frac{2b}{3} = 1 \implies b = \frac{3}{2}$$

Hence $$b = \frac{3}{2}$$, $$c = \frac{3}{2}$$, and $$a = \frac{3}{4}$$.

$$a^2 + b^2 + c^2 = \frac{9}{16} + \frac{9}{4} + \frac{9}{4} = \frac{9}{16} + \frac{36}{16} + \frac{36}{16} = \frac{81}{16}$$

$$16(a^2 + b^2 + c^2) = 16 \times \frac{81}{16} = 81$$

Using L'Hôpital or Taylor series for the limit as x→1:

Let f(x) = $$(5x+1)^(1/3)-(x+5)^(1/3) $$ and $$g(x)=(2x+3)^(1/2)-(x+4)^(1/2)$$

f(1) = $$6^(1/3)-6^(1/3)=0,g(1)=5^(1/2)-5^(1/2)$$= 0

f'(x) =$$5/(3(5x+1)^(2/3))-1/(3(x+5)^(2/3))$$

f'(1) = $$5/(3·6^(2/3))-1/(3·6^(2/3))=4/(3·6^(2/3))$$

g'(x) =$$1/\sqrt{(}2x+3)-1/(2\sqrt{(}x+4))$$

g'(1) = $$1/\sqrt{5}-1/(2\sqrt{5})=1/(2\sqrt{5})$$

Limit = $$f'(1)/g'(1)=[4/(3·6^(2/3))]·[2\sqrt{5}/1]=8\sqrt{5}/(3·6^(2/3))$$

= $$8\sqrt{5}/(3·(2·3)^(2/3))=8\sqrt{5}/(3·2^(2/3)·3^(2/3))=8\sqrt{5}/(3^(5/3)·2^(2/3))$$

Comparing with$$m\sqrt{5}/(n·(2n)^(2/3)$$): we need n such that $$n·(2n)^(2/3)=3^(5/3)·2^(2/3)$$

Try n=3: 3·6^(2/3) = 3·6^(2/3) = 3^(5/3)·2^(2/3) ✓ (since 3·(2·3)^(2/3) = 3·2^(2/3)·3^(2/3) = 3^(5/3)·2^(2/3))

So m = 8, n = 3, $$\gcd(8,3)=1✓$$

8m+12n = 64+36 = 100

We have two limits to evaluate.

$$\alpha = \lim_{x \to 0^{+}}\frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}$$

Write $$f(u)=e^{u}$$. Near $$x=0$$, both $$u_1=\sqrt{\tan x}$$ and $$u_2=\sqrt{x}$$ approach $$0$$. For a differentiable function, $$\displaystyle\lim_{u_1,u_2\to 0}\frac{f(u_1)-f(u_2)}{u_1-u_2}=f'(0)$$, because the expression is the symmetric derivative at the point $$0$$.

Here $$f'(u)=e^{u}$$, so $$f'(0)=1$$. Hence

$$\alpha = 1$$.

For the second limit

$$\beta = \lim_{x \to 0}\left(1+\sin x\right)^{\frac12\cot x}$$

take natural logarithm: $$\ln\beta = \lim_{x\to 0}\left(\frac12\cot x\right)\ln(1+\sin x)$$.

Series expansions about $$x=0$$:

$$\sin x = x-\frac{x^3}{6}+O(x^5)$$, $$\cos x = 1-\frac{x^2}{2}+O(x^4)$$, $$\cot x = \frac{\cos x}{\sin x} = \frac1x-\frac{x}{3}+O(x^3)$$.

Thus $$\frac12\cot x=\frac1{2x}-\frac{x}{6}+O(x^3)$$.

Also $$\ln(1+\sin x)=\sin x-\frac{\sin^2 x}{2}+O(x^3) = x-\frac{x^2}{2}-\frac{x^3}{6}+O(x^3).$$

Multiply the two series:

$$\Bigl(\frac1{2x}-\frac{x}{6}+O(x^3)\Bigr) \Bigl(x-\frac{x^2}{2}+O(x^3)\Bigr) = \frac12 + O(x).$$

Therefore $$\ln\beta \to \frac12$$ and

$$\beta = e^{1/2} = \sqrt{e}.$$

The numbers $$\alpha=1$$ and $$\beta=\sqrt{e}$$ are roots of the quadratic

$$ax^2 + bx - \sqrt{e}=0.$$

Using the relations for roots of $$Ax^2+Bx+C=0$$:
Sum of roots $$= -\frac{B}{A}$$, product of roots $$=\frac{C}{A}$$.

Sum: $$1+\sqrt{e} = -\frac{b}{a} \quad -(1)$$
Product: $$1\cdot\sqrt{e} = \frac{-\sqrt{e}}{a} \quad -(2)$$.

From $$(2)$$, $$a = -1.$$ Substituting in $$(1)$$ gives $$b = 1+\sqrt{e}.$$

Hence $$a+b = -1 + 1+\sqrt{e} = \sqrt{e}.$$

Finally,

$$12\log_e(a+b) = 12\log_e\!\bigl(\sqrt{e}\bigr) = 12\cdot\frac12 = 6.$$

So the required value is $$6$$.

We need to find $$\alpha + \beta$$ where $$\lim_{x \to 1/a} \frac{16(1 - \cos(2 + x - 2x^2))}{(1 - ax)^2} = \alpha + \beta\sqrt{17}$$.

Find $$a$$. From $$2x^2 + x - 2 = 0$$ with $$a > 0$$:

$$ a = \frac{-1 + \sqrt{1 + 16}}{4} = \frac{-1 + \sqrt{17}}{4} $$

As $$x \to \frac{1}{a}$$, we need $$2 + x - 2x^2 \to 0$$ (since $$\frac{1}{a}$$ is a root of $$2x^2 - x - 2 = 0$$, i.e., $$2x^2 = x + 2$$, so $$2 + x - 2x^2 = 2 + x - (x+2) = 0$$). ✓

So both numerator and denominator approach 0. Use the limit $$\lim_{u \to 0} \frac{1 - \cos u}{u^2} = \frac{1}{2}$$.

$$ \lim_{x \to 1/a} \frac{16(1 - \cos(2 + x - 2x^2))}{(1 - ax)^2} = 16 \cdot \frac{1}{2} \cdot \lim_{x \to 1/a} \frac{(2 + x - 2x^2)^2}{(1 - ax)^2} $$

$$ = 8 \cdot \left(\lim_{x \to 1/a} \frac{2 + x - 2x^2}{1 - ax}\right)^2 $$

Evaluate the limit. Factor $$2 + x - 2x^2 = -2(x^2 - x/2 - 1) = -2(x - 1/a)(x + 1/(2\cdot something))$$.

$$2 + x - 2x^2 = -(2x^2 - x - 2)$$. The roots of $$2x^2 - x - 2 = 0$$ are $$x = \frac{1 \pm \sqrt{17}}{4}$$.

So $$2 + x - 2x^2 = -2\left(x - \frac{1+\sqrt{17}}{4}\right)\left(x - \frac{1-\sqrt{17}}{4}\right)$$

And $$\frac{1}{a} = \frac{4}{-1+\sqrt{17}} = \frac{4(\sqrt{17}+1)}{16} = \frac{\sqrt{17}+1}{4} = \frac{1+\sqrt{17}}{4}$$.

Also $$1 - ax = -a(x - 1/a)$$.

$$ \frac{2 + x - 2x^2}{1 - ax} = \frac{-2(x - \frac{1+\sqrt{17}}{4})(x - \frac{1-\sqrt{17}}{4})}{-a(x - \frac{1+\sqrt{17}}{4})} = \frac{2}{a}\left(x - \frac{1-\sqrt{17}}{4}\right) $$

At $$x = \frac{1+\sqrt{17}}{4}$$:

$$ = \frac{2}{a} \cdot \left(\frac{1+\sqrt{17}}{4} - \frac{1-\sqrt{17}}{4}\right) = \frac{2}{a} \cdot \frac{2\sqrt{17}}{4} = \frac{\sqrt{17}}{a} $$

With $$a = \frac{-1+\sqrt{17}}{4}$$:

$$ = \frac{\sqrt{17}}{\frac{\sqrt{17}-1}{4}} = \frac{4\sqrt{17}}{\sqrt{17}-1} = \frac{4\sqrt{17}(\sqrt{17}+1)}{16} = \frac{\sqrt{17}(\sqrt{17}+1)}{4} = \frac{17+\sqrt{17}}{4} $$

$$ \text{Limit} = 8 \cdot \left(\frac{17+\sqrt{17}}{4}\right)^2 = 8 \cdot \frac{(17+\sqrt{17})^2}{16} = \frac{(17+\sqrt{17})^2}{2} $$

$$ = \frac{289 + 34\sqrt{17} + 17}{2} = \frac{306 + 34\sqrt{17}}{2} = 153 + 17\sqrt{17} $$

So $$\alpha = 153$$, $$\beta = 17$$, and $$\alpha + \beta = 170$$.

The answer is 170.

Given $$f(1) = 1$$ and $$\lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1$$.

The limit is of the form $$\frac{0}{0}$$ when $$t = x$$. Using L'Hôpital's rule (differentiating with respect to $$t$$):

$$ \lim_{t \to x} \frac{2tf(x) - x^2 f'(t)}{1} = 2xf(x) - x^2f'(x) = 1 $$

So: $$x^2f'(x) - 2xf(x) = -1$$.

Dividing by $$x^2$$: $$f'(x) - \frac{2}{x}f(x) = -\frac{1}{x^2}$$.

I.F. = $$e^{\int -2/x \, dx} = e^{-2\ln x} = x^{-2}$$.

$$ \frac{f(x)}{x^2} = \int \frac{-1}{x^2} \cdot \frac{1}{x^2} dx = \int -x^{-4} dx = \frac{1}{3x^3} + C $$

$$ f(x) = \frac{x^2}{3x^3} + Cx^2 = \frac{1}{3x} + Cx^2 $$

Using $$f(1) = 1$$: $$\frac{1}{3} + C = 1 \Rightarrow C = \frac{2}{3}$$.

$$ f(x) = \frac{1}{3x} + \frac{2x^2}{3} $$

$$ f(2) = \frac{1}{6} + \frac{8}{3} = \frac{1 + 16}{6} = \frac{17}{6} $$

$$ f(3) = \frac{1}{9} + \frac{18}{3} = \frac{1}{9} + 6 = \frac{55}{9} $$

$$ 2f(2) + 3f(3) = 2 \times \frac{17}{6} + 3 \times \frac{55}{9} = \frac{17}{3} + \frac{55}{3} = \frac{72}{3} = 24 $$

The answer is 24.

We want to find $$\lim_{x\to0}2\left(\frac{1-\cos x\sqrt{\cos2x}\sqrt[3]{\cos3x}\cdots\sqrt[10]{\cos10x}}{x^2}\right).$$

Set $$P=\prod_{k=1}^{10}(\cos kx)^{1/k},$$ so that the expression inside the limit becomes $$2\frac{1-P}{x^2}.$$ Taking natural logarithms gives $$\ln P=\sum_{k=1}^{10}\frac{1}{k}\ln(\cos kx).$$

For small $$x$$ one has the approximation $$\ln(\cos kx)\approx-\frac{k^2x^2}{2},$$ which implies
$$\ln P\approx -\frac{x^2}{2}\sum_{k=1}^{10}\frac{k^2}{k}=-\frac{x^2}{2}\sum_{k=1}^{10}k=-\frac{x^2}{2}\cdot55=-\frac{55x^2}{2}.$$

Exponentiating this approximation yields $$P\approx e^{-55x^2/2}\approx1-\frac{55x^2}{2}$$ for small $$x$$, and therefore
$$1-P\approx\frac{55x^2}{2}.$$

Substituting into the limit gives
$$2\cdot\frac{1-P}{x^2}\approx2\cdot\frac{\frac{55x^2}{2}}{x^2}=55.$$

Hence, the value of the limit is 55.

Squaring the given limit definition equation:

$$(f(x))^2 = \frac{2x^2 f(x) f'(x)}{2x} - x^3 e^{\frac{f(x)}{x}} \implies (f(x))^2 = x f(x) f'(x) - x^3 e^{\frac{f(x)}{x}}$$

Divide by $$x^2$$ and substitute $$f(x) = vx$$:

$$v^2 = v\left(v + x\frac{dv}{dx}\right) - xe^v \implies v e^{-v} dv = dx$$

Integrating both sides:

$$-e^{-v}(v+1) = x + C$$

Using $$f(1) = 1 \implies v = 1$$ at $$x = 1$$:

$$-e^{-1}(2) = 1 + C \implies C = -1 - \frac{2}{e}$$

 $$f(a) = 0 \implies v = 0$$:

$$-e^0(0+1) = a - 1 - \frac{2}{e} \implies -1 = a - 1 - \frac{2}{e} \implies a = \frac{2}{e}$$

$$ae = \left(\frac{2}{e}\right)e = 2$$.

The points $$1,\frac12,\frac13,\ldots$$ split the interval $$(0,1)$$ into the half-open intervals $$\left[\frac1{n+1},\frac1n\right)\;(n\in\mathbb N)$$. For $$x\in\left[\frac1{n+1},\frac1n\right)$$ we are told $$f(x)=\sqrt n.$$

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1. A relation between $$n$$ and $$x$$.
Because $$x\in\left[\frac1{n+1},\frac1n\right)$$ we have $$\frac1{n+1}\le x&lt\frac1n\; \Longrightarrow\; n\le\frac1x&lt n+1.$$ Multiplying by $$x$$ gives $$n\,x\le1&lt(n+1)\,x.$$ Hence $$1-\frac1{n+1}&lt n\,x\le1\qquad -(1)$$ and therefore $$\sqrt{1-\frac1{n+1}}&lt\sqrt{n\,x}\le1.\qquad -(2)$$

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2. Squeezing $$g(x)$$ as $$x\to0^+$$.
Define $$I(x)=\int_{x^{2}}^{x}\sqrt{\frac{1-t}{t}}\;dt,$$ so that $$I(x)&lt g(x)&lt2\sqrt x.$$ For $$0&lt t\le x&lt1$$ we have $$1-t\ge1-x,$$ hence $$\sqrt{\frac{1-t}{t}}\ge\sqrt{1-x}\,\frac1{\sqrt t}.$$ Therefore $$I(x)\ge\sqrt{1-x}\int_{x^{2}}^{x}\frac{dt}{\sqrt t} =\sqrt{1-x}\,[2\sqrt t]_{x^{2}}^{x} =2\sqrt{1-x}\,(\sqrt x-x).$$ Divide this and the upper bound by $$\sqrt x$$: $$2\sqrt{1-x}\,(1-\sqrt x)\;&lt\;\frac{g(x)}{\sqrt x}\;&lt\;2.$$ When $$x\to0^{+}$$, both the left and right bounds tend to $$2$$, so by the squeeze theorem $$\lim_{x\to0^{+}}\frac{g(x)}{\sqrt x}=2\;\;\Longrightarrow\;\; g(x)=2\sqrt x\,[1+\varepsilon(x)]$$ with $$\varepsilon(x)\to0$$ as $$x\to0^{+}.$$

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3. Limit of the product $$f(x)g(x)$$.
Using $$f(x)=\sqrt n$$ and the form of $$g(x)$$ just obtained, $$f(x)g(x)=\sqrt n\,g(x)=2\sqrt n\,\sqrt x\,[1+\varepsilon(x)] =2\sqrt{n\,x}\,[1+\varepsilon(x)].$$ From inequality $$(2)$$ we have $$\sqrt{1-\dfrac1{n+1}}\;\le\;\sqrt{n\,x}\;\le\;1.$$ Hence $$2\sqrt{1-\dfrac1{n+1}}\,[1+\varepsilon(x)] \;\le\;f(x)g(x) \;\le\;2[1+\varepsilon(x)].$$ Now let $$x\to0^{+}$$. Then $$n\to\infty$$ (because $$x\sim1/n$$) so the factor $$\sqrt{1-\dfrac1{n+1}}\to1,$$ and simultaneously $$\varepsilon(x)\to0.$$ Therefore both the lower and upper bounds tend to $$2$$, giving $$\boxed{\displaystyle\lim_{x\to0^{+}}f(x)g(x)=2}.$$

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Option C is the correct choice.

We need to evaluate $$\lim_{t \to 0} \left(1^{1/\sin^2 t} + 2^{1/\sin^2 t} + 3^{1/\sin^2 t} + \cdots + n^{1/\sin^2 t}\right)^{\sin^2 t}$$.

As $$t \to 0$$, one has $$\sin^2 t \to 0$$ and hence $$\frac{1}{\sin^2 t} \to \infty$$. By setting $$s = \frac{1}{\sin^2 t}$$, the problem reduces to evaluating the limit $$\lim_{s \to \infty} \left(1^s + 2^s + 3^s + \cdots + n^s\right)^{1/s}$$.

For large $$s$$, the term $$n^s$$ dominates the sum since it grows faster than any other $$k^s$$ with $$k < n$$. Factoring out $$n^s$$ from inside the parentheses yields

$$\left(n^s\left[\left(\frac{1}{n}\right)^s + \left(\frac{2}{n}\right)^s + \cdots + 1\right]\right)^{1/s} = n \cdot \left[\left(\frac{1}{n}\right)^s + \cdots + 1\right]^{1/s}$$.

As $$s \to \infty$$, each term $$\left(\frac{k}{n}\right)^s \to 0$$ for $$k < n$$, so the bracketed sum approaches $$1$$, and the limit simplifies to $$n \cdot 1 = n$$.

The answer is Option 2: $$n$$.

$$S_n = 1 + 2 - 3 + 4 + 5 - 6 + \dots + (3n-2) + (3n-1) - 3n$$

Each triplet $$(3k-2) + (3k-1) - 3k$$ simplifies to $$3k - 3$$

$$S_n = \sum_{k=1}^n (3k - 3) = 3 \sum_{k=1}^n (k-1) = 3 \left( \frac{(n-1)n}{2} \right) = \frac{3n(n-1)}{2}$$

The denominator is $$D_n = \sqrt{2n^4 + 4n + 3} - \sqrt{n^4 + 5n + 4}$$.

As $$n \to \infty$$, we can factor out $$n^2$$ from each square root:
$$D_n = n^2 \sqrt{2 + \frac{4}{n^3} + \frac{3}{n^4}} - n^2 \sqrt{1 + \frac{5}{n^3} + \frac{4}{n^4}}$$
$$D_n \approx n^2 (\sqrt{2} - 1)$$

$$\lim_{n\to\infty} \frac{\frac{3n^2 - 3n}{2}}{n^2(\sqrt{2} - 1)}$$

$$\lim_{n\to\infty} \frac{\frac{3}{2} - \frac{3}{2n}}{\sqrt{2} - 1} = \frac{3}{2(\sqrt{2} - 1)}$$

$$\frac{3(\sqrt{2} + 1)}{2(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{3(\sqrt{2} + 1)}{2(2 - 1)} = \frac{3}{2}(\sqrt{2} + 1)$$

To find the value of the limit, we first simplify the expression inside the cosine function using the given root condition.

Since $$x = 2$$ is a root of the quadratic equation $$x^2 + px + q = 0$$, substituting $$x = 2$$ gives:

$$2^2 + p(2) + q = 0 \implies 4 + 2p + q = 0 \implies q + 4 = -2p$$

---

Step 1: Simplify the argument of the cosine function

Let the algebraic expression inside the cosine be $$g(x)$$:

$$g(x) = x^2 - 4px + q^2 + 8q + 16$$

Recognizing the perfect square formula for the last three terms:

$$q^2 + 8q + 16 = (q + 4)^2$$

Substituting $$(q + 4) = -2p$$ into this perfect square:

$$(q + 4)^2 = (-2p)^2 = 4p^2$$

Now substitute this back into the full expression for $$g(x)$$:

$$g(x) = x^2 - 4px + 4p^2 = (x - 2p)^2$$

---

Step 2: Rewrite the function $$f(x)$$ for $$x \neq 2p$$

Substituting the simplified argument back into the definition of $$f(x)$$:

$$f(x) = \frac{1 - \cos((x - 2p)^2)}{(x - 2p)^4}$$

Let $$u = (x - 2p)^2$$. As $$x \to 2p^+$$, we have $$u \to 0^+$$. The function becomes:

$$f(x) = \frac{1 - \cos u}{u^2}$$

---

Step 3: Analyze the behavior of $$f(x)$$ near $$u = 0$$

Using the Taylor series expansion for $$\cos u$$ around 0:

$$\cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \dots = 1 - \frac{u^2}{2} + \frac{u^4}{24} - \dots$$

Substituting this series into the function expression:

$$f(x) = \frac{1 - \left(1 - \frac{u^2}{2} + \frac{u^4}{24} - \dots\right)}{u^2}$$

$$f(x) = \frac{\frac{u^2}{2} - \frac{u^4}{24} + \dots}{u^2} = \frac{1}{2} - \frac{u^2}{24} + \dots$$

For small positive values of $$u$$ (since $$u = (x-2p)^2 > 0$$ as $$x \to 2p^+$$), the term $$-\frac{u^2}{24}$$ is strictly negative. This means that as $$x$$ approaches $$2p^+$$ from the right, the value of $$f(x)$$ approaches $$\frac{1}{2}$$ from below:

$$0 < f(x) < \frac{1}{2}$$

---

Step 4: Evaluate the limit with the greatest integer function

Since $$f(x)$$ lies strictly between 0 and $$\frac{1}{2}$$ in the immediate right-neighborhood of $$2p$$, applying the greatest integer function $$[\cdot]$$ yields:

$$[f(x)] = 0$$

Taking the limit of this constant value:

$$\lim_{x \to 2p^+} [f(x)] = 0$$

Therefore, the value of the limit is equal to 0.

Each term in the product looks like $$(\sqrt{2} - 2^{1/2^{k+1}})$$.

As $$k$$ increases, $$2^{1/2^{k+1}}$$ gets closer and closer to $$2^0$$, which is 1.

So, every term in the sequence eventually settles at approximately $$\sqrt{2} - 1 \approx \mathbf{0.414}$$.

When you multiply a number less than 1 (like 0.414) by itself infinitely many times, the result always shrinks to 0.

Mathematically, if $$|a| < 1$$, then $$\lim_{n\to\infty} a^n = 0$$. Since all terms here are positive and less than 1, the product must converge to zero.

$$[x - 5] = [x] - 5$$

$$[2x + 2] = [2x] + 2$$

$$f(x) = ([x] - 5) - ([2x] + 2)$$

$$f(x) = [x] - [2x] - 7$$

For this limit to exist and equal 0, the function must be exactly 0 in an open interval around $$a$$.

So, we need to solve the equation: $$[x] - [2x] - 7 = 0$$ 

$$[2x] - [x] = -7$$

$$x = n + f$$ ($$x$$ is being written as the combination of integral and fractional part)

$$[2(n + f)] - n = -7$$

$$[2n + 2f] - n = -7$$

$$(2n + [2f]) - n = -7$$ (Since $$2n$$ is an integer)

$$n + [2f] = -7$$

Since $$0 \le f < 1$$, the value of $$2f$$ lies in the range $$0 \le 2f < 2$$. Thus, $$[2f]$$ can only be either $$0$$ or $$1$$

Case 1: ($$[2f] = 0$$): $$0 \le f < 0.5 \implies n = -7 \implies x \in [-7, -6.5)$$

Case 2: ($$[2f] = 1$$): $$0.5 \le f < 1 \implies n = -8 \implies x \in [-7.5, -7)$$

Combining these, $$f(x) = 0$$ on the interval $$[-7.5, -6.5)$$.

For the two-sided limit $$\lim_{x \to a} f(x) = 0$$ to exist, $$f(x)$$ must be $$0$$ on both sides of $$a$$. At the endpoints ($$-7.5$$ and $$-6.5$$), the function jumps to a non-zero value, meaning the limit does not exist there. Therefore, $$a$$ must lie strictly inside the open interval.

To solve this limit quickly, we use the standard limits as $$x \to 0$$:

• $$\sin(kx) \approx kx$$
• $$\ln(1 + kx) \approx kx$$
• $$\cos(kx) \approx 1$$

The expression is:

$$L = \lim_{x \to 0} \left( \frac{1 - \cos^2(3x)}{\cos^3(4x)} \right) \left( \frac{\sin^3(4x)}{(\ln(2x + 1))^5} \right)$$

1. • $$1 - \cos^2(3x) = \sin^2(3x) \approx (3x)^2 = \mathbf{9x^2}$$
   • $$\cos^3(4x) \approx 1^3 = \mathbf{1}$$
   • $$\sin^3(4x) \approx (4x)^3 = \mathbf{64x^3}$$
   • $$(\ln(2x + 1))^5 \approx (2x)^5 = \mathbf{32x^5}$$
2. $$L = \lim_{x \to 0} \left( \frac{9x^2}{1} \right) \left( \frac{64x^3}{32x^5} \right)$$
3. $$L = 9x^2 \cdot \frac{64x^3}{32x^5} = 9 \cdot \frac{64}{32} \cdot \frac{x^5}{x^5}$$
4. $$L = 9 \cdot 2 = \mathbf{18}$$

We are given that $$\alpha > \beta > 0$$ are the roots of the equation $$ax^2 + bx + 1 = 0$$, and we need to find the value of $$k$$.

To begin,

Since $$\alpha$$ and $$\beta$$ are roots of $$ax^2 + bx + 1 = 0$$, we can write:

$$ax^2 + bx + 1 = a(x - \alpha)(x - \beta)$$

Dividing both sides by $$a$$:

$$x^2 + \frac{b}{a}x + \frac{1}{a} = (x - \alpha)(x - \beta)$$

Now consider $$x^2 + bx + a$$. The roots of $$x^2 + bx + a = 0$$ can be found by substituting $$x \to \frac{1}{x}$$ in the original equation. From $$ax^2 + bx + 1 = 0$$, replacing $$x$$ with $$\frac{1}{x}$$:

$$\frac{a}{x^2} + \frac{b}{x} + 1 = 0 \implies a + bx + x^2 = 0$$

So the roots of $$x^2 + bx + a = 0$$ are $$\frac{1}{\alpha}$$ and $$\frac{1}{\beta}$$. Therefore:

$$x^2 + bx + a = \left(x - \frac{1}{\alpha}\right)\left(x - \frac{1}{\beta}\right)$$

Next,

As $$x \to \frac{1}{\alpha}$$, we have $$x^2 + bx + a \to 0$$. Using the approximation $$1 - \cos\theta \approx \frac{\theta^2}{2}$$ for small $$\theta$$:

$$\sqrt{\frac{1 - \cos(x^2 + bx + a)}{2(1 - \alpha x)^2}} \approx \sqrt{\frac{(x^2 + bx + a)^2}{4(1 - \alpha x)^2}} = \frac{|x^2 + bx + a|}{2|1 - \alpha x|}$$

From this,

We have $$x^2 + bx + a = \left(x - \frac{1}{\alpha}\right)\left(x - \frac{1}{\beta}\right)$$ and $$1 - \alpha x = -\alpha\left(x - \frac{1}{\alpha}\right)$$.

$$\frac{\left|x - \frac{1}{\alpha}\right| \cdot \left|x - \frac{1}{\beta}\right|}{2\alpha\left|x - \frac{1}{\alpha}\right|} = \frac{\left|x - \frac{1}{\beta}\right|}{2\alpha}$$

Continuing,

As $$x \to \frac{1}{\alpha}$$:

$$\frac{\left|\frac{1}{\alpha} - \frac{1}{\beta}\right|}{2\alpha}$$

Since $$\alpha > \beta > 0$$, we have $$\frac{1}{\alpha} < \frac{1}{\beta}$$, so $$\left|\frac{1}{\alpha} - \frac{1}{\beta}\right| = \frac{1}{\beta} - \frac{1}{\alpha}$$.

$$\text{Limit} = \frac{1}{2\alpha}\left(\frac{1}{\beta} - \frac{1}{\alpha}\right)$$

Now,

We are told the limit equals $$\frac{1}{k}\left(\frac{1}{\beta} - \frac{1}{\alpha}\right)$$. Comparing:

$$\frac{1}{k} = \frac{1}{2\alpha} \implies k = 2\alpha$$

The answer is Option C: $$2\alpha$$.

We need to evaluate $$\lim_{n \to \infty} \left(\frac{1}{1+n} + \frac{1}{2+n} + \frac{1}{3+n} + \ldots + \frac{1}{2n}\right)$$.

Write the sum in sigma notation.

$$ S_n = \sum_{k=1}^{n} \frac{1}{k+n} $$

Convert to a Riemann sum.

Factor out $$\frac{1}{n}$$ from each term:

$$ S_n = \sum_{k=1}^{n} \frac{1}{n} \cdot \frac{1}{\frac{k}{n} + 1} $$

This is a Riemann sum for the integral $$\int_0^1 \frac{1}{x+1}\, dx$$, where the interval $$[0, 1]$$ is divided into $$n$$ equal subintervals of width $$\Delta x = 1/n$$, and the sample points are $$x_k = k/n$$.

As $$n \to \infty$$, the Riemann sum converges to the definite integral:

$$ \lim_{n \to \infty} S_n = \int_0^1 \frac{dx}{x+1} $$

Evaluate the integral.

Using the standard integral $$\int \frac{dx}{x+1} = \ln|x+1| + C$$:

$$ \int_0^1 \frac{dx}{x+1} = \left[\ln(x+1)\right]_0^1 = \ln(1+1) - \ln(0+1) = \ln 2 - \ln 1 = \ln 2 $$

Therefore, the limit is $$\log_e 2$$.

The correct answer is Option 2: $$\log_e 2$$.

We need to evaluate $$\lim_{x \to 0} \frac{48}{x^4} \int_0^x \frac{t^3}{t^6+1} dt$$. As $$x \to 0$$, both $$\int_0^x \frac{t^3}{t^6+1}dt \to 0$$ and $$x^4 \to 0$$, so the expression is of the indeterminate form $$\frac{0}{0}$$.

To resolve this, we rewrite the limit as $$\lim_{x \to 0} \frac{48 \int_0^x \frac{t^3}{t^6+1}dt}{x^4}$$ and apply L'Hopital's rule, differentiating numerator and denominator with respect to $$x$$. By the Fundamental Theorem of Calculus, $$\frac{d}{dx}\int_0^x f(t)dt = f(x)$$, hence the limit becomes $$\lim_{x \to 0} \frac{48 \cdot \frac{x^3}{x^6+1}}{4x^3}$$.

Finally, simplifying inside the limit yields $$\lim_{x \to 0} \frac{48}{4(x^6+1)} = \frac{48}{4(0+1)} = \frac{48}{4} = 12$$, so the value of the original limit is 12.

We need to find $$ 8(\alpha + \beta) $$ given that $$ \lim_{n \to \infty} \left(\sqrt{n^2 - n - 1} + n\alpha + \beta\right) = 0 $$.

As $$ n \to \infty $$, $$ \sqrt{n^2 - n - 1} \approx n\sqrt{1 - \frac{1}{n} - \frac{1}{n^2}} \approx n $$.

For the limit to be 0 (and not $$ \pm\infty $$), we need the coefficient of $$ n $$ in the expression to vanish, which requires $$ \alpha = -1 $$.

$$\sqrt{n^2 - n - 1} - n + \beta$$

Rationalize:

$$\sqrt{n^2 - n - 1} - n = \frac{(n^2 - n - 1) - n^2}{\sqrt{n^2 - n - 1} + n} = \frac{-n - 1}{\sqrt{n^2 - n - 1} + n}$$

$$\frac{-n - 1}{\sqrt{n^2 - n - 1} + n} = \frac{-n(1 + \frac{1}{n})}{n(\sqrt{1 - \frac{1}{n} - \frac{1}{n^2}} + 1)}$$

As $$ n \to \infty $$:

$$\to \frac{-1}{1 + 1} = -\frac{1}{2}$$

$$-\frac{1}{2} + \beta = 0 \implies \beta = \frac{1}{2}$$

$$8(\alpha + \beta) = 8\left(-1 + \frac{1}{2}\right) = 8 \times \left(-\frac{1}{2}\right) = -4$$

Therefore, $$ 8(\alpha + \beta) = -4 $$, which corresponds to Option C.

We need to find $$\alpha, \beta, \gamma$$ such that $$\displaystyle\lim_{x \to 0} \frac{\alpha e^x + \beta e^{-x} + \gamma \sin x}{x \sin^2 x} = \frac{2}{3}$$.

We use Taylor expansions about $$x = 0$$. We have $$e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots$$, and $$e^{-x} = 1 - x + \dfrac{x^2}{2} - \dfrac{x^3}{6} + \cdots$$, and $$\sin x = x - \dfrac{x^3}{6} + \cdots$$.

The numerator becomes $$\alpha\!\left(1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots\right) + \beta\!\left(1 - x + \dfrac{x^2}{2} - \dfrac{x^3}{6} + \cdots\right) + \gamma\!\left(x - \dfrac{x^3}{6} + \cdots\right)$$.

Collecting by powers of $$x$$: the constant term is $$\alpha + \beta$$, the coefficient of $$x$$ is $$\alpha - \beta + \gamma$$, the coefficient of $$x^2$$ is $$\dfrac{\alpha + \beta}{2}$$, and the coefficient of $$x^3$$ is $$\dfrac{\alpha - \beta}{6} - \dfrac{\gamma}{6} = \dfrac{\alpha - \beta - \gamma}{6}$$.

The denominator is $$x \sin^2 x \approx x \cdot x^2 = x^3$$ as $$x \to 0$$. For the limit to be finite and non-zero, the numerator must begin at $$x^3$$, so all lower-order coefficients must vanish.

Setting the constant term to zero: $$\alpha + \beta = 0$$ ... (i). This also makes the $$x^2$$ coefficient vanish automatically. Setting the coefficient of $$x$$ to zero: $$\alpha - \beta + \gamma = 0$$ ... (ii).

From (i), $$\beta = -\alpha$$. Substituting into (ii): $$\alpha + \alpha + \gamma = 0$$, so $$\gamma = -2\alpha$$.

Now the leading term of the numerator is $$\dfrac{\alpha - \beta - \gamma}{6} \cdot x^3 = \dfrac{\alpha + \alpha + 2\alpha}{6} \cdot x^3 = \dfrac{4\alpha}{6} \cdot x^3 = \dfrac{2\alpha}{3} \cdot x^3$$.

Hence the limit equals $$\dfrac{2\alpha/3}{1} = \dfrac{2\alpha}{3}$$. Setting this equal to $$\dfrac{2}{3}$$ gives $$\alpha = 1$$. Therefore $$\beta = -1$$ and $$\gamma = -2$$.

Now we check each option:

Option A: $$\alpha^2 + \beta^2 + \gamma^2 = 1 + 1 + 4 = 6$$ ✓ (correct statement)

Option B: $$\alpha\beta + \beta\gamma + \gamma\alpha + 1 = (1)(-1) + (-1)(-2) + (-2)(1) + 1 = -1 + 2 - 2 + 1 = 0$$ ✓ (correct statement)

Option C: $$\alpha\beta^2 + \beta\gamma^2 + \gamma\alpha^2 + 3 = (1)(1) + (-1)(4) + (-2)(1) + 3 = 1 - 4 - 2 + 3 = -2 \neq 0$$ ✗ (NOT correct)

Option D: $$\alpha^2 - \beta^2 + \gamma^2 = 1 - 1 + 4 = 4$$ ✓ (correct statement)

Hence, the correct answer is Option C.

We need to find the integer $$a$$ such that $$\displaystyle\lim_{x \to 7} \frac{18 - [1-x]}{[x-3a]}$$ exists, where $$[t]$$ denotes the greatest integer function (floor function).

As $$x \to 7^-$$, we have $$1 - x \to -6^+$$, so $$[1-x] = -6$$ and the numerator becomes $$18 - (-6) = 24$$. As $$x \to 7^+$$, $$1 - x \to -6^-$$, giving $$[1-x] = -7$$ and the numerator $$18 - (-7) = 25$$. Since the numerator has different left-hand and right-hand values (24 and 25), for the limit to exist, the denominator must also have a jump at $$x = 7$$ that compensates.

At $$x = 7$$ the expression in the denominator is $$x - 3a = 7 - 3a$$. If $$7 - 3a$$ is not an integer, then $$[x - 3a]$$ remains constant in a neighborhood of $$x = 7$$ and the limit would be $$\frac{24}{[7-3a]} \neq \frac{25}{[7-3a]}$$, so the limit would not exist. Therefore, $$7 - 3a$$ must be an integer (which is automatically satisfied since $$a$$ is an integer).

When $$7 - 3a$$ is an integer, as $$x \to 7^-$$ we have $$x - 3a \to (7-3a)^-$$ so $$[x - 3a] = 7 - 3a - 1 = 6 - 3a$$, and as $$x \to 7^+$$ we have $$x - 3a \to (7-3a)^+$$ so $$[x - 3a] = 7 - 3a$$.

Equating the left-hand limit and the right-hand limit yields $$\frac{24}{6 - 3a} = \frac{25}{7 - 3a}$$. Cross-multiplying gives $$24(7 - 3a) = 25(6 - 3a)$$, which simplifies to $$168 - 72a = 150 - 75a$$ and then to $$18 = -3a$$, so $$a = -6$$.

Verification: With $$a = -6$$, the denominator values are $$6 - 3(-6) = 24$$ and $$7 - 3(-6) = 25$$. Hence LHL = $$\frac{24}{24} = 1$$ and RHL = $$\frac{25}{25} = 1$$. The limit exists and equals 1.

The correct answer is Option C: $$-6$$.

We need to find $$\lim_{x \to 1} \frac{f(x)}{x - 1}$$ where $$f(x) + f'(x) + f''(x) = x^5 + 64$$. Since the right-hand side is a degree 5 polynomial, it follows that $$f(x)$$ must also be of degree 5; accordingly, let $$f(x) = x^5 + a x^4 + b x^3 + c x^2 + d x + e$$ so that $$f'(x) = 5x^4 + 4a x^3 + 3b x^2 + 2c x + d$$ and $$f''(x) = 20x^3 + 12a x^2 + 6b x + 2c$$.

By equating coefficients of like powers of $$x$$ one obtains:
x^5: 1 = 1 ✓
x^4: a + 5 = 0 ⇒ a = -5
x^3: b + 4a + 20 = 0 ⇒ b + 4(-5) + 20 = 0 ⇒ b = 0
x^2: c + 3b + 12a = 0 ⇒ c + 0 - 60 = 0 ⇒ c = 60
x^1: d + 2c + 6b = 0 ⇒ d + 120 + 0 = 0 ⇒ d = -120
x^0: e + d + 2c = 64 ⇒ e - 120 + 120 = 64 ⇒ e = 64

Hence $$f(x) = x^5 - 5x^4 + 60x^2 - 120x + 64$$, and substituting $$x = 1$$ gives $$f(1) = 1 - 5 + 60 - 120 + 64 = 0$$, so the limit is of the form $$\frac{0}{0}$$. Thus one can apply L'Hôpital's rule or observe that $$\lim_{x \to 1} \frac{f(x)}{x - 1} = f'(1)$$.

Since $$f'(x) = 5x^4 - 20x^3 + 120x - 120$$, one finds $$f'(1) = 5 - 20 + 120 - 120 = -15$$.

Option A: $$-15$$.

We need to evaluate $$\displaystyle\lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4}$$.

Recall that $$\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right).$$

Let $$A = \sin x$$ and $$B = x$$, so that $$\cos(\sin x) - \cos x = -2\sin\left(\frac{\sin x + x}{2}\right)\sin\left(\frac{\sin x - x}{2}\right).$$

Using the Taylor series expansion for small x, $$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots.$$

It follows that $$\sin x + x = 2x - \frac{x^3}{6} + O(x^5).$$

Therefore, $$\frac{\sin x + x}{2} = x - \frac{x^3}{12} + O(x^5).$$

Similarly, $$\sin x - x = -\frac{x^3}{6} + O(x^5),$$ so $$\frac{\sin x - x}{2} = -\frac{x^3}{12} + O(x^5).$$

As x → 0, both arguments of sine are small, so one may approximate $$\sin u \approx u$$ for each.

Thus $$\sin\left(\frac{\sin x + x}{2}\right) \approx x - \frac{x^3}{12},$$ whose leading term is x.

Also, $$\sin\left(\frac{\sin x - x}{2}\right) \approx -\frac{x^3}{12}.$$

Substituting these approximations into the difference of cosines gives $$\cos(\sin x) - \cos x \approx -2 \left(x - \frac{x^3}{12}\right)\left(-\frac{x^3}{12}\right).$$

Retaining terms up to order $$x^4$$ yields $$-2x\left(-\frac{x^3}{12}\right) + O(x^6) = \frac{2x^4}{12} = \frac{x^4}{6}.$$

Therefore, $$\lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4} = \lim_{x \to 0} \frac{x^4/6}{x^4} = \frac{1}{6}.$$

Thus the limit is $$\dfrac{1}{6}$$.

We need to find $$\alpha + \beta$$ where $$\beta = \displaystyle\lim_{x \to 0} \dfrac{\alpha x - (e^{3x} - 1)}{\alpha x(e^{3x} - 1)}$$ for some $$\alpha \in \mathbb{R}$$.

As $$x \to 0$$: $$e^{3x} - 1 \to 0$$ and $$\alpha x \to 0$$.

The denominator $$\alpha x(e^{3x}-1) \to 0$$.

For the limit to be finite, the numerator must also approach 0:

$$\lim_{x \to 0} [\alpha x - (e^{3x} - 1)] = 0$$

$$\alpha \cdot 0 - (1 - 1) = 0$$. This is $$0$$ regardless of $$\alpha$$.

So we have a $$\frac{0}{0}$$ form. We use Taylor expansion.

$$e^{3x} = 1 + 3x + \frac{(3x)^2}{2} + \frac{(3x)^3}{6} + \cdots = 1 + 3x + \frac{9x^2}{2} + \frac{27x^3}{6} + \cdots$$

$$e^{3x} - 1 = 3x + \frac{9x^2}{2} + O(x^3)$$

$$\alpha x - (e^{3x} - 1) = \alpha x - 3x - \frac{9x^2}{2} - O(x^3) = (\alpha - 3)x - \frac{9x^2}{2} - O(x^3)$$

If $$\alpha \neq 3$$, the numerator is $$O(x)$$ and the denominator is $$O(x^2)$$, giving an infinite limit.

So we need $$\alpha = 3$$.

Numerator: $$-\frac{9x^2}{2} - O(x^3)$$

Denominator: $$3x(3x + \frac{9x^2}{2} + \cdots) = 9x^2 + \frac{27x^3}{2} + \cdots$$

$$\beta = \lim_{x \to 0} \frac{-\frac{9x^2}{2}}{9x^2} = \frac{-9/2}{9} = -\frac{1}{2}$$

$$\alpha + \beta = 3 + \left(-\frac{1}{2}\right) = \frac{5}{2}$$

Therefore, the correct answer is Option C: $$\dfrac{5}{2}$$.

We need to find $$\lim_{x \to 1} \frac{(x^2-1)\sin^2(\pi x)}{x^4-2x^3+2x-1}$$. Numerator: $$(x^2-1)\sin^2(\pi x) = (x-1)(x+1)\sin^2(\pi x)$$. For the denominator, since $$x = 1$$ is a root, we factor out $$(x-1)$$: $$x^4 - 2x^3 + 2x - 1 = (x-1)(x^3 - x^2 - x + 1)$$, and as $$x = 1$$ is again a root of $$x^3 - x^2 - x + 1$$: $$x^3 - x^2 - x + 1 = (x-1)(x^2 - 1) = (x-1)^2(x+1)$$, so finally $$x^4 - 2x^3 + 2x - 1 = (x-1)^3(x+1)$$.

Hence the expression simplifies to $$\frac{(x-1)(x+1)\sin^2(\pi x)}{(x-1)^3(x+1)} = \frac{\sin^2(\pi x)}{(x-1)^2}$$.

Let $$t = x - 1$$, so as $$x \to 1$$, $$t \to 0$$. Then $$\sin(\pi x) = \sin(\pi + \pi t) = -\sin(\pi t)$$ and hence $$\sin^2(\pi x) = \sin^2(\pi t)$$. Therefore, $$\lim_{t \to 0} \frac{\sin^2(\pi t)}{t^2} = \left(\lim_{t \to 0} \frac{\sin(\pi t)}{t}\right)^2 = \pi^2$$ (using the standard limit $$\lim_{u \to 0} \frac{\sin u}{u} = 1$$ with $$u = \pi t$$).

Therefore, the answer is Option D: $$\pi^2$$.

We wish to evaluate the limit $$\displaystyle\lim_{x \to \frac{\pi}{4}} \dfrac{8\sqrt{2} - (\cos x + \sin x)^7}{\sqrt{2} - \sqrt{2}\sin 2x}\,. $$

Setting $$t = x - \dfrac{\pi}{4}$$ so that $$t \to 0$$, we observe that $$ \cos x + \sin x = \sqrt{2}\cos\!\Bigl(x - \tfrac{\pi}{4}\Bigr) = \sqrt{2}\cos t\,, $$ and hence $$ (\cos x + \sin x)^7 = (\sqrt{2})^7 \cos^7 t = 8\sqrt{2}\cos^7 t\,. $$

It follows that the numerator becomes $$ 8\sqrt{2} - 8\sqrt{2}\cos^7 t = 8\sqrt{2}\bigl(1 - \cos^7 t\bigr)\,. $$

Meanwhile, since $$ \sin 2x = \sin\!\Bigl(2t + \tfrac{\pi}{2}\Bigr) = \cos 2t\,, $$ the denominator simplifies as $$ \sqrt{2} - \sqrt{2}\sin 2x = \sqrt{2}\bigl(1 - \cos 2t\bigr) = \sqrt{2}\cdot 2\sin^2 t = 2\sqrt{2}\sin^2 t\,. $$

Combining these results gives $$ L = \lim_{t \to 0} \dfrac{8\sqrt{2}(1 - \cos^7 t)}{2\sqrt{2}\sin^2 t} = 4 \lim_{t \to 0} \dfrac{1 - \cos^7 t}{\sin^2 t}\,. $$

We factor the numerator using $$1 - \cos^7 t = (1 - \cos t)\bigl(1 + \cos t + \cos^2 t + \cdots + \cos^6 t\bigr)\,, $$ so that $$ L = 4 \lim_{t \to 0} \dfrac{(1 - \cos t)\,(1 + \cos t + \cos^2 t + \cdots + \cos^6 t)}{\sin^2 t}\,. $$

As $$t \to 0$$, we know $$\lim_{t \to 0} \dfrac{1 - \cos t}{\sin^2 t} = \lim_{t \to 0} \dfrac{1 - \cos t}{1 - \cos^2 t} = \lim_{t \to 0} \dfrac{1}{1 + \cos t} = \tfrac12$$ and $$\lim_{t \to 0} \bigl(1 + \cos t + \cos^2 t + \cdots + \cos^6 t\bigr) = 7\,. $$

Therefore, $$ L = 4 \times \tfrac12 \times 7 = 14\,. $$

The correct answer is Option A: $$14$$.

Given,

$$\lim_{x\to\frac{\pi}{2}}\tan^2x\left(\sqrt{2\sin^2x+3\sin x+4}-\sqrt{\sin^2x+6\sin x+2}\right)$$

Let

$$u=\sin x$$

As

$$x\to\frac{\pi}{2}$$

we have

$$u\to1$$

Also,

$$\tan^2x=\frac{\sin^2x}{1-\sin^2x}=\frac{u^2}{1-u^2}$$

Hence, the limit becomes

$$\lim_{u\to1}\frac{u^2}{1-u^2}\left(\sqrt{2u^2+3u+4}-\sqrt{u^2+6u+2}\right)$$

Rationalizing,

$$=\lim_{u\to1}\frac{u^2\left[(2u^2+3u+4)-(u^2+6u+2)\right]}{(1-u^2)\left(\sqrt{2u^2+3u+4}+\sqrt{u^2+6u+2}\right)}$$

$$=\lim_{u\to1}\frac{u^2(u^2-3u+2)}{(1-u)(1+u)\left(\sqrt{2u^2+3u+4}+\sqrt{u^2+6u+2}\right)}$$

$$=\lim_{u\to1}\frac{u^2(u-1)(u-2)}{(1-u)(1+u)\left(\sqrt{2u^2+3u+4}+\sqrt{u^2+6u+2}\right)}$$

Since,

$$1-u=-(u-1)$$

we get

$$=\lim_{u\to1}\frac{-u^2(u-2)}{(1+u)\left(\sqrt{2u^2+3u+4}+\sqrt{u^2+6u+2}\right)}$$

Substituting

$$u=1$$

$$=\frac{-1(1-2)}{2(3+3)}$$

$$=\frac1{12}$$

Hence, $$\boxed{\frac1{12}}$$.

We need to evaluate:

$$\lim_{n \to \infty} \sum_{r=1}^{n} \frac{n^2}{(n^2+r^2)(n+r)}$$

We rewrite the general term by factoring out $$n^3$$ from the denominator:

$$\frac{n^2}{(n^2+r^2)(n+r)} = \frac{n^2}{n^2\left(1+\frac{r^2}{n^2}\right) \cdot n\left(1+\frac{r}{n}\right)} = \frac{1}{n} \cdot \frac{1}{\left(1+\left(\frac{r}{n}\right)^2\right)\left(1+\frac{r}{n}\right)}$$

Let $$x = \frac{r}{n}$$. As $$n \to \infty$$, the Riemann sum becomes:

$$I = \int_0^1 \frac{dx}{(1+x^2)(1+x)}$$

We perform partial fraction decomposition. We write:

$$\frac{1}{(1+x)(1+x^2)} = \frac{A}{1+x} + \frac{Bx+C}{1+x^2}$$

Multiplying both sides by $$(1+x)(1+x^2)$$:

$$1 = A(1+x^2) + (Bx+C)(1+x)$$

Substituting $$x = -1$$: $$1 = A(1+1) + 0 = 2A$$, so $$A = \frac{1}{2}$$.

Expanding the right side: $$1 = A + Ax^2 + Bx + Bx^2 + C + Cx$$

$$= (A+C) + (B+C)x + (A+B)x^2$$

Comparing coefficients of $$x^2$$: $$0 = A + B = \frac{1}{2} + B$$, so $$B = -\frac{1}{2}$$.

Comparing constant terms: $$1 = A + C = \frac{1}{2} + C$$, so $$C = \frac{1}{2}$$.

Therefore:

$$\frac{1}{(1+x)(1+x^2)} = \frac{1}{2} \cdot \frac{1}{1+x} + \frac{1}{2} \cdot \frac{1-x}{1+x^2}$$

We can verify: $$\frac{1}{2} \cdot \frac{1-x}{1+x^2} = \frac{1}{2} \cdot \frac{1}{1+x^2} - \frac{1}{2} \cdot \frac{x}{1+x^2}$$

Now integrating each term from 0 to 1:

First integral: $$\frac{1}{2}\int_0^1 \frac{dx}{1+x} = \frac{1}{2}[\ln(1+x)]_0^1 = \frac{1}{2}(\ln 2 - \ln 1) = \frac{1}{2}\ln 2$$

Second integral: $$\frac{1}{2}\int_0^1 \frac{dx}{1+x^2} = \frac{1}{2}[\tan^{-1}x]_0^1 = \frac{1}{2}\left(\frac{\pi}{4} - 0\right) = \frac{\pi}{8}$$

Third integral: $$-\frac{1}{2}\int_0^1 \frac{x\,dx}{1+x^2}$$

Let $$u = 1+x^2$$, then $$du = 2x\,dx$$:

$$= -\frac{1}{2} \cdot \frac{1}{2}[\ln(1+x^2)]_0^1 = -\frac{1}{4}(\ln 2 - \ln 1) = -\frac{1}{4}\ln 2$$

Adding all three parts:

$$I = \frac{1}{2}\ln 2 + \frac{\pi}{8} - \frac{1}{4}\ln 2 = \frac{\pi}{8} + \frac{1}{4}\ln 2$$

The correct answer is Option A.

We need to find $$(a - b)$$ given $$\lim_{x \to 1}\frac{\sin(3x^2 - 4x + 1) - x^2 + 1}{2x^3 - 7x^2 + ax + b} = -2$$.

Check that numerator and denominator both vanish at $$x = 1$$:

Numerator at $$x = 1$$: $$\sin(3 - 4 + 1) - 1 + 1 = \sin(0) = 0$$ $$\checkmark$$

Denominator at $$x = 1$$: $$2 - 7 + a + b = a + b - 5$$

For the limit to exist and be finite, we need $$a + b - 5 = 0$$, i.e., $$a + b = 5 \quad \cdots (1)$$.

Apply L'Hopital's rule:

Numerator derivative: $$\cos(3x^2 - 4x + 1)(6x - 4) - 2x$$

At $$x = 1$$: $$\cos(0)(2) - 2 = 2 - 2 = 0$$

Denominator derivative: $$6x^2 - 14x + a$$

At $$x = 1$$: $$6 - 14 + a = a - 8$$

For 0/0 form again, we need $$a - 8 = 0$$, so $$a = 8$$ and from (1), $$b = -3$$.

Apply L'Hopital's rule again:

Numerator second derivative:

$$\frac{d}{dx}[\cos(3x^2 - 4x + 1)(6x-4) - 2x]$$

$$= -\sin(3x^2 - 4x + 1)(6x-4)^2 + \cos(3x^2 - 4x + 1) \cdot 6 - 2$$

At $$x = 1$$: $$-\sin(0)(4) + \cos(0)(6) - 2 = 0 + 6 - 2 = 4$$

Denominator second derivative: $$12x - 14$$

At $$x = 1$$: $$12 - 14 = -2$$

Compute the limit:

$$\lim_{x \to 1} = \frac{4}{-2} = -2$$ $$\checkmark$$

Find $$a - b$$:

$$a - b = 8 - (-3) = 11$$

The answer is $$\boxed{11}$$.

We need to find the integral value of $$\alpha$$ for which the left-hand limit of $$f(x) = [1 + x] + \frac{\alpha^{2[x] + \{x\}} + [x] - 1}{2[x] + \{x\}}$$ at $$x = 0$$ equals $$\alpha - \frac{4}{3}$$.

For $$x$$ slightly less than 0, we have $$[x] = -1$$, $$\{x\} = x - [x] = x + 1$$, and $$[1 + x] = 0$$ since $$1 + x \to 1^-$$.

In this neighborhood, $$2[x] + \{x\} = 2(-1) + (x + 1) = x - 1 \to -1$$, $$\alpha^{2[x] + \{x\}} = \alpha^{x - 1} \to \alpha^{-1} = \frac{1}{\alpha}$$, and $$[x] - 1 = -1 - 1 = -2$$.

Substituting these into the definition of $$f$$ gives $$\lim_{x \to 0^-} f(x) = 0 + \frac{\frac{1}{\alpha} + (-2)}{-1} = 2 - \frac{1}{\alpha}$$.

Requiring this limit to equal $$\alpha - \frac{4}{3}$$ leads to the equation $$2 - \frac{1}{\alpha} = \alpha - \frac{4}{3}$$, which can be rewritten as $$\alpha + \frac{1}{\alpha} = \frac{10}{3}$$.

Multiplying through by $$\alpha$$ and rearranging yields the quadratic $$3\alpha^2 - 10\alpha + 3 = 0$$.

The solutions of this quadratic are $$\alpha = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm 8}{6}$$, namely $$\alpha = 3$$ or $$\alpha = \frac{1}{3}$$.

Since we are looking for an integral value, the answer is $$\alpha = 3$$.

The correct answer is $$3$$.

We need to evaluate $$\lim_{x \to 0} \left(\frac{(x+2\cos x)^3 + 2(x+2\cos x)^2 + 3\sin(x+2\cos x)}{(x+2)^3 + 2(x+2)^2 + 3\sin(x+2)}\right)^{\frac{100}{x}}$$.

Let $$g(u) = u^3 + 2u^2 + 3\sin u$$.

The expression becomes $$\left(\frac{g(x + 2\cos x)}{g(x + 2)}\right)^{100/x}$$.

At $$x = 0$$: $$x + 2\cos x = 0 + 2(1) = 2$$ and $$x + 2 = 2$$.

So the base $$\to \frac{g(2)}{g(2)} = 1$$ and the exponent $$\to \infty$$. This is a $$1^\infty$$ indeterminate form.

For a limit of the form $$\lim(1 + h)^{1/\epsilon}$$ where $$h \to 0$$ and $$\epsilon \to 0$$, we use:

$$L = e^{\lim h/\epsilon}$$

Let $$R = \frac{g(x+2\cos x)}{g(x+2)}$$. Then $$R - 1 = \frac{g(x+2\cos x) - g(x+2)}{g(x+2)}$$.

We need $$\ln L = \lim_{x \to 0} \frac{100}{x} \cdot (R - 1)$$ (since $$\ln(1+t) \approx t$$ when $$t \to 0$$, and $$R - 1 \to 0$$).

Let $$u = x + 2$$ and $$v = x + 2\cos x$$.

$$v - u = 2\cos x - 2$$

Using the Taylor expansion $$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$$:

$$v - u = 2\left(1 - \frac{x^2}{2} + \cdots\right) - 2 = -x^2 + O(x^4)$$

By the Mean Value Theorem / first-order Taylor expansion: when $$v - u$$ is small,

$$g(v) - g(u) \approx g'(u) \cdot (v - u)$$

This approximation is valid because $$g$$ is smooth and $$v - u \to 0$$ as $$x \to 0$$.

$$g'(u) = 3u^2 + 4u + 3\cos u$$

$$g'(2) = 12 + 8 + 3\cos 2 = 20 + 3\cos 2$$

$$g(2) = 8 + 8 + 3\sin 2 = 16 + 3\sin 2$$

$$R - 1 \approx \frac{g'(2) \cdot (-x^2)}{g(2)} = \frac{-(20 + 3\cos 2)}{16 + 3\sin 2} \cdot x^2$$

$$\ln L = \lim_{x \to 0} \frac{100}{x} \cdot \frac{-(20 + 3\cos 2)}{16 + 3\sin 2} \cdot x^2 = \lim_{x \to 0} \frac{-100(20 + 3\cos 2)}{16 + 3\sin 2} \cdot x = 0$$

Since the expression is proportional to $$x$$ which goes to $$0$$, we get $$\ln L = 0$$.

$$L = e^0 = 1$$

The answer is 1.

We have to evaluate the limit

$$\lim_{x \to \infty}\Bigl(\sqrt{x^2 - x + 1}\;-\;a\,x\Bigr)=b.$$

For very large positive $$x$$, the expression inside the square-root is dominated by $$x^2$$, so we first factor $$x^2$$ out of the radical:

$$\sqrt{x^2 - x + 1}=\sqrt{x^2\Bigl(1-\frac{1}{x}+\frac{1}{x^2}\Bigr)}.$$

Using the property $$\sqrt{m\,n}=\sqrt{m}\,\sqrt{n}$$, we take $$\sqrt{x^2}=x$$ outside the radical:

$$\sqrt{x^2 - x + 1}=x\;\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}.$$

Now we expand the square-root of a quantity close to 1. The Taylor (binomial) expansion about $$u=0$$ is

$$\sqrt{1+u}=1+\frac{u}{2}-\frac{u^2}{8}+O(u^3).$$

In our case

$$u=-\frac{1}{x}+\frac{1}{x^2}.$$

Substituting this $$u$$ into the series gives

$$\sqrt{1-\frac{1}{x}+\frac{1}{x^2}} \;=\;1+\frac{1}{2}\!\Bigl(-\frac{1}{x}+\frac{1}{x^2}\Bigr) -\frac{1}{8}\!\Bigl(-\frac{1}{x}+\frac{1}{x^2}\Bigr)^{\!2}+O\!\Bigl(\frac{1}{x^3}\Bigr).$$

We now simplify each part. First the linear term:

$$\frac{1}{2}\!\Bigl(-\frac{1}{x}+\frac{1}{x^2}\Bigr)=-\,\frac{1}{2x}+\frac{1}{2x^2}.$$

Next the quadratic term. Compute the square:

$$\Bigl(-\frac{1}{x}+\frac{1}{x^2}\Bigr)^{\!2} =\frac{1}{x^2}-\frac{2}{x^3}+\frac{1}{x^4}.$$

Multiplying by $$-\dfrac{1}{8}$$ gives

$$-\frac{1}{8}\Bigl(-\frac{1}{x}+\frac{1}{x^2}\Bigr)^{\!2} =-\frac{1}{8x^2}+\frac{1}{4x^3}-\frac{1}{8x^4}.$$

Adding the linear and quadratic pieces we obtain, up to order $$\dfrac{1}{x^2}$$:

$$\sqrt{1-\frac{1}{x}+\frac{1}{x^2}} =1-\frac{1}{2x}+\frac{1}{2x^2}-\frac{1}{8x^2}+O\!\Bigl(\frac{1}{x^3}\Bigr).$$

Combine the $$\dfrac{1}{x^2}$$ coefficients:

$$\frac{1}{2x^2}-\frac{1}{8x^2}=\frac{4-1}{8x^2}=\frac{3}{8x^2}.$$

Thus

$$\sqrt{1-\frac{1}{x}+\frac{1}{x^2}} =1-\frac{1}{2x}+\frac{3}{8x^2}+O\!\Bigl(\frac{1}{x^3}\Bigr).$$

Multiplying this by the prefactor $$x$$ we found earlier, we get

$$\sqrt{x^2 - x + 1} =x\Bigl[1-\frac{1}{2x}+\frac{3}{8x^2}+O\!\Bigl(\frac{1}{x^3}\Bigr)\Bigr] =x-\frac{1}{2}+\frac{3}{8x}+O\!\Bigl(\frac{1}{x^2}\Bigr).$$

So for large $$x$$ the entire expression behaves as

$$\sqrt{x^2 - x + 1}=x-\frac{1}{2}+O\!\Bigl(\frac{1}{x}\Bigr).$$

Returning to our limit, we substitute this approximation:

$$\sqrt{x^2 - x + 1}-a\,x =\Bigl(x-\frac{1}{2}+O\!\bigl(\tfrac{1}{x}\bigr)\Bigr)-a\,x =\bigl(1-a\bigr)x-\frac{1}{2}+O\!\Bigl(\frac{1}{x}\Bigr).$$

For the limit as $$x \to \infty$$ to settle at a finite number $$b$$, the coefficient of the unbounded term $$x$$ must vanish. Hence we require

$$1-a=0 \quad\Longrightarrow\quad a=1.$$

With $$a=1$$, the remaining constant term is clearly

$$b=-\frac{1}{2}.$$

Therefore the ordered pair $$(a,b)$$ equals $$\left(1,-\dfrac{1}{2}\right).$$

Hence, the correct answer is Option A.

We need to evaluate $$\displaystyle\lim_{x \to a} \dfrac{x f(a) - a f(x)}{x - a}$$, given that $$f'(a) = 2$$ and $$f(a) = 4$$.

We rewrite the numerator by adding and subtracting $$af(a)$$: $$xf(a) - af(x) = f(a)(x - a) - a\big(f(x) - f(a)\big)$$.

Therefore $$\dfrac{xf(a) - af(x)}{x - a} = f(a) - a \cdot \dfrac{f(x) - f(a)}{x - a}$$.

Taking the limit as $$x \to a$$: $$\displaystyle\lim_{x \to a}\left[f(a) - a \cdot \dfrac{f(x) - f(a)}{x - a}\right] = f(a) - a \cdot f'(a) = 4 - 2a$$.

The answer is $$4 - 2a$$.

The limit is given by:

$$\alpha = \lim_{x \to \pi/4} \frac{\tan^3 x - \tan x}{\cos(x + \pi/4)}$$

Substituting $$x = \pi/4$$ gives the indeterminate form $$\frac{1-1}{0} = \frac{0}{0}$$. We can apply L'Hôpital's Rule:

$$\alpha = \lim_{x \to \pi/4} \frac{3\tan^2 x \sec^2 x - \sec^2 x}{-\sin(x + \pi/4)}$$

Now, substitute $$x = \pi/4$$:

$$\alpha = \frac{3(1)^2(\sqrt{2})^2 - (\sqrt{2})^2}{-\sin(\pi/4 + \pi/4)} = \frac{3(2) - 2}{-\sin(\pi/2)} = \frac{6 - 2}{-1} = -4$$

The limit is given by:

$$\beta = \lim_{x \to 0} (\cos x)^{\cot x}$$

This is of the form $$1^\infty$$. We use the formula $$\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x)[f(x) - 1]}$$:

$$\beta = e^{\lim_{x \to 0} \cot x (\cos x - 1)} = e^{\lim_{x \to 0} \frac{\cos x - 1}{\tan x}}$$

Applying L'Hôpital's Rule to the exponent:

$$\lim_{x \to 0} \frac{-\sin x}{\sec^2 x} = \frac{0}{1} = 0$$

Thus, $$\beta = e^0 = 1$$.

The roots of the equation $$ax^2 + bx - 4 = 0$$ are $$\alpha = -4$$ and $$\beta = 1$$.

Using the properties of roots:

• Product of roots: $$\alpha \beta = \frac{-4}{a}$$ $$(-4)(1) = \frac{-4}{a} \implies -4 = \frac{-4}{a} \implies a = 1$$
• Sum of roots: $$\alpha + \beta = -\frac{b}{a}$$ $$-4 + 1 = -\frac{b}{1} \implies -3 = -b \implies b = 3$$

The values are $$a = 1$$ and $$b = 3$$. The ordered pair $$(a, b)$$ is (1, 3).

Correct Option: C

We have to evaluate the limit

$$\lim_{x \to 0}\dfrac{\sin^{2}\!\bigl(\pi\cos^{4}x\bigr)}{x^{4}}\;.$$

Because the variable $$x$$ is tending to zero, it is natural to replace every transcendental function by its Maclaurin (Taylor) expansion around $$x=0$$. We first expand the cosine function itself.

The standard Maclaurin expansion is stated as

$$\cos x \;=\;1-\dfrac{x^{2}}{2}+\dfrac{x^{4}}{24}+O(x^{6}).$$

Now we need $$\cos^{4}x$$, so we raise the above series to the fourth power. To do this smoothly, we write

$$\cos x \;=\;1+\delta,$$

where

$$\delta \;=\;-\dfrac{x^{2}}{2}+\dfrac{x^{4}}{24}+O(x^{6}).$$

Using the binomial theorem

$$\bigl(1+\delta\bigr)^{4} \;=\;1+4\delta+6\delta^{2}+4\delta^{3}+\delta^{4},$$

and remembering that we only need terms up to $$x^{4}$$ (since the denominator already contains $$x^{4}$$), we calculate term by term.

First,

$$4\delta \;=\;4\left(-\dfrac{x^{2}}{2}+\dfrac{x^{4}}{24}\right) \;=\;-2x^{2}+\dfrac{x^{4}}{6}.$$

Next,

$$\delta^{2} \;=\;\left(-\dfrac{x^{2}}{2}\right)^{2}+O(x^{6}) \;=\;\dfrac{x^{4}}{4}+O(x^{6}),$$

and so

$$6\delta^{2} \;=\;6\cdot\dfrac{x^{4}}{4}+O(x^{6}) \;=\;\dfrac{3x^{4}}{2}+O(x^{6}).$$

The terms $$\delta^{3}$$ and $$\delta^{4}$$ start with powers $$x^{6}$$ and higher, so they are beyond the accuracy we require and can be written collectively as $$O(x^{6})$$.

Collecting the pieces, we arrive at

$$\cos^{4}x \;=\;1-2x^{2}+\left(\dfrac{1}{6}+\dfrac{3}{2}\right)x^{4}+O(x^{6}) \;=\;1-2x^{2}+\dfrac{5}{3}x^{4}+O(x^{6}).$$

Multiplying by $$\pi$$ gives

$$\pi\cos^{4}x \;=\;\pi\bigl(1-2x^{2}+\tfrac{5}{3}x^{4}+O(x^{6})\bigr) \;=\;\pi-2\pi x^{2}+\dfrac{5\pi}{3}x^{4}+O(x^{6}).$$

It is convenient to denote the small deviation from $$\pi$$ by

$$y \;=\;-2\pi x^{2}+\dfrac{5\pi}{3}x^{4}+O(x^{6}),$$

so that

$$\pi\cos^{4}x \;=\;\pi+y.$$

Now we turn to the sine term. A fundamental trigonometric identity tells us

$$\sin(\pi+y) \;=\;-\sin y.$$

Because we shall finally square the sine, the minus sign will disappear, and we may write

$$\sin^{2}\!\bigl(\pi\cos^{4}x\bigr) \;=\;\sin^{2}(\pi+y) \;=\;\sin^{2}y.$$

Next we expand the sine of a small argument. The Maclaurin expansion reads

$$\sin y \;=\;y-\dfrac{y^{3}}{6}+O(y^{5}).$$

Squaring both sides, we obtain

$$\sin^{2}y \;=\;y^{2}-\dfrac{y^{4}}{3}+O(y^{6}).$$

Because $$y$$ itself is proportional to $$x^{2}$$, the term $$y^{4}$$ is already of order $$x^{8}$$, which is far smaller than we need. Thus, for the present limit, it suffices to keep only the leading piece:

$$\sin^{2}y \;=\;y^{2}+O(x^{6}).$$

Let us compute $$y^{2}$$ up to the required accuracy:

$$y^{2} \;=\;\Bigl(-2\pi x^{2}+\dfrac{5\pi}{3}x^{4}+O(x^{6})\Bigr)^{2} \;=\;4\pi^{2}x^{4}+O(x^{6}).$$

Therefore,

$$\sin^{2}\!\bigl(\pi\cos^{4}x\bigr) \;=\;4\pi^{2}x^{4}+O(x^{6}).$$

Finally we divide this by $$x^{4}$$, exactly as dictated by the original limit:

$$\dfrac{\sin^{2}\!\bigl(\pi\cos^{4}x\bigr)}{x^{4}} \;=\;\dfrac{4\pi^{2}x^{4}+O(x^{6})}{x^{4}} \;=\;4\pi^{2}+O(x^{2}).$$

As $$x \to 0,$$ the error term $$O(x^{2})$$ vanishes, leaving

$$\lim_{x \to 0}\dfrac{\sin^{2}\!\bigl(\pi\cos^{4}x\bigr)}{x^{4}} \;=\;4\pi^{2}.$$

Hence, the correct answer is Option C.

We have to evaluate the limit

$$\displaystyle \lim_{x\to 2}\left(\sum_{n=1}^{9}\frac{x}{n(n+1)x^{2}+2(2n+1)x+4}\right).$$

The first task is to understand the algebraic structure of the denominator. Observe that

$$n(n+1)x^{2}+2(2n+1)x+4$$

resembles the expansion of a product of two linear factors in $$x$$. Let us check whether it actually factors as $$(nx+2)\bigl((n+1)x+2\bigr)$$:

$$\bigl(nx+2\bigr)\bigl((n+1)x+2\bigr)=n(n+1)x^{2}+2n\,x+2(n+1)\,x+4 =n(n+1)x^{2}+2(2n+1)x+4.$$

The match is exact, so we may replace the quadratic by the product of two binomials:

$$n(n+1)x^{2}+2(2n+1)x+4=(nx+2)\bigl((n+1)x+2\bigr).$$

Substituting this factorization into the summand yields

$$\frac{x}{n(n+1)x^{2}+2(2n+1)x+4}=\frac{x}{(nx+2)\bigl((n+1)x+2\bigr)}.$$

Now we apply the standard partial-fraction decomposition for a difference of reciprocals. The useful identity is

$$\frac{1}{\alpha}-\frac{1}{\beta}=\frac{\beta-\alpha}{\alpha\beta}.$$ For our case, choose $$\alpha=nx+2,\qquad \beta=(n+1)x+2.$$ Then

$$\frac{1}{nx+2}-\frac{1}{(n+1)x+2}=\frac{(n+1)x+2-\bigl(nx+2\bigr)}{(nx+2)\bigl((n+1)x+2\bigr)} =\frac{x}{(nx+2)\bigl((n+1)x+2\bigr)}.$$

Comparing with the fraction we started from, we see that

$$\frac{x}{(nx+2)\bigl((n+1)x+2\bigr)}=\frac{1}{nx+2}-\frac{1}{(n+1)x+2}.$$

Hence every term in the summation can be rewritten as a simple difference:

$$\sum_{n=1}^{9}\frac{x}{(nx+2)\bigl((n+1)x+2\bigr)} =\sum_{n=1}^{9}\left(\frac{1}{nx+2}-\frac{1}{(n+1)x+2}\right).$$

The expression inside the large parentheses is telescopic; consecutive positive and negative terms cancel when expanded. Writing the first few and the last few terms makes the pattern clear:

$$\Bigl(\frac{1}{x+2}-\frac{1}{2x+2}\Bigr) +\Bigl(\frac{1}{2x+2}-\frac{1}{3x+2}\Bigr) +\dots +\Bigl(\frac{1}{9x+2}-\frac{1}{10x+2}\Bigr).$$

Every intermediate $$\displaystyle \frac{1}{kx+2}$$ for $$k=2,3,\dots,9$$ appears once with a plus sign and once with a minus sign, so they cancel pairwise. Only the very first positive term and the very last negative term remain:

$$\sum_{n=1}^{9}\left(\frac{1}{nx+2}-\frac{1}{(n+1)x+2}\right) =\frac{1}{x+2}-\frac{1}{10x+2}.$$

Therefore the whole limit problem reduces to substituting $$x=2$$ in this much simpler expression. Because both denominators stay non-zero at $$x=2$$, no indeterminate form arises, and we simply plug in:

$$\lim_{x\to 2}\left(\sum_{n=1}^{9}\frac{x}{n(n+1)x^{2}+2(2n+1)x+4}\right) =\frac{1}{2+2}-\frac{1}{10\cdot 2+2} =\frac{1}{4}-\frac{1}{22}.$$

To combine the fractions, use a common denominator $$4\times 22 = 88$$:

$$\frac{1}{4}-\frac{1}{22} =\frac{22-4}{88} =\frac{18}{88} =\frac{9}{44}.$$

Hence, the correct answer is Option D.

We have a quadratic polynomial $$f(x)=x^{2}+bx+c$$ whose distinct real roots are $$\alpha$$ and $$\beta$$, so that $$f(\beta)=0$$.

We must evaluate the limit

$$\displaystyle L=\lim_{x\to\beta}\dfrac{e^{2\bigl(x^{2}+bx+c\bigr)}-1-2\bigl(x^{2}+bx+c\bigr)}{(x-\beta)^{2}}.$$

First recall the Maclaurin expansion of the exponential function. For any real number $$u$$,

$$e^{2u}=1+2u+\dfrac{(2u)^{2}}{2!}+\dfrac{(2u)^{3}}{3!}+\dots$$

Simplifying the first few terms we obtain

$$e^{2u}=1+2u+2u^{2}+\dfrac{4u^{3}}{3!}+\dots.$$

Subtracting $$1+2u$$ from both sides gives

$$e^{2u}-1-2u=2u^{2}+\dfrac{4u^{3}}{3!}+\dots.$$

Near $$u=0$$ the higher-order terms (those containing $$u^{3},u^{4},\dots$$) are negligible compared with $$u^{2}$$, so the dominant behaviour is

$$e^{2u}-1-2u\;=\;2u^{2}+O(u^{3}).$$

In the given limit the role of $$u$$ is played by $$f(x)=x^{2}+bx+c$$. Therefore, when $$x$$ is very close to $$\beta$$, we may write

$$e^{2f(x)}-1-2f(x)=2\bigl(f(x)\bigr)^{2}+O\!\bigl((f(x))^{3}\bigr).$$

Next we express $$f(x)$$ itself in terms of $$x-\beta$$. Because $$f(x)$$ is differentiable, its first-order Taylor expansion about $$x=\beta$$ is

$$f(x)=f(\beta)+f'(\beta)\,(x-\beta)+O\bigl((x-\beta)^{2}\bigr).$$

But $$f(\beta)=0$$ (since $$\beta$$ is a root), so

$$f(x)=f'(\beta)\,(x-\beta)+O\bigl((x-\beta)^{2}\bigr).$$

Keeping only the leading term, we have

$$f(x)\approx f'(\beta)\,(x-\beta).$$

Substituting this approximation into $$2\bigl(f(x)\bigr)^{2}$$ yields

$$e^{2f(x)}-1-2f(x)\;\approx\;2\bigl(f'(\beta)\,(x-\beta)\bigr)^{2}=2\bigl(f'(\beta)\bigr)^{2}(x-\beta)^{2}.$$

Now insert this into the required limit:

$$L=\lim_{x\to\beta}\dfrac{e^{2f(x)}-1-2f(x)}{(x-\beta)^{2}} \;=\;\lim_{x\to\beta}\dfrac{2\bigl(f'(\beta)\bigr)^{2}(x-\beta)^{2}}{(x-\beta)^{2}} \;=\;2\bigl(f'(\beta)\bigr)^{2}.$$

Thus the entire task reduces to finding the derivative $$f'(x)$$ at $$x=\beta$$. Since

$$f(x)=x^{2}+bx+c,\qquad f'(x)=2x+b,$$

we get

$$f'(\beta)=2\beta+b.$$

Therefore

$$L=2\,(2\beta+b)^{2}.$$

To express this purely in terms of the coefficients $$b$$ and $$c$$, we eliminate $$\beta$$ using the relationships between the roots and coefficients. For the quadratic $$x^{2}+bx+c=0$$ with roots $$\alpha,\beta$$ we have

$$\alpha+\beta=-b\quad\text{and}\quad\alpha\beta=c.$$

The root $$\beta$$ itself satisfies its own equation:

$$\beta^{2}+b\beta+c=0\;\Longrightarrow\;\beta^{2}=-b\beta-c.$$

Compute $$(2\beta+b)^{2}$$ explicitly:

$$\begin{aligned} (2\beta+b)^{2}&=4\beta^{2}+4b\beta+b^{2}.\\ \text{Substituting }\beta^{2}&=-b\beta-c,\\ (2\beta+b)^{2}&=4(-b\beta-c)+4b\beta+b^{2}\\ &=(-4b\beta-4c)+4b\beta+b^{2}\\ &=b^{2}-4c. \end{aligned}$$

Finally, multiply by the factor $$2$$ obtained earlier:

$$L=2\,(2\beta+b)^{2}=2\,(b^{2}-4c)=2(b^{2}-4c).$$

Hence, the correct answer is Option C.

We need to find $$\lim_{n \to \infty} \frac{[r] + [2r] + \ldots + [nr]}{n^2}$$, where $$[x]$$ denotes the greatest integer function.

We use the property that for any real number $$x$$, we have $$x - 1 < [x] \leq x$$. Applying this to each term $$[kr]$$ for $$k = 1, 2, \ldots, n$$:

$$kr - 1 < [kr] \leq kr$$.

Summing from $$k = 1$$ to $$n$$: $$\sum_{k=1}^{n}(kr - 1) < \sum_{k=1}^{n}[kr] \leq \sum_{k=1}^{n}kr$$.

The right side equals $$r \cdot \frac{n(n+1)}{2}$$, and the left side equals $$r \cdot \frac{n(n+1)}{2} - n$$.

Dividing everything by $$n^2$$: $$\frac{r \cdot \frac{n(n+1)}{2} - n}{n^2} < \frac{\sum_{k=1}^{n}[kr]}{n^2} \leq \frac{r \cdot \frac{n(n+1)}{2}}{n^2}$$.

As $$n \to \infty$$, the left bound becomes $$\frac{r}{2} \cdot \frac{n+1}{n} - \frac{1}{n} \to \frac{r}{2}$$, and the right bound becomes $$\frac{r}{2} \cdot \frac{n+1}{n} \to \frac{r}{2}$$.

By the Squeeze Theorem, the limit equals $$\frac{r}{2}$$, which is Option A.

We need $$\lim_{x \to 0^+} \frac{\cos^{-1}(x - [x]^2) \cdot \sin^{-1}(x - [x]^2)}{x - x^3}$$, where $$[x]$$ is the greatest integer function.

As $$x \to 0^+$$, we have $$0 < x < 1$$, so $$[x] = 0$$. Therefore $$[x]^2 = 0$$.

The expression $$x - [x]^2 = x - 0 = x$$.

So the limit becomes $$\lim_{x \to 0^+} \frac{\cos^{-1}(x) \cdot \sin^{-1}(x)}{x - x^3}$$.

We can factor the denominator: $$x - x^3 = x(1 - x^2)$$.

So the limit is $$\lim_{x \to 0^+} \frac{\cos^{-1}(x) \cdot \sin^{-1}(x)}{x(1 - x^2)}$$.

We know that as $$x \to 0$$, $$\sin^{-1}(x) \approx x$$, so $$\frac{\sin^{-1}(x)}{x} \to 1$$.

Also as $$x \to 0$$, $$\cos^{-1}(x) \to \cos^{-1}(0) = \frac{\pi}{2}$$, and $$(1 - x^2) \to 1$$.

Therefore the limit equals $$\frac{\pi}{2} \cdot 1 \cdot \frac{1}{1} = \frac{\pi}{2}$$.

This matches Option D: $$\frac{\pi}{2}$$.

We have to evaluate the following limit as $$x$$ tends to $$0$$

$$\displaystyle L=\lim_{x\rightarrow 0}\left(\frac{x}{\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}}\right)\;.-(1)$$

Because $$x\rightarrow 0$$, the quantity $$\sin x$$ is also very small. For any small number $$u$$ we may employ the binomial (or Taylor) expansion of $$(1+u)^{\alpha}$$:

$$\displaystyle (1+u)^{\alpha}=1+\alpha\,u+\frac{\alpha(\alpha-1)}{2}\,u^{2}+O(u^{3}).-(2)$$

Here $$\alpha=\dfrac18$$. Let us set

$$u_{1}=-\sin x,\qquad u_{2}=+\sin x.$$(3)

Applying (2) individually to the two eighth-root expressions gives

$$\sqrt[8]{1-\sin x}=(1+u_{1})^{1/8}=1+\frac18\,u_{1}+\frac{\dfrac18\!\left(\dfrac18-1\right)}2\,u_{1}^{2}+O(u_{1}^{3}),-(4)$$

$$\sqrt[8]{1+\sin x}=(1+u_{2})^{1/8}=1+\frac18\,u_{2}+\frac{\dfrac18\!\left(\dfrac18-1\right)}2\,u_{2}^{2}+O(u_{2}^{3}).-(5)$$

Because $$u_{1}=-\sin x$$ and $$u_{2}=+\sin x$$, their second powers $$u_{1}^{2}$$ and $$u_{2}^{2}$$ are identical; hence those quadratic terms will cancel when we subtract (5) from (4). Subtracting (5) from (4) we obtain

$$\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}=\left[1+\frac18(-\sin x)\right]-\left[1+\frac18(\sin x)\right]+O\!\left((\sin x)^{3}\right).-(6)$$

Simplifying (6):

$$\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}=-\frac18\sin x-\frac18\sin x+O\!\left((\sin x)^{3}\right)= -\frac14\,\sin x+O\!\left((\sin x)^{3}\right).-(7)$$

Near $$x=0$$ the dominant part of the denominator is therefore $$-\dfrac14\sin x$$. Insert (7) into the original fraction (1):

$$\frac{x}{\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}}=\frac{x}{-\dfrac14\sin x+O\!\left((\sin x)^{3}\right)}. -(8)$$

For very small $$x$$ we may divide numerator and denominator by $$x$$, recalling that $$\displaystyle\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$$ :

$$\frac{x}{-\dfrac14\sin x+O\!\left((\sin x)^{3}\right)}=\frac{x}{-\dfrac14\sin x}\left[1+O\!\left((\sin x)^{2}\right)\right].-(9)$$

The factor inside the square brackets in (9) approaches $$1$$ as $$x\rightarrow 0$$, so the limit becomes

$$L=\lim_{x\rightarrow 0}\left[\frac{x}{-\dfrac14\sin x}\right]=\lim_{x\rightarrow 0}\left[-4\cdot\frac{x}{\sin x}\right]= -4\cdot 1 = -4.\;(10)$$

Hence, the correct answer is Option 3.

We have to evaluate the limit

$$\lim_{x\to 2}\frac{x^{2}\,f(2)-4\,f(x)}{x-2}.$$

The data given in the question are

$$f(2)=4\qquad\text{and}\qquad f'(2)=1.$$

First, substitute the value of $$f(2)$$ in the numerator:

$$x^{2}f(2)-4f(x)=x^{2}\cdot 4-4f(x)=4x^{2}-4f(x).$$

So the limit becomes

$$\lim_{x\to 2}\frac{4x^{2}-4f(x)}{x-2} =4\,\lim_{x\to 2}\frac{x^{2}-f(x)}{x-2}.$$

Now split the fraction in the bracket into two separate differences by adding and subtracting the same quantity 4 inside the numerator:

$$x^{2}-f(x)=\bigl(x^{2}-4\bigr)+\bigl(4-f(x)\bigr).$$

Hence,

$$\frac{x^{2}-f(x)}{x-2} =\frac{x^{2}-4}{x-2}+\frac{4-f(x)}{x-2}.$$

We shall evaluate the two limits that appear on the right separately.

First limit. Using the algebraic identity $$x^{2}-4=(x-2)(x+2),$$ we have

$$\frac{x^{2}-4}{x-2}=x+2,$$

which tends to

$$\lim_{x\to 2}(x+2)=4.$$

Second limit. Observe that

$$\frac{4-f(x)}{x-2}=-\frac{f(x)-4}{x-2}.$$

By the definition of the derivative,

$$\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}=f'(2)=1.$$

Replacing $$f(2)$$ by 4, we get

$$\lim_{x\to 2}\frac{f(x)-4}{x-2}=1.$$

Therefore,

$$\lim_{x\to 2}\frac{4-f(x)}{x-2}=-1.$$

Putting the two limits back together, we obtain

$$\lim_{x\to 2}\frac{x^{2}-f(x)}{x-2} =\;4+(-1)=3.$$

Finally, multiply by the factor 4 that we had taken out earlier:

$$\lim_{x\to 2}\frac{x^{2}f(2)-4f(x)}{x-2} =4\times 3=12.$$

Hence, the correct answer is Option D.

Given,

$$\lim_{n\to\infty}\left(1+\frac{1+\frac12+\cdots+\frac1n}{n^2}\right)^n$$

Let

$$H_n=1+\frac12+\cdots+\frac1n$$

Then expression becomes

$$\lim_{n\to\infty}\left(1+\frac{H_n}{n^2}\right)^n$$

Now,

$$H_n\sim\log n$$

Hence,

$$\frac{H_n}{n^2}\to0$$

Using standard result,

$$\left(1+\frac{a_n}{n}\right)^n\to e^L \quad \text{if } a_n\to L$$

Here,

$$a_n=\frac{H_n}{n}\to0$$

Therefore,

$$\left(1+\frac{H_n}{n^2}\right)^n\to e^0$$

$$=1$$

Hence, the required limit is

$$\boxed{1}$$

We need to evaluate $$\lim_{\theta \to 0} \frac{\tan(\pi\cos^2\theta)}{\sin(2\pi\sin^2\theta)}$$.

As $$\theta \to 0$$, we have $$\cos^2\theta \to 1$$ and $$\sin^2\theta \to 0$$. So the argument of $$\tan$$ approaches $$\pi$$ and the argument of $$\sin$$ approaches $$0$$.

We write $$\cos^2\theta = 1 - \sin^2\theta$$, so $$\pi\cos^2\theta = \pi - \pi\sin^2\theta$$.

Using the identity $$\tan(\pi - x) = -\tan(x)$$, we get $$\tan(\pi\cos^2\theta) = \tan(\pi - \pi\sin^2\theta) = -\tan(\pi\sin^2\theta)$$.

The limit becomes $$\lim_{\theta \to 0} \frac{-\tan(\pi\sin^2\theta)}{\sin(2\pi\sin^2\theta)}$$.

Let $$t = \sin^2\theta$$. As $$\theta \to 0$$, $$t \to 0$$. The expression becomes $$\lim_{t \to 0} \frac{-\tan(\pi t)}{\sin(2\pi t)}$$.

For small $$t$$, $$\tan(\pi t) \approx \pi t$$ and $$\sin(2\pi t) \approx 2\pi t$$. Therefore:

$$\lim_{t \to 0} \frac{-\tan(\pi t)}{\sin(2\pi t)} = \lim_{t \to 0} \frac{-\pi t}{2\pi t} = -\frac{1}{2}$$.

The answer is $$-\frac{1}{2}$$, which is Option A.

Given,

$$\lim_{x\to0}\frac{\int_0^{x^2}\sin\sqrt t\,dt}{x^3}$$

Put

$$t=u^2$$

Then,

$$dt=2u\,du$$

When

$$t=0,\quad u=0$$

and when

$$t=x^2,\quad u=x$$

Therefore,

$$\int_0^{x^2}\sin\sqrt t\,dt=\int_0^x\sin u\cdot2u\,du$$

Hence limit becomes

$$\lim_{x\to0}\frac{2\int_0^x u\sin u\,du}{x^3}$$

Using

$$\sin u\sim u\quad \text{as }u\to0,$$

we get

$$u\sin u\sim u^2$$

Thus,

$$2\int_0^x u\sin u\,du\sim2\int_0^x u^2\,du$$

$$=2\left[\frac{u^3}{3}\right]_0^x$$

$$=\frac{2x^3}{3}$$

Therefore,

$$\lim_{x\to0}\frac{\frac{2x^3}{3}}{x^3}$$

$$=\frac23$$

Hence, the required limit is

$$\boxed{\frac23}$$

The given expression is $$\lim_{n \to \infty} \left[\frac{1}{n} + \frac{n}{(n+1)^2} + \frac{n}{(n+2)^2} + \ldots + \frac{n}{(2n-1)^2}\right]$$.

The first term $$\frac{1}{n} = \frac{n}{n^2} = \frac{n}{(n+0)^2}$$, so the entire sum can be written as $$\sum_{k=0}^{n-1} \frac{n}{(n+k)^2}$$.

Factoring out $$\frac{1}{n}$$ from each term: $$\sum_{k=0}^{n-1} \frac{n}{(n+k)^2} = \sum_{k=0}^{n-1} \frac{1}{n} \cdot \frac{n^2}{(n+k)^2} = \sum_{k=0}^{n-1} \frac{1}{n} \cdot \frac{1}{\left(1 + \frac{k}{n}\right)^2}$$.

This is a Riemann sum of the form $$\sum_{k=0}^{n-1} \frac{1}{n} \cdot f\left(\frac{k}{n}\right)$$ where $$f(x) = \frac{1}{(1+x)^2}$$, corresponding to the integral $$\int_0^1 \frac{1}{(1+x)^2}\,dx$$ over the partition $$x_k = \frac{k}{n}$$ for $$k = 0, 1, \ldots, n-1$$.

Evaluating the integral: $$\int_0^1 \frac{1}{(1+x)^2}\,dx = \left[-\frac{1}{1+x}\right]_0^1 = \left(-\frac{1}{2}\right) - \left(-1\right) = -\frac{1}{2} + 1 = \frac{1}{2}$$.

Therefore, the limit equals $$\frac{1}{2}$$.

We need to find $$\lim_{x \to 0} \frac{ae^x - b\cos x + ce^{-x}}{x \sin x} = 2$$.

As $$x \to 0$$, the denominator $$x\sin x \to 0$$. For the limit to exist and be finite, the numerator must also vanish at $$x = 0$$. Substituting $$x = 0$$ in the numerator: $$a \cdot 1 - b \cdot 1 + c \cdot 1 = a - b + c = 0$$. This gives us our first condition: $$a - b + c = 0$$ ... (i).

Since we have a $$\frac{0}{0}$$ form, we apply L'Hopital's rule. Differentiating the numerator gives $$ae^x + b\sin x - ce^{-x}$$, and differentiating the denominator gives $$\sin x + x\cos x$$. Evaluating at $$x = 0$$: the numerator becomes $$a \cdot 1 + b \cdot 0 - c \cdot 1 = a - c$$, and the denominator becomes $$0 + 0 = 0$$. For the limit to still exist as a finite value, we again need the numerator to vanish: $$a - c = 0$$ ... (ii).

We still have a $$\frac{0}{0}$$ form, so we apply L'Hopital's rule a second time. Differentiating the numerator again gives $$ae^x + b\cos x + ce^{-x}$$, and differentiating the denominator again gives $$\cos x + \cos x - x\sin x = 2\cos x - x\sin x$$. Evaluating at $$x = 0$$: the numerator becomes $$a + b + c$$, and the denominator becomes $$2$$. Therefore the limit equals $$\frac{a + b + c}{2} = 2$$, which gives $$a + b + c = 4$$ ... (iii).

From equation (i): $$b = a + c$$. From equation (ii): $$a = c$$. Substituting $$a = c$$ into $$b = a + c$$: $$b = 2a$$. Now substituting into equation (iii): $$a + 2a + a = 4$$, so $$4a = 4$$, which gives $$a = 1$$. Therefore $$c = 1$$ and $$b = 2$$.

Hence $$a + b + c = 1 + 2 + 1 = 4$$.

We need $$\lim_{x \to 0} \frac{ax - (e^{4x} - 1)}{ax(e^{4x} - 1)}$$ to exist. Using the Taylor expansion $$e^{4x} - 1 = 4x + 8x^2 + \frac{32x^3}{3} + \cdots$$, the numerator becomes $$ax - 4x - 8x^2 - \cdots = (a - 4)x - 8x^2 - \cdots$$.

The denominator is $$ax(4x + 8x^2 + \cdots) = 4ax^2 + 8ax^3 + \cdots$$.

For the limit to exist (and be finite), the numerator must vanish to at least order $$x^2$$, so we need the coefficient of $$x$$ to be zero: $$a - 4 = 0$$, giving $$a = 4$$.

With $$a = 4$$, the numerator becomes $$-8x^2 - \frac{32x^3}{3} - \cdots$$ and the denominator becomes $$16x^2 + 32x^3 + \cdots$$.

Therefore, $$b = \lim_{x \to 0} \frac{-8x^2}{16x^2} = -\frac{1}{2}$$.

The value of $$a - 2b = 4 - 2\left(-\frac{1}{2}\right) = 4 + 1 = 5$$.

We need $$\lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4}$$ so that $$f(0) = \frac{1}{k}$$.

We use Taylor series. Recall $$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$$ and $$\cos u = 1 - \frac{u^2}{2} + \frac{u^4}{24} - \cdots$$

First, expand $$\cos x$$ up to $$x^4$$: $$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$$ $$-(1)$$

Next, expand $$\cos(\sin x)$$. We have $$\sin x = x - \frac{x^3}{6} + \cdots$$, so $$(\sin x)^2 = \left(x - \frac{x^3}{6}\right)^2 = x^2 - \frac{x^4}{3} + \cdots$$ (keeping terms up to $$x^4$$).

Also $$(\sin x)^4 = x^4 + \cdots$$ (higher-order terms are $$x^6$$ and above).

Therefore $$\cos(\sin x) = 1 - \frac{(\sin x)^2}{2} + \frac{(\sin x)^4}{24} - \cdots = 1 - \frac{x^2 - x^4/3}{2} + \frac{x^4}{24} - \cdots$$

$$= 1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24} + \cdots = 1 - \frac{x^2}{2} + \frac{4x^4 + x^4}{24} + \cdots = 1 - \frac{x^2}{2} + \frac{5x^4}{24} + \cdots$$ $$-(2)$$

Subtracting $$(1)$$ from $$(2)$$: $$\cos(\sin x) - \cos x = \left(1 - \frac{x^2}{2} + \frac{5x^4}{24}\right) - \left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) + \cdots = \frac{5x^4}{24} - \frac{x^4}{24} + \cdots = \frac{4x^4}{24} + \cdots = \frac{x^4}{6} + \cdots$$

Therefore $$\lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4} = \frac{1}{6}$$.

Since $$f(0) = \frac{1}{k} = \frac{1}{6}$$, we get $$k = 6$$.

We need to evaluate $$L = \lim_{x \to 0}\left(2 - \cos x\sqrt{\cos 2x}\right)^{(x+2)/x^2}$$. As $$x \to 0$$, the base approaches $$2 - 1 \cdot 1 = 1$$ and the exponent approaches $$\infty$$, giving the indeterminate form $$1^\infty$$.

We write $$L = e^A$$ where $$A = \lim_{x \to 0} \frac{x+2}{x^2}\left(2 - \cos x\sqrt{\cos 2x} - 1\right) = \lim_{x \to 0} \frac{x+2}{x^2}\left(1 - \cos x\sqrt{\cos 2x}\right)$$.

Using Taylor series: $$\cos x = 1 - \frac{x^2}{2} + O(x^4)$$ and $$\cos 2x = 1 - 2x^2 + O(x^4)$$. For the square root: $$\sqrt{\cos 2x} = (1 - 2x^2)^{1/2} \approx 1 - x^2 + O(x^4)$$. Multiplying:

$$\cos x \cdot \sqrt{\cos 2x} \approx \left(1 - \frac{x^2}{2}\right)\left(1 - x^2\right) = 1 - x^2 - \frac{x^2}{2} + \frac{x^4}{2} \approx 1 - \frac{3x^2}{2}$$

Therefore $$1 - \cos x\sqrt{\cos 2x} \approx \frac{3x^2}{2}$$ for small $$x$$. Substituting into the exponent:

$$A = \lim_{x \to 0} \frac{x+2}{x^2} \cdot \frac{3x^2}{2} = \lim_{x \to 0} \frac{3(x+2)}{2} = \frac{3 \cdot 2}{2} = 3$$

Therefore $$L = e^3$$, so $$a = 3$$.

We need $$\lim_{x \to 0} \frac{\alpha x e^x - \beta \ln(1+x) + \gamma x^2 e^{-x}}{x \sin^2 x} = 10$$.

Since $$\sin^2 x \approx x^2$$ near $$x = 0$$, the denominator behaves like $$x^3$$ for small $$x$$. For the limit to be finite and nonzero, the numerator must behave like $$x^3$$ near $$x = 0$$, meaning the coefficients of $$x^0$$, $$x^1$$, and $$x^2$$ in the numerator must vanish.

Expanding using Taylor series: $$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$$ $$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$$ $$e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots$$

So: $$\alpha x e^x = \alpha\left(x + x^2 + \frac{x^3}{2} + \cdots\right)$$ $$\beta \ln(1+x) = \beta\left(x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\right)$$ $$\gamma x^2 e^{-x} = \gamma\left(x^2 - x^3 + \cdots\right)$$

The numerator expanded is: $$(\alpha - \beta)x + \left(\alpha + \frac{\beta}{2} + \gamma\right)x^2 + \left(\frac{\alpha}{2} - \frac{\beta}{3} - \gamma\right)x^3 + \cdots$$

For the limit to exist (and equal 10), the coefficients of $$x^1$$ and $$x^2$$ must be zero: $$\alpha - \beta = 0 \implies \alpha = \beta \quad \cdots (i)$$ $$\alpha + \frac{\beta}{2} + \gamma = 0 \quad \cdots (ii)$$

Using $$\alpha = \beta$$ in (ii): $$\alpha + \frac{\alpha}{2} + \gamma = 0 \implies \gamma = -\frac{3\alpha}{2}$$.

The limit then equals: $$\lim_{x \to 0} \frac{\left(\frac{\alpha}{2} - \frac{\beta}{3} - \gamma\right)x^3}{x \cdot x^2} = \frac{\alpha}{2} - \frac{\alpha}{3} + \frac{3\alpha}{2} = \frac{\alpha}{2} - \frac{\alpha}{3} + \frac{3\alpha}{2} = 2\alpha - \frac{\alpha}{3} = \frac{5\alpha}{3}$$

Setting this equal to 10: $$\frac{5\alpha}{3} = 10 \implies \alpha = 6$$. Then $$\beta = 6$$ and $$\gamma = -9$$.

Therefore $$\alpha + \beta + \gamma = 6 + 6 - 9 = 3$$.

The functional equation $$f(x+y) = f(x) \cdot f(y)$$ with $$f(x) \neq 0$$ for all $$x$$ tells us that $$f$$ is an exponential function. Setting $$x = y = 0$$ gives $$f(0) = f(0)^2$$, so $$f(0) = 1$$ (since $$f(0) \neq 0$$).

Now consider the limit $$\lim_{h \to 0}\dfrac{f(h) - 1}{h}$$. Since $$f$$ is differentiable at $$x = 0$$ with $$f'(0) = 3$$, we use the definition of the derivative: $$f'(0) = \lim_{h \to 0}\dfrac{f(0+h) - f(0)}{h} = \lim_{h \to 0}\dfrac{f(h) - 1}{h} = 3$$.

Therefore $$\lim_{h \to 0}\dfrac{1}{h}(f(h) - 1) = 3$$.

We have the polynomial function $$f(x)=x^{6}+2x^{4}+x^{3}+2x+3$$ defined for all real numbers.

First we evaluate $$f(1)$$, because it will appear repeatedly.

Substituting $$x=1$$ gives

$$f(1)=1^{6}+2\cdot1^{4}+1^{3}+2\cdot1+3 =1+2+1+2+3 =9.$$

The limit in the question is

$$\lim_{x\to1} \frac{x^{n}\,f(1)-f(x)}{x-1}.$$

Since we already know that $$f(1)=9$$, we rewrite the expression inside the limit as

$$\frac{9x^{n}-f(x)}{x-1}.$$

Notice that the form $$\displaystyle\lim_{x\to a}\frac{g(x)-g(a)}{x-a}$$, provided the limit exists, equals the derivative of $$g(x)$$ at $$x=a$$. This is just the definition of the derivative:

$$g'(a)=\lim_{x\to a}\frac{g(x)-g(a)}{x-a}.$$

Here we may set

$$g(x)=9x^{n}-f(x).$$

Then the given limit is precisely $$g'(1)$$. Therefore we must compute the derivative of $$g(x)$$ and then evaluate it at $$x=1$$.

Differentiating term by term, we find

$$g'(x)=\frac{d}{dx}\!\bigl(9x^{n}\bigr)-\frac{d}{dx}\!\bigl(f(x)\bigr) =9n\,x^{\,n-1}-f'(x).$$

So we next need $$f'(x)$$, the derivative of the original polynomial. Differentiating each power of $$x$$ gives

$$f'(x)=\frac{d}{dx}\!\bigl(x^{6}\bigr)+\frac{d}{dx}\!\bigl(2x^{4}\bigr)+\frac{d}{dx}\!\bigl(x^{3}\bigr)+\frac{d}{dx}\!\bigl(2x\bigr)+\frac{d}{dx}\!\bigl(3\bigr).$$

Using the power rule $$\dfrac{d}{dx}(x^{k})=k\,x^{k-1}$$ we obtain

$$f'(x)=6x^{5}+8x^{3}+3x^{2}+2.$$

Now evaluate this derivative at $$x=1$$:

$$f'(1)=6\cdot1^{5}+8\cdot1^{3}+3\cdot1^{2}+2 =6+8+3+2 =19.$$

Returning to $$g'(x)=9n\,x^{\,n-1}-f'(x)$$, we substitute $$x=1$$ and the just-found value of $$f'(1)$$:

$$g'(1)=9n\cdot1^{\,n-1}-19=9n-19.$$

By the definition of $$g'(1)$$, this must equal the limit given in the problem, which is $$44$$. Hence we set

$$9n-19=44.$$

Now we solve this simple linear equation for $$n$$:

$$9n=44+19=63,$$

$$n=\frac{63}{9}=7.$$

Thus the required natural number is $$n=7$$.

Hence, the correct answer is Option 7.

We need to evaluate $$\lim_{n \to \infty} \tan\left(\sum_{r=1}^{n} \tan^{-1}\frac{1}{1 + r + r^2}\right)$$.

We observe that $$\frac{1}{1 + r + r^2} = \frac{1}{1 + r(r+1)} = \frac{(r+1) - r}{1 + r(r+1)}$$.

Using the identity $$\tan^{-1} A - \tan^{-1} B = \tan^{-1}\frac{A - B}{1 + AB}$$, we can write $$\tan^{-1}\frac{(r+1) - r}{1 + r(r+1)} = \tan^{-1}(r+1) - \tan^{-1}(r)$$.

So the sum telescopes: $$\sum_{r=1}^{n} \left[\tan^{-1}(r+1) - \tan^{-1}(r)\right] = \tan^{-1}(n+1) - \tan^{-1}(1)$$.

As $$n \to \infty$$, $$\tan^{-1}(n+1) \to \frac{\pi}{2}$$ and $$\tan^{-1}(1) = \frac{\pi}{4}$$.

So the sum approaches $$\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$.

Therefore, $$\lim_{n \to \infty} \tan\left(\sum_{r=1}^{n} \tan^{-1}\frac{1}{1 + r + r^2}\right) = \tan\frac{\pi}{4} = 1$$.

Hence, the answer is $$1$$.

First, we observe the quadratic equation $$p(x)=x^2-x-2=0$$ whose positive root has been denoted by $$\alpha$$. To find this root we apply the quadratic‐formula

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

with $$a=1,\; b=-1,\; c=-2$$. Substituting, we get

$$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-2)}}{2(1)} =\frac{1\pm\sqrt{1+8}}{2} =\frac{1\pm3}{2}.$$

The two values are $$\frac{1+3}{2}=2$$ and $$\frac{1-3}{2}=-1$$. Since we need the positive root, we take

$$\alpha=2.$$

The limit to be evaluated is

$$L=\lim_{x\to\alpha^{+}}\frac{\sqrt{1-\cos\!\bigl(p(x)\bigr)}}{x+\alpha-4}.$$

Because $$\alpha=2$$, the denominator simplifies neatly:

$$x+\alpha-4=x+2-4=x-2.$$

Hence

$$L=\lim_{x\to2^{+}}\frac{\sqrt{1-\cos\!\bigl(p(x)\bigr)}}{x-2}.$$

Now we rewrite $$p(x)$$ in a factorised form to understand its behaviour near $$x=2$$:

$$p(x)=x^2-x-2 =x^2-x-2+2-2 =(x^2-4)+(-x+2) =(x-2)(x+2)-1(x-2) =(x-2)(x+2-1) =(x-2)(x+1).$$

So, very compactly,

$$p(x)=(x-2)(x+1).$$

Notice that $$p(2)=0$$, and for $$x>2$$ we have $$x-2>0$$ and $$x+1>0$$, giving $$p(x)>0$$ just to the right of $$x=2$$. Therefore $$p(x)$$ itself (not its absolute value) will be small and positive as $$x\to2^{+}$$.

Whenever an angle $$\theta$$ approaches $$0$$, we use the standard trigonometric approximation

$$1-\cos\theta\;\approx\;\frac{\theta^{2}}{2}\quad\text{as}\quad\theta\to0.$$

Here our “angle” is $$\theta=p(x)$$, which indeed tends to $$0$$ when $$x\to2^{+}$$. Applying the approximation gives

$$1-\cos\!\bigl(p(x)\bigr)\;\approx\;\frac{p(x)^2}{2}.$$

Taking square roots on both sides, we obtain

$$\sqrt{1-\cos\!\bigl(p(x)\bigr)}\;\approx\;\sqrt{\frac{p(x)^2}{2}} =\frac{|p(x)|}{\sqrt{2}}.$$

But, as discussed, $$p(x)>0$$ when $$x>2$$, so $$|p(x)|=p(x)$$. Thus for $$x$$ close to but greater than $$2$$ we may replace the numerator by $$\dfrac{p(x)}{\sqrt{2}}$$, giving

$$\frac{\sqrt{1-\cos\!\bigl(p(x)\bigr)}}{x-2}\;\approx\;\frac{\dfrac{p(x)}{\sqrt{2}}}{x-2} =\frac{p(x)}{\sqrt{2}\,(x-2)}.$$

Hence our limit becomes

$$L=\lim_{x\to2^{+}}\frac{p(x)}{\sqrt{2}\,(x-2)}.$$

We have already factored $$p(x)$$, so we can simplify the fraction algebraically before taking the limit:

$$\frac{p(x)}{x-2} =\frac{(x-2)(x+1)}{x-2} =x+1,\qquad x\neq2.$$

Therefore

$$L=\lim_{x\to2^{+}}\frac{x+1}{\sqrt{2}} =\frac{\;\lim_{x\to2^{+}}(x+1)}{\sqrt{2}} =\frac{2+1}{\sqrt{2}} =\frac{3}{\sqrt{2}}.$$

Hence, the correct answer is Option B.

We are given $$[t]$$ denotes the greatest integer $$\leq t$$. We need to find:

$$L = \lim_{x \to 0} \left|\frac{1 - x + |x|}{\lambda - x + [x]}\right|$$

where the outer bars denote absolute value, the numerator contains $$|x|$$ (absolute value), and the denominator contains $$[x]$$ (greatest integer function).

Case 1: $$x \to 0^+$$

When $$x > 0$$ and close to 0: $$|x| = x$$ and $$[x] = 0$$.

Numerator $$= 1 - x + x = 1$$

Denominator $$= \lambda - x + 0 = \lambda - x$$

So the expression $$= \left|\frac{1}{\lambda - x}\right| \to \left|\frac{1}{\lambda}\right| = \frac{1}{|\lambda|}$$ as $$x \to 0^+$$.

Case 2: $$x \to 0^-$$

When $$-1 < x < 0$$: $$|x| = -x$$ and $$[x] = -1$$.

Numerator $$= 1 - x + (-x) = 1 - 2x$$

Denominator $$= \lambda - x + (-1) = (\lambda - 1) - x$$

So the expression $$= \left|\frac{1 - 2x}{(\lambda - 1) - x}\right| \to \left|\frac{1}{\lambda - 1}\right| = \frac{1}{|\lambda - 1|}$$ as $$x \to 0^-$$.

For the limit $$L$$ to exist, both one-sided limits must be equal:

$$\frac{1}{|\lambda|} = \frac{1}{|\lambda - 1|}$$

This gives $$|\lambda| = |\lambda - 1|$$.

Squaring both sides: $$\lambda^2 = (\lambda - 1)^2$$

$$\lambda^2 = \lambda^2 - 2\lambda + 1$$

$$0 = -2\lambda + 1$$

$$\lambda = \frac{1}{2}$$

Since $$\lambda = \frac{1}{2} \in \mathbb{R} - \{0, 1\}$$, this is a valid value.

Substituting $$\lambda = \frac{1}{2}$$:

$$L = \frac{1}{|\lambda|} = \frac{1}{1/2} = 2$$

Therefore, $$L = 2$$, which corresponds to Option B.

We have to find the limit

$$\displaystyle \lim_{x \to 0}\left(\frac{3x^{2}+2}{7x^{2}+2}\right)^{\tfrac{1}{x^{2}}}.$$

Whenever a limit is of the indeterminate form $$1^{\infty},$$ a standard method is to write the inner expression in the form $$1+u(x)$$ where $$u(x)\to 0$$ as $$x\to 0,$$ and then recall the well-known formula

$$\lim_{u\to 0}\left(1+u\right)^{\tfrac{1}{u}} = e.$$

For our problem, let us define

$$A(x)=\frac{3x^{2}+2}{7x^{2}+2}.$$

First we check the value of $$A(x)$$ at $$x=0$$:

$$A(0)=\frac{3\cdot 0^{2}+2}{7\cdot 0^{2}+2}=\frac{2}{2}=1.$$

So indeed $$A(x)\to 1$$ as $$x\to 0,$$ and the whole expression is of the desired $$1^{\infty}$$ type. Now we write

$$A(x)=1+u(x),\quad\text{where}\quad u(x)=\frac{3x^{2}+2}{7x^{2}+2}-1.$$

We simplify $$u(x)$$ step by step:

$$u(x)=\frac{3x^{2}+2-(7x^{2}+2)}{7x^{2}+2}.$$

Combining like terms in the numerator, we obtain

$$u(x)=\frac{3x^{2}+2-7x^{2}-2}{7x^{2}+2} =\frac{-4x^{2}}{7x^{2}+2}.$$

Clearly, $$u(x)\to 0$$ as $$x\to 0,$$ so the expression

$$\left(1+u(x)\right)^{\tfrac{1}{x^{2}}}$$

is suitable for the exponential trick. The limit we seek can be rewritten as

$$\lim_{x\to 0}\left(1+u(x)\right)^{\tfrac{1}{x^{2}}} =\exp\!\Bigl(\,\lim_{x\to 0}\tfrac{u(x)}{x^{2}}\Bigr).$$

Thus we must compute

$$\lim_{x\to 0}\frac{u(x)}{x^{2}} =\lim_{x\to 0}\frac{-4x^{2}}{x^{2}\bigl(7x^{2}+2\bigr)}.$$

The factor $$x^{2}$$ cancels:

$$\frac{u(x)}{x^{2}} =\frac{-4}{7x^{2}+2}.$$

Now let $$x\to 0$$ in the remaining expression:

$$\lim_{x\to 0}\frac{-4}{7x^{2}+2} =\frac{-4}{2} =-2.$$

Substituting this result back into the exponential, we get

$$\exp\!\left(-2\right)=e^{-2}=\frac{1}{e^{2}}.$$

Hence, the correct answer is Option B.

We have to evaluate the limit

$$L=\lim_{x \to a}\dfrac{(a+2x)^{\frac13}-(3x)^{\frac13}}{(3a+x)^{\frac13}-(4x)^{\frac13}}, \qquad (a\neq 0).$$

First we check what happens when we directly put $$x=a$$:

$$\text{Numerator at }x=a=(a+2a)^{\frac13}-(3a)^{\frac13}=(3a)^{\frac13}-(3a)^{\frac13}=0,$$

$$\text{Denominator at }x=a=(3a+a)^{\frac13}-(4a)^{\frac13}=(4a)^{\frac13}-(4a)^{\frac13}=0.$$

Thus the limit is of the indeterminate form $$0/0$$. Consequently we can apply L’Hospital’s Rule, which states:

$$\text{If }\displaystyle\lim_{x\to c}\dfrac{f(x)}{g(x)}=\dfrac{0}{0}\text{ or }\dfrac{\infty}{\infty},\ \text{then }\displaystyle\lim_{x\to c}\dfrac{f(x)}{g(x)}=\lim_{x\to c}\dfrac{f'(x)}{g'(x)},$$

provided the latter limit exists.

So we differentiate the numerator and the denominator with respect to $$x$$. We recall the derivative formula

$$\dfrac{d}{dx}\bigl(u^{1/3}\bigr)=\dfrac{1}{3}\,u^{-2/3}\,\dfrac{du}{dx}.$$

Numerator:

$$f(x)=(a+2x)^{\frac13}-(3x)^{\frac13}.$$

Using the above formula,

$$f'(x)=\dfrac{1}{3}(a+2x)^{-\frac23}\cdot 2-\dfrac{1}{3}(3x)^{-\frac23}\cdot 3.$$

Simplifying each term gives

$$f'(x)=\dfrac{2}{3\,(a+2x)^{\frac23}}-(3x)^{-\frac23}.$$

Now we evaluate at $$x=a$$:

$$f'(a)=\dfrac{2}{3\,(a+2a)^{\frac23}}-(3a)^{-\frac23} =\dfrac{2}{3\,(3a)^{\frac23}}-(3a)^{-\frac23}.$$

Writing the second term with the same base factor,

$$f'(a)=\dfrac{2}{3}(3a)^{-\frac23}-(3a)^{-\frac23} =\left(\dfrac{2}{3}-1\right)(3a)^{-\frac23} =-\dfrac{1}{3}(3a)^{-\frac23}.$$

So

$$f'(a)=-\dfrac{1}{3(3a)^{\frac23}}.$$

Denominator:

$$g(x)=(3a+x)^{\frac13}-(4x)^{\frac13}.$$

Differentiating,

$$g'(x)=\dfrac{1}{3}(3a+x)^{-\frac23}\cdot 1-\dfrac{1}{3}(4x)^{-\frac23}\cdot 4.$$

Hence

$$g'(x)=\dfrac{1}{3(3a+x)^{\frac23}}-\dfrac{4}{3(4x)^{\frac23}}.$$

Putting $$x=a$$ gives

$$g'(a)=\dfrac{1}{3(3a+a)^{\frac23}}-\dfrac{4}{3(4a)^{\frac23}} =\dfrac{1}{3(4a)^{\frac23}}-\dfrac{4}{3(4a)^{\frac23}} =\dfrac{1-4}{3(4a)^{\frac23}} =-\dfrac{1}{(4a)^{\frac23}}.$$

Applying L’Hospital’s Rule:

$$L=\dfrac{f'(a)}{g'(a)} =\dfrac{-\dfrac{1}{3(3a)^{\frac23}}}{-\dfrac{1}{(4a)^{\frac23}}} =\dfrac{(4a)^{\frac23}}{3(3a)^{\frac23}}.$$

We separate the powers of $$a$$ and simplify the numerical ratio:

$$L=\dfrac{1}{3}\,\dfrac{(4a)^{\frac23}}{(3a)^{\frac23}} =\dfrac{1}{3}\left(\dfrac{4a}{3a}\right)^{\frac23} =\dfrac{1}{3}\left(\dfrac{4}{3}\right)^{\frac23}.$$

Since $$4=2^{2}$$, we write

$$\left(\dfrac{4}{3}\right)^{\frac23} =\left(\dfrac{2^{2}}{3}\right)^{\frac23} =\dfrac{2^{\frac43}}{3^{\frac23}}.$$

Therefore

$$L=\dfrac{1}{3}\cdot\dfrac{2^{\frac43}}{3^{\frac23}} =\dfrac{2^{\frac43}}{3^{1+\frac23}} =\dfrac{2^{\frac43}}{3^{\frac53}}.$$

We now express this in the form appearing in the options. Notice that

$$\dfrac{2^{\frac43}}{3^{\frac53}} =\dfrac{2}{3}\cdot\dfrac{2^{\frac13}}{3^{\frac23}} =\left(\dfrac{2}{3}\right)\left(\dfrac{2}{9}\right)^{\frac13}.$$

This matches Option D.

Hence, the correct answer is Option D.

1. Identify the Limit and Define a Substitution

We are given the limit:

$$\lim_{x \to 0} \frac{x \left( e^{\frac{\sqrt{1+x^2+x^4}-1}{x}} - 1 \right)}{\sqrt{1+x^2+x^4}-1}$$

Let's define a new variable for the exponent of $$e$$ :

$$y = \frac{\sqrt{1+x^2+x^4}-1}{x}$$

From this substitution, we can express the denominator of our original limit as:

$$x \cdot y = \sqrt{1+x^2+x^4}-1$$

2. Evaluate the Limit of the Exponent as

$$x \to 0$$

Before substituting back into the main equation, let's find what $$y$$

approaches as $$x$$ approaches $$0$$ :

$$\lim_{x \to 0} y = \lim_{x \to 0} \frac{\sqrt{1+x^2+x^4}-1}{x}$$

To evaluate this, we multiply the numerator and denominator by the conjugate of the numerator: 

$$\lim_{x \to 0} \frac{\left(\sqrt{1+x^2+x^4}-1\right)\left(\sqrt{1+x^2+x^4}+1\right)}{x\left(\sqrt{1+x^2+x^4}+1\right)}$$

Simplify the numerator using the difference of squares
$$(a-b)(a+b) = a^2 - b^2$$ :

$$\lim_{x \to 0} \frac{1+x^2+x^4-1}{x\left(\sqrt{1+x^2+x^4}+1\right)}$$

$$\lim_{x \to 0} \frac{x^2+x^4}{x\left(\sqrt{1+x^2+x^4}+1\right)}$$

Factor out $$x$$

from the numerator to cancel with the denominator: 

$$\lim_{x \to 0} \frac{x(x+x^3)}{\sqrt{1+x^2+x^4}+1}$$

Now, substitute $$x = 0$$ :

$$\frac{0+0}{\sqrt{1+0+0}+1} = \frac{0}{2} = 0$$

So, as $$x \to 0$$ , we have $$y \to 0$$.

3. Substitute and Evaluate the Main Limit

Now, substitute $$y$$ and $$x \cdot y$$

back into the original limit expression:

$$\lim_{x \to 0} \frac{x \left( e^y - 1 \right)}{x \cdot y}$$

Cancel the $$x$$ terms:

$$\lim_{y \to 0} \frac{e^y - 1}{y}$$

This is a standard fundamental limit in calculus:

$$\lim_{y \to 0} \frac{e^y - 1}{y} = 1$$

Final Answer

The limit evaluates to $$1$$ , which corresponds to Option B.

We have to evaluate the limit

$$\lim_{x \to 0}\left(\tan\!\left(\frac{\pi}{4}+x\right)\right)^{\;1/x}.$$

First, recall the compound-angle identity for tangent:

$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\;\tan B}.$$

Using this with $$A=\frac{\pi}{4}$$ and $$B=x,$$ we get

$$\tan\!\left(\frac{\pi}{4}+x\right)=\frac{\tan\!\left(\frac{\pi}{4}\right)+\tan x}{1-\tan\!\left(\frac{\pi}{4}\right)\tan x}.$$

Because $$\tan\!\left(\frac{\pi}{4}\right)=1,$$ the expression simplifies to

$$\tan\!\left(\frac{\pi}{4}+x\right)=\frac{1+\tan x}{1-\tan x}.$$

Next, as $$x\to 0,$$ we know the standard small-angle approximation:

$$\tan x = x + \dfrac{x^3}{3}+\dots.$$

Keeping only the first-order term, we may write $$\tan x \approx x.$$ Substituting this approximation gives

$$\tan\!\left(\frac{\pi}{4}+x\right)\approx\frac{1+x}{1-x}.$$

Now we manipulate the fraction algebraically:

$$\frac{1+x}{1-x}=\left(1+x\right)\left(\frac{1}{1-x}\right).$$

Using the binomial expansion $$\frac{1}{1-x}=1+x+x^{2}+\dots,$$ the product becomes

$$\left(1+x\right)\left(1+x+x^{2}+\dots\right)=1+2x+\mathcal O(x^{2}).$$

Thus, very close to $$x=0$$ we may write

$$\tan\!\left(\frac{\pi}{4}+x\right)\approx 1+2x.$$

Returning to the original limit, we now have

$$\left(\tan\!\left(\frac{\pi}{4}+x\right)\right)^{1/x}\approx\left(1+2x\right)^{1/x}.$$

Let us rewrite the right-hand side so that the familiar exponential limit appears clearly. Observe that

$$\left(1+2x\right)^{1/x}=\left[\left(1+2x\right)^{1/(2x)}\right]^{2}.$$

We will now use the well-known limit

$$\lim_{t\to 0}\left(1+t\right)^{1/t}=e.$$

In our expression, set $$t=2x.$$ Then as $$x\to 0,$$ we also have $$t\to 0.$$ Therefore,

$$\lim_{x\to 0}\left(1+2x\right)^{1/(2x)}=e.$$

Substituting this result back, we obtain

$$\lim_{x\to 0}\left(1+2x\right)^{1/x}=\left(\lim_{x\to 0}\left(1+2x\right)^{1/(2x)}\right)^{2}=e^{2}.$$

Hence, the required limit equals $$e^{2}.$$

Hence, the correct answer is Option D.

First we evaluate the constant $$A$$ that appears in the statement. The limit to be found is

$$A=\lim_{x\to 0}\;x\left[\frac{4}{x}\right].$$

To recognise how the greatest-integer (floor) function behaves, recall the identity

$$t=\,[t]+\{t\},\quad\text{where }0\le\{t\}<1.$$

Applying this to $$t=\dfrac4x$$ gives

$$\left[\dfrac4x\right]=\dfrac4x-\left\{\dfrac4x\right\}.$$

Hence

$$x\left[\dfrac4x\right]=x\left(\dfrac4x-\left\{\dfrac4x\right\}\right)=4-x\left\{\dfrac4x\right\}.$$

The fractional part $$\displaystyle\left\{\dfrac4x\right\}$$ is always bounded between $$0$$ and $$1$$, so that the product $$x\left\{\dfrac4x\right\}$$ is squeezed between $$-\,|x|$$ and $$|x|$$. Therefore

$$\lim_{x\to 0}x\left\{\dfrac4x\right\}=0,$$

and we immediately get

$$\lim_{x\to 0}x\left[\dfrac4x\right]=\lim_{x\to 0}\left(4-x\left\{\dfrac4x\right\}\right)=4-0=4.$$

So we have established

$$A=4.$$

Now we turn to the function whose points of discontinuity are to be located:

$$f(x)=\,[x^{2}]\,\sin(\pi x).$$

The greatest-integer function $$[x^{2}]$$ can change its value only when $$x^{2}$$ itself passes through an integer. Hence potential trouble points satisfy

$$x^{2}=n,\quad n\in\{0,1,2,3,\dots\}\quad\Longrightarrow\quad x=\pm\sqrt n.$$

At such points $$[x^{2}]$$ has a jump. However, the overall product may still be continuous if the factor $$\sin(\pi x)$$ happens simultaneously to be zero, because

$$\sin(\pi x)=0\quad\Longleftrightarrow\quad x\in\mathbb Z.$$

Putting the two observations together:

• If $$x=\sqrt n$$ but $$\sqrt n\notin\mathbb Z,$$ then $$[x^{2}]$$ jumps while $$\sin(\pi x)\neq 0$$, so $$f(x)$$ is discontinuous.
• If $$x=\sqrt n$$ and in addition $$\sqrt n\in\mathbb Z$$ (namely $$x=0,\,\pm1,\,\pm2,\,\pm3,\dots$$), then $$\sin(\pi x)=0$$ on both sides and the product tends to $$0$$ from either side, making $$f(x)$$ continuous.

Thus the function is discontinuous precisely at all non-integral square roots of natural numbers.

With $$A=4$$ already found, the option list yields

$$\sqrt{A+1}=\sqrt5,\qquad \sqrt{A+5}=\sqrt9=3,\qquad \sqrt{A+21}=\sqrt{25}=5,\qquad \sqrt A=\sqrt4=2.$$

Among these, $$\sqrt5$$ is not an integer, so it satisfies the discontinuity condition, whereas $$2,3,5$$ are integers and correspond to points where $$f(x)$$ remains continuous.

Hence, the correct answer is Option A.

We have to evaluate the limit

$$\lim_{x \to 0} \dfrac{\displaystyle \int_{0}^{x} t \sin(10t)\,dt}{x}.$$

At first glance both the numerator and the denominator vanish when $$x = 0$$, giving an indeterminate form $$\dfrac{0}{0}$$. A convenient way to handle such a form is L’Hospital’s Rule.

L’Hospital’s Rule: If $$\displaystyle\lim_{x\to a} \dfrac{f(x)}{g(x)} = \dfrac{0}{0}$$ or $$\dfrac{\pm\infty}{\pm\infty}$$ and the derivatives exist near $$a$$, then

$$\lim_{x\to a} \dfrac{f(x)}{g(x)} = \lim_{x\to a} \dfrac{f'(x)}{g'(x)}.$$

In our problem we identify

$$f(x)=\int_{0}^{x} t\sin(10t)\,dt, \qquad g(x)=x.$$

We differentiate each with respect to $$x$$.

First, by the Fundamental Theorem of Calculus, the derivative of the integral is simply the integrand evaluated at the upper limit:

$$f'(x)=x\sin(10x).$$

Next, the derivative of $$g(x)=x$$ is

$$g'(x)=1.$$

Applying L’Hospital’s Rule, the original limit equals the new limit

$$\lim_{x\to 0} \dfrac{f'(x)}{g'(x)}=\lim_{x\to 0} \dfrac{x\sin(10x)}{1}= \lim_{x\to 0} x\sin(10x).$$

Now we examine $$x\sin(10x)$$ as $$x\to 0$$. We recall the standard small-angle fact:

$$\sin\theta \approx \theta \text{ when } \theta\to 0.$$

Here $$\theta = 10x$$, so $$\sin(10x) \approx 10x$$ for very small $$x$$. Substituting this approximation we get

$$x\sin(10x) \approx x\,(10x)=10x^{2}.$$

Clearly $$10x^{2}\to 0$$ as $$x\to 0$$, and therefore

$$\lim_{x\to 0} x\sin(10x)=0.$$

Hence the required limit is $$0$$.

Hence, the correct answer is Option A.

We are told that the real-valued function $$f:(0,\infty)\to(0,\infty)$$ is differentiable, that it satisfies $$f(1)=e,$$ and that the limit

$$\lim_{t\to x}\frac{t^{2}f^{2}(x)\;-\;x^{2}f^{2}(t)}{t-x}=0$$

exists and is equal to zero for every positive real number $$x$$ in the domain. We wish to find the point $$x$$ (if any) at which $$f(x)=1$$ holds.

Because the limit involves the difference quotient of an expression that vanishes when $$t=x,$$ we recognise it as the derivative (with respect to $$t$$) of the numerator, evaluated at $$t=x.$$ To formalise this observation we proceed step by step.

Let us define a new auxiliary function

$$k(t)=t^{2}f^{2}(x)-x^{2}f^{2}(t),$$

where $$x>0$$ is fixed and $$t$$ is the variable that approaches $$x.$$ Notice first that

$$k(x)=x^{2}f^{2}(x)-x^{2}f^{2}(x)=0,$$

so both the numerator and the denominator of the given difference quotient vanish when $$t=x,$$ exactly the situation in the usual definition of a derivative. By that definition, for a differentiable function $$k,$$ we have

$$\lim_{t\to x}\frac{k(t)-k(x)}{t-x}=k'(x).$$

Because $$k(x)=0,$$ the given limit becomes simply $$k'(x).$$ Therefore the condition in the problem statement may be rewritten as

$$k'(x)=0 \quad\text{for every } x>0.$$

We now differentiate $$k(t)$$ with respect to its variable $$t.$$ The first term, $$t^{2}f^{2}(x),$$ involves $$t$$ explicitly but contains the constant factor $$f^{2}(x)$$ (constant because $$x$$ is fixed while we differentiate with respect to $$t$$). Using the elementary rule $$\dfrac{d}{dt}\bigl(t^{n}\bigr)=nt^{\,n-1},$$ we obtain

$$\frac{d}{dt}\Bigl[t^{2}f^{2}(x)\Bigr]=2t\,f^{2}(x).$$

The second term, $$-x^{2}f^{2}(t),$$ contains $$t$$ only inside $$f(t).$$ First we differentiate $$f^{2}(t)$$ using the chain rule. Stating the chain rule explicitly: if $$u(t)=f(t),$$ then $$\dfrac{d}{dt}\bigl[u^{2}\bigr]=2\,u\,\dfrac{du}{dt}=2f(t)f'(t).$$ Multiplying by the constant factor $$-x^{2},$$ we get

$$\frac{d}{dt}\Bigl[-x^{2}f^{2}(t)\Bigr]=-x^{2}\cdot 2f(t)f'(t)=-2x^{2}f(t)f'(t).$$

Adding the two differentiated parts, the derivative of $$k(t)$$ is

$$k'(t)=2t\,f^{2}(x)\;-\;2x^{2}f(t)f'(t).$$

We now evaluate this derivative at the point $$t=x$$ (because the limit involves $$t\to x$$):

$$k'(x)=2x\,f^{2}(x)\;-\;2x^{2}f(x)f'(x).$$

According to the earlier deduction, the given limit equalling zero implies

$$k'(x)=0.$$

Hence

$$2x\,f^{2}(x)-2x^{2}f(x)f'(x)=0.$$

We can divide the entire equality by the common positive factor $$2x\;(x>0)$$ to simplify it:

$$f^{2}(x)-x\,f(x)f'(x)=0.$$

Since $$f(x)>0$$ for all $$x$$ in the domain (as the codomain is $$(0,\infty)$$), we may divide once more by the positive quantity $$f(x)$$ to arrive at a very clean first-order differential equation:

$$f(x)-x\,f'(x)=0.$$

We rewrite this as

$$x\,f'(x)=f(x).$$

To make the separation of variables absolutely explicit, we bring all occurrences of $$f$$ to one side and all occurrences of $$x$$ to the other:

$$\frac{f'(x)}{f(x)}=\frac{1}{x}.$$

We now integrate both sides with respect to $$x.$$ Stating the standard integrals: $$\displaystyle\int \frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C_{1}$$ and $$\displaystyle\int\frac{1}{x}\,dx=\ln|x|+C_{2}.$$ Combining the constants into one overall constant $$C,$$ we have

$$\ln f(x)=\ln x + C.$$

Exponentiating both sides to remove the natural logarithm, we obtain

$$f(x)=e^{C}\,x.$$

Let us denote the positive constant $$e^{C}$$ by the single symbol $$K.$$

So the general solution satisfying the differential equation and remaining positive is

$$f(x)=Kx,\qquad K>0.$$

We now employ the initial condition supplied in the problem: $$f(1)=e.$$ Substituting $$x=1$$ into the expression $$f(x)=Kx$$ gives

$$f(1)=K\cdot 1=K.$$

But we are told that $$f(1)=e.$$ Therefore

$$K=e.$$

Consequently the explicit form of our function is

$$f(x)=ex.$$

We are finally ready to answer the specific question, namely to find the positive number $$x$$ that satisfies $$f(x)=1.$$ Substituting the concrete formula for $$f$$ we have derived, we set

$$ex=1.$$

Solving for $$x$$ gives

$$x=\frac{1}{e}.$$

This value coincides with Option A in the list provided.

Hence, the correct answer is Option A.

Let us write $$x-1=h$$ so that $$x\to 1$$ is equivalent to $$h\to 0.$$ Re-expressing everything in terms of the new variable $$h$$ gives

$$\lim_{h\to 0}\left(\dfrac{\displaystyle\int_{0}^{h^{2}}\;t\cos t^{2}\,dt}{\;h\sin h\;}\right).$$

We now approximate the numerator. For small arguments we may use the Maclaurin series $$\cos z=1-\dfrac{z^{2}}{2}+O(z^{4}).$$ Putting $$z=t^{2}$$ we obtain

$$\cos t^{2}=1-\dfrac{(t^{2})^{2}}{2}+O(t^{8})=1-\dfrac{t^{4}}{2}+O(t^{8}).$$

Hence

$$t\cos t^{2}=t-\dfrac{t^{5}}{2}+O(t^{9}).$$

We integrate this term by term from $$0$$ to $$h^{2}:$$

$$\int_{0}^{h^{2}}t\,dt=\left[\dfrac{t^{2}}{2}\right]_{0}^{h^{2}}=\dfrac{(h^{2})^{2}}{2}=\dfrac{h^{4}}{2},$$

$$\int_{0}^{h^{2}}-\dfrac{t^{5}}{2}\,dt=-\dfrac{1}{2}\left[\dfrac{t^{6}}{6}\right]_{0}^{h^{2}}=-\dfrac{(h^{2})^{6}}{12}=-\dfrac{h^{12}}{12},$$

and the next terms are of still higher order. Therefore

$$\int_{0}^{h^{2}}t\cos t^{2}\,dt=\dfrac{h^{4}}{2}+O(h^{12}).$$

Now we study the denominator. The standard Maclaurin expansion for sine is

$$\sin h=h-\dfrac{h^{3}}{6}+O(h^{5}).$$

Hence

$$h\sin h = h\left(h-\dfrac{h^{3}}{6}+O(h^{5})\right)=h^{2}-\dfrac{h^{4}}{6}+O(h^{6}).$$

Putting the two approximations together, our limit becomes

$$\lim_{h\to 0}\dfrac{\dfrac{h^{4}}{2}+O(h^{12})}{h^{2}-\dfrac{h^{4}}{6}+O(h^{6})}.$$

First we divide numerator and denominator by $$h^{2}:$$

$$=\lim_{h\to 0}\dfrac{\dfrac{h^{2}}{2}+O(h^{10})}{1-\dfrac{h^{2}}{6}+O(h^{4})}.$$

Next, using the algebraic identity $$\dfrac{1}{1-a}=1+a+O(a^{2})$$ valid for small $$a,$$ and taking $$a=\dfrac{h^{2}}{6}+O(h^{4}),$$ we have

$$\dfrac{1}{1-\dfrac{h^{2}}{6}+O(h^{4})}=1+\dfrac{h^{2}}{6}+O(h^{4}).$$

Multiplying, we get

$$\left(\dfrac{h^{2}}{2}+O(h^{10})\right)\left(1+\dfrac{h^{2}}{6}+O(h^{4})\right)=\dfrac{h^{2}}{2}+O(h^{4}).$$

As $$h\to 0,$$ the expression $$\dfrac{h^{2}}{2}+O(h^{4})$$ clearly tends to $$0.$$ Therefore the desired limit is $$0.$$

Hence, the correct answer is Option 4.

We need to evaluate the limit

$$\lim_{x \to 1}\frac{x + x^{2} + x^{3} + \ldots + x^{n} - n}{\,x - 1\,}$$

The numerator is the series $$x^{1} + x^{2} + \ldots + x^{n}$$ minus $$n$$, so when $$x = 1$$ it becomes $$1 + 1 + \ldots + 1 - n = n - n = 0$$. Similarly the denominator $$x-1$$ also becomes $$0$$ at $$x = 1$$. Thus we have an indeterminate form $$\dfrac{0}{0}$$, and we can apply L’Hospital’s Rule.

L’Hospital’s Rule states: if $$\displaystyle\lim_{x \to a} \dfrac{f(x)}{g(x)}$$ gives $$\dfrac{0}{0}$$ or $$\dfrac{\infty}{\infty}$$, then, provided derivatives exist and the second limit exists,

$$\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}.$$

Here we take

$$f(x)=x + x^{2} + x^{3} + \ldots + x^{n}-n,\qquad g(x)=x-1.$$

Differentiating term by term:

$$f'(x)=\frac{d}{dx}\bigl(x + x^{2} + x^{3} + \ldots + x^{n}-n\bigr) = 1 + 2x + 3x^{2} + \ldots + n x^{\,n-1},$$

because $$\dfrac{d}{dx}(x^{k}) = k x^{k-1}$$ and $$\dfrac{d}{dx}(-n)=0$$. Also

$$g'(x)=\frac{d}{dx}(x-1)=1.$$

Therefore, by L’Hospital’s Rule,

$$\lim_{x \to 1}\frac{x + x^{2} + x^{3} + \ldots + x^{n} - n}{x - 1} = \lim_{x \to 1} \frac{1 + 2x + 3x^{2} + \ldots + n x^{\,n-1}}{1}.$$

Now put $$x = 1$$ directly into the differentiated expression:

$$1 + 2(1) + 3(1)^{2} + \ldots + n(1)^{\,n-1} = 1 + 2 + 3 + \ldots + n.$$

We recognize this as the sum of the first $$n$$ natural numbers. The standard formula is

$$1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}.$$

According to the question this limit equals $$820$$, so

$$\frac{n(n+1)}{2}=820.$$

Multiplying both sides by $$2$$:

$$n(n+1)=1640.$$

Expanding gives the quadratic equation

$$n^{2}+n-1640=0.$$

We solve using the quadratic formula $$n=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$ for $$ax^{2}+bx+c=0$$. Here $$a=1,\; b=1,\; c=-1640$$, so

$$n=\frac{-1 \pm \sqrt{1^{2}-4(1)(-1640)}}{2} =\frac{-1 \pm \sqrt{1+6560}}{2} =\frac{-1 \pm \sqrt{6561}}{2}.$$

Since $$6561=81^{2}$$, we have

$$n=\frac{-1 \pm 81}{2}.$$

This gives two numerical values:

$$n=\frac{-1+81}{2}=\frac{80}{2}=40,\qquad n=\frac{-1-81}{2}=\frac{-82}{2}=-41.$$

We discard $$-41$$ because $$n$$ must be a natural number $$(n\in\mathbb N)$$. Thus $$n=40.$$

Hence, the correct answer is Option 40.

We have to evaluate the limit

$$$\displaystyle \lim_{x \to 2}\frac{3^{x}+3^{\,3-x}-12}{3^{-x/2}-3^{\,1-x}}.$$$

First we check the value of the numerator and the denominator at $$x=2$$ to identify the indeterminate form.

For the numerator,

$$$3^{x}+3^{\,3-x}-12 \; \Big|_{x=2}=3^{2}+3^{\,3-2}-12=9+3-12=0.$$$

For the denominator,

$$$3^{-x/2}-3^{\,1-x}\; \Big|_{x=2}=3^{-2/2}-3^{\,1-2}=3^{-1}-3^{-1}=0.$$$

So the limit is of the indeterminate type $$0/0$$. We therefore need to simplify the expression. A convenient way is to shift the variable so that the point $$x=2$$ becomes $$y=0$$.

Put $$x=2+y,$$ where $$y \to 0$$ as $$x \to 2$$. We now express every power of 3 in terms of $$y$$.

Using the law $$3^{a+b}=3^{a}\,3^{b},$$ the numerator becomes

$$$\begin{aligned} 3^{x}+3^{\,3-x}-12 &=3^{\,2+y}+3^{\,3-(2+y)}-12 \\ &=3^{2}\,3^{y}+3^{1}\,3^{-y}-12 \\ &=9\,3^{y}+3\,3^{-y}-12. \end{aligned}$$$

The denominator is

$$$\begin{aligned} 3^{-x/2}-3^{\,1-x} &=3^{-(2+y)/2}-3^{\,1-(2+y)} \\ &=3^{-1-y/2}-3^{-1-y} \\ &=3^{-1}\,3^{-y/2}-3^{-1}\,3^{-y} \\ &=\frac13\bigl(3^{-y/2}-3^{-y}\bigr). \end{aligned}$$$

Dividing by $$\frac13$$ is equivalent to multiplying the whole fraction by 3, so the complete expression becomes

$$$\begin{aligned} \frac{3^{x}+3^{\,3-x}-12}{3^{-x/2}-3^{\,1-x}} &=\frac{9\,3^{y}+3\,3^{-y}-12}{\frac13\bigl(3^{-y/2}-3^{-y}\bigr)} \\ &=3\,\frac{9\,3^{y}+3\,3^{-y}-12}{3^{-y/2}-3^{-y}}. \end{aligned}$$$

Now we use the exponential expansion around $$y=0$$: for a small $$y$$,

$$3^{y}=1+y\ln3+\frac{y^{2}(\ln3)^{2}}{2}+\cdots,$$

$$3^{-y}=1-y\ln3+\frac{y^{2}(\ln3)^{2}}{2}+\cdots,$$

$$3^{-y/2}=1-\frac{y}{2}\ln3+\frac{y^{2}(\ln3)^{2}}{8}+\cdots.$$

Substituting these in the numerator:

$$$\begin{aligned} 9\,3^{y}+3\,3^{-y}-12 &=9\!\left(1+y\ln3+\frac{y^{2}(\ln3)^{2}}{2}\right) +3\!\left(1-y\ln3+\frac{y^{2}(\ln3)^{2}}{2}\right) -12 \\ &=(9+3-12) \;+\; (9-3)\,y\ln3 \;+\; \left(\frac{9}{2}+\frac{3}{2}\right)\!y^{2}(\ln3)^{2}+\cdots \\ &=0+6y\ln3+6y^{2}(\ln3)^{2}+\cdots. \end{aligned}$$$

So, to first order,

$$\text{Numerator}=6y\ln3+O(y^{2}).$$

For the denominator, we expand similarly:

$$$\begin{aligned} 3^{-y/2}-3^{-y} &=\left(1-\frac{y}{2}\ln3+\frac{y^{2}(\ln3)^{2}}{8}\right) -\left(1-y\ln3+\frac{y^{2}(\ln3)^{2}}{2}\right) \\ &=\left(-\frac{y}{2}\ln3\right)+y\ln3 +\left(\frac{1}{8}-\frac12\right)y^{2}(\ln3)^{2}+\cdots \\ &=\frac{y\ln3}{2}-\frac{3y^{2}(\ln3)^{2}}{8}+\cdots. \end{aligned}$$$

Thus, to first order,

$$\text{Denominator}=\frac{y\ln3}{2}+O(y^{2}).$$

Forming their ratio we get

$$$\frac{9\,3^{y}+3\,3^{-y}-12}{3^{-y/2}-3^{-y}} =\frac{6y\ln3+O(y^{2})}{\dfrac{y\ln3}{2}+O(y^{2})} =\frac{6+O(y)}{\dfrac12+O(y)} \longrightarrow\frac{6}{1/2}=12 \quad\text{as }y\to0.$$$

Remember we still have the factor 3 outside, hence

$$$\lim_{x \to 2}\frac{3^{x}+3^{\,3-x}-12}{3^{-x/2}-3^{\,1-x}} =3\times 12=36.$$$

So, the answer is $$36$$.

We have to evaluate the limit

$$$L=\lim_{x\to 0}\left\{\dfrac{1}{x^{8}}\Bigl(1-\cos\dfrac{x^{2}}{2}-\cos\dfrac{x^{2}}{4}+\cos\dfrac{x^{2}}{2}\,\cos\dfrac{x^{2}}{4}\Bigr)\right\}.$$$

Whenever a limit contains trigonometric functions near zero, the easiest path is to expand every cosine in its Maclaurin (Taylor about $$0$$) series. The standard series we shall use is

$$\cos z \;=\;1-\dfrac{z^{2}}{2!}+\dfrac{z^{4}}{4!}-\dfrac{z^{6}}{6!}+\dots$$

Because the final expression will later be divided by $$x^{8}$$, terms up to the power $$x^{8}$$ must be retained. Higher-degree terms will vanish in the limit.

Let us introduce the abbreviations

$$a=\dfrac{x^{2}}{2},\qquad b=\dfrac{x^{2}}{4}.$$

Plainly $$b=\dfrac{a}{2}$$. We shall now expand each cosine.

First, for $$\cos a$$:

$$$ \begin{aligned} \cos a&=1-\dfrac{a^{2}}{2}+\dfrac{a^{4}}{24}+O(a^{6})\\ &=1-\dfrac{(x^{2}/2)^{2}}{2}+\dfrac{(x^{2}/2)^{4}}{24}+O(x^{10})\\ &=1-\dfrac{x^{4}}{8}+\dfrac{x^{8}}{384}+O(x^{10}). \end{aligned} $$$

Next, for $$\cos b$$:

$$$ \begin{aligned} \cos b&=1-\dfrac{b^{2}}{2}+\dfrac{b^{4}}{24}+O(b^{6})\\ &=1-\dfrac{(x^{2}/4)^{2}}{2}+\dfrac{(x^{2}/4)^{4}}{24}+O(x^{10})\\ &=1-\dfrac{x^{4}}{32}+\dfrac{x^{8}}{6144}+O(x^{10}). \end{aligned} $$$

The next task is to obtain the product $$\cos a\,\cos b$$. Multiplying the two truncated series and keeping all terms through $$x^{8}$$ gives

$$$ \begin{aligned} \cos a\,\cos b &=\Bigl(1-\dfrac{x^{4}}{8}+\dfrac{x^{8}}{384}\Bigr) \Bigl(1-\dfrac{x^{4}}{32}+\dfrac{x^{8}}{6144}\Bigr)+O(x^{10})\\ &=1-\dfrac{x^{4}}{32}-\dfrac{x^{4}}{8}+\dfrac{x^{8}}{256} +\dfrac{x^{8}}{6144}+\dfrac{x^{8}}{384}+O(x^{10})\\ &=1-\dfrac{5x^{4}}{32}+\dfrac{41x^{8}}{6144}+O(x^{10}). \end{aligned} $$$

Now the entire numerator in the given limit can be assembled. We write

$$$N(x)=1-\cos\dfrac{x^{2}}{2}-\cos\dfrac{x^{2}}{4}+\cos\dfrac{x^{2}}{2}\cos\dfrac{x^{2}}{4},$$$

and substitute the three expansions computed above:

$$$ \begin{aligned} N(x)=&\;1-\Bigl(1-\dfrac{x^{4}}{8}+\dfrac{x^{8}}{384}\Bigr) -\Bigl(1-\dfrac{x^{4}}{32}+\dfrac{x^{8}}{6144}\Bigr) +\Bigl(1-\dfrac{5x^{4}}{32}+\dfrac{41x^{8}}{6144}\Bigr)+O(x^{10}). \end{aligned} $$$

We carefully combine like terms.

• The constant terms: $$1-1-1+1 = 0$$.
• The $$x^{4}$$ terms: $$\dfrac{x^{4}}{8}+\dfrac{x^{4}}{32}-\dfrac{5x^{4}}{32}=0$$.
• The $$x^{8}$$ terms: $$$ -\dfrac{x^{8}}{384}-\dfrac{x^{8}}{6144}+\dfrac{41x^{8}}{6144} = -\dfrac{16x^{8}}{6144}-\dfrac{x^{8}}{6144}+\dfrac{41x^{8}}{6144} =\dfrac{24x^{8}}{6144} =\dfrac{x^{8}}{256}. $$$

Thus

$$N(x)=\dfrac{x^{8}}{256}+O(x^{10}).$$

Returning to the original limit, we divide by $$x^{8}$$:

$$$ \begin{aligned} L&=\lim_{x\to 0}\dfrac{N(x)}{x^{8}} =\lim_{x\to 0}\dfrac{\dfrac{x^{8}}{256}+O(x^{10})}{x^{8}} =\lim_{x\to 0}\left(\dfrac{1}{256}+O(x^{2})\right) =\dfrac{1}{256}. \end{aligned} $$$

We recognize that

$$\dfrac{1}{256}=2^{-8},$$

so the limit equals $$2^{-k}$$ with $$k=8$$.

So, the answer is $$8$$.

We start with the given limit

$$L=\lim_{x\to 0}\frac{x\cot(4x)}{\sin^{2}x\;\cot^{2}(2x)}.$$

First we express every $$\cot$$ in terms of $$\sin$$ and $$\cos$$:

$$\cot(4x)=\frac{\cos(4x)}{\sin(4x)},\qquad \cot(2x)=\frac{\cos(2x)}{\sin(2x)}.$$

Substituting these in $$L$$ we obtain

$$L=\lim_{x\to 0}\frac{x\displaystyle\frac{\cos(4x)}{\sin(4x)}}{\sin^{2}x \left(\displaystyle\frac{\cos(2x)}{\sin(2x)}\right)^{2}} =\lim_{x\to 0}\frac{x\cos(4x)}{\sin(4x)} \;\frac{\sin^{2}(2x)}{\sin^{2}x\,\cos^{2}(2x)}.$$

Now we use the double-angle identity $$\sin(2x)=2\sin x\cos x$$, so

$$\sin^{2}(2x)=4\sin^{2}x\cos^{2}x.$$

Substituting this form of $$\sin^{2}(2x)$$ gives

$$L=\lim_{x\to 0}\frac{x\cos(4x)}{\sin(4x)} \;\frac{4\sin^{2}x\cos^{2}x}{\sin^{2}x\cos^{2}(2x)} =\lim_{x\to 0}\frac{4x\cos(4x)\cos^{2}x}{\sin(4x)\cos^{2}(2x)}.$$

The $$\sin^{2}x$$ terms have cancelled, leaving

$$L=\lim_{x\to 0}\Bigl(\frac{4x}{\sin(4x)}\Bigr)\; \Bigl(\cos(4x)\Bigr)\; \Bigl(\frac{\cos^{2}x}{\cos^{2}(2x)}\Bigr).$$

We can now evaluate each factor separately by using the standard limits

$$\lim_{t\to 0}\frac{\sin t}{t}=1\quad\text{and}\quad \lim_{t\to 0}\cos t=1.$$

For the first factor we put $$t=4x$$, so when $$x\to 0$$, $$t\to 0$$ and

$$\frac{4x}{\sin(4x)}=\frac{t}{\sin t}\longrightarrow 1.$$

The second factor tends to

$$\cos(4x)\longrightarrow 1.$$

For the third factor we use $$\cos x\longrightarrow 1$$ and $$\cos(2x)\longrightarrow 1$$, hence

$$\frac{\cos^{2}x}{\cos^{2}(2x)}\longrightarrow \frac{1^{2}}{1^{2}}=1.$$

Multiplying these three limiting values we get

$$L=1\times 1\times 1=1.$$

Hence, the correct answer is Option D.

We have to equate two limits:

$$\lim_{x \to 1}\frac{x^4 - 1}{x - 1}= \lim_{x \to k}\frac{x^3 - k^3}{x^2 - k^2}.$$

We first evaluate the limit on the left. A standard algebraic technique for limits of the type $$\frac{f(x)-f(a)}{x-a}$$ is to factor the numerator. We factor $$x^4-1$$ using the identity $$A^4-B^4=(A-B)(A^3+A^2B+AB^2+B^3).$$ Putting $$A=x$$ and $$B=1$$, we get

$$x^4-1=(x-1)\bigl(x^3+x^2+x+1\bigr).$$

Hence

$$\frac{x^4-1}{x-1}=x^3+x^2+x+1.$$

Now we let $$x \to 1$$:

$$x^3+x^2+x+1 \;\xrightarrow[x\to 1]{}\;1^3+1^2+1+1=1+1+1+1=4.$$

So

$$\lim_{x \to 1}\frac{x^4 - 1}{x - 1}=4.$$

Next we evaluate the limit on the right:

$$\lim_{x \to k}\frac{x^3 - k^3}{x^2 - k^2}.$$

Again we factor numerator and denominator. For the numerator we use $$A^3-B^3=(A-B)(A^2+AB+B^2).$$ With $$A=x$$ and $$B=k$$,

$$x^3-k^3=(x-k)\bigl(x^2+kx+k^2\bigr).$$

The denominator factors by the difference of squares formula $$A^2-B^2=(A-B)(A+B):$$

$$x^2-k^2=(x-k)(x+k).$$

Substituting these factorizations,

$$\frac{x^3-k^3}{x^2-k^2}=\frac{(x-k)(x^2+kx+k^2)}{(x-k)(x+k)}.$$

Because $$x \to k,$$ but $$x \neq k$$ inside the limit, we may cancel the common factor $$x-k$$:

$$\frac{x^3-k^3}{x^2-k^2}=\frac{x^2+kx+k^2}{x+k}.$$

Now we let $$x \to k$$. Replace every occurrence of $$x$$ by $$k$$:

$$\frac{k^2+kk+k^2}{k+k}=\frac{k^2+k^2+k^2}{2k}=\frac{3k^2}{2k}=\frac{3k}{2}.$$

Therefore

$$\lim_{x \to k}\frac{x^3 - k^3}{x^2 - k^2}=\frac{3k}{2}.$$

By the given equality of the two limits, we set

$$4=\frac{3k}{2}.$$

We solve for $$k$$ step by step. Multiply both sides by 2:

$$8=3k.$$

Now divide by 3:

$$k=\frac{8}{3}.$$

Hence, the correct answer is Option 4.

We have a differentiable function $$f:\; \mathbb R \to \mathbb R$$ which satisfies $$f'(3)+f'(2)=0.$$

We are asked to evaluate

$$\displaystyle \lim_{x\to 0}\Bigg(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\Bigg)^{\frac1x}.$$

Because the limit concerns a quotient whose terms approach $$1$$, it is natural to expand each term in the numerator and the denominator by the first-order Taylor formula.

Formula stated: For a differentiable function $$g$$ at a point $$a$$,

$$g(a+h)=g(a)+g'(a)\,h+o(h)\quad\text{as }h\to 0.$$

Applying this to $$f(3+x)$$ about $$x=0$$, we get

$$f(3+x)=f(3)+f'(3)\,x+o(x).$$

Substituting into the expression in the numerator,

$$1+f(3+x)-f(3)=1+\bigl(f(3)+f'(3)x+o(x)\bigr)-f(3)=1+f'(3)\,x+o(x).$$

Next, apply the same Taylor expansion to $$f(2-x)$$ about $$x=0$$:

$$f(2-x)=f(2)+f'(2)\,(-x)+o(x)=f(2)-f'(2)\,x+o(x).$$

So the denominator becomes

$$1+f(2-x)-f(2)=1+\bigl(f(2)-f'(2)x+o(x)\bigr)-f(2)=1-f'(2)\,x+o(x).$$

Because the given information tells us $$f'(3)+f'(2)=0$$, we can set

$$a=f'(3)=-f'(2).$$

Now rewrite both factors with this common constant $$a$$:

Numerator: $$1+f'(3)\,x+o(x)=1+a\,x+o(x).$$

Denominator: $$1-f'(2)\,x+o(x)=1+a\,x+o(x).$$

Hence the entire fraction is

$$\frac{1+a\,x+o(x)}{1+a\,x+o(x)}.$$

To simplify, recall that for small $$u$$,

$$\frac1{1+u}=1-u+o(u).$$

Using this,

$$\frac{1+a\,x+o(x)}{1+a\,x+o(x)}=(1+a\,x+o(x))\bigl(1-a\,x+o(x)\bigr)=1+o(x).$$

Thus we have shown that

$$\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}=1+o(x)\quad\text{as }x\to 0.$$

Let us call this quotient $$1+\phi(x)$$, where $$\phi(x)=o(x)$$, i.e. $$\dfrac{\phi(x)}{x}\to 0$$ when $$x\to 0$$.

Our limit becomes

$$\lim_{x\to 0}\bigl(1+\phi(x)\bigr)^{\frac1x}.$$

Take natural logarithms to convert the exponential form into a product:

$$\ln\!\Bigl[\bigl(1+\phi(x)\bigr)^{\frac1x}\Bigr]=\frac1x\,\ln\bigl(1+\phi(x)\bigr).$$

Again, for small $$u,$$ the logarithmic expansion is $$\ln(1+u)=u-\dfrac{u^{2}}{2}+o(u^{2}).$$ Setting $$u=\phi(x),$$ we have

$$\ln\bigl(1+\phi(x)\bigr)=\phi(x)+o(\phi(x)),$$

and therefore

$$\frac1x\,\ln\bigl(1+\phi(x)\bigr)=\frac{\phi(x)}{x}+o\!\Bigl(\frac{\phi(x)}{x}\Bigr).$$

But by construction $$\dfrac{\phi(x)}{x}\to 0,$$ so the entire right-hand side tends to $$0$$.

Consequently,

$$\lim_{x\to 0}\ln\!\Bigl[\bigl(1+\phi(x)\bigr)^{\frac1x}\Bigr]=0,$$

and exponentiating both sides gives

$$\lim_{x\to 0}\Bigl(1+\phi(x)\Bigr)^{\frac1x}=e^{\,0}=1.$$

Since this limit is exactly the one required, we conclude that

$$\lim_{x\to 0}\Bigg(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\Bigg)^{\frac1x}=1.$$

Hence, the correct answer is Option A.

We wish to evaluate the limit

$$\lim_{x \to 0} \dfrac{\sin^{2}x}{\sqrt{2} - \sqrt{1 + \cos x}}.$$

First, to remove the square-root expression from the denominator we rationalise. The standard algebraic identity we use is

$$(a-b)(a+b)=a^{2}-b^{2}.$$

Here we take $$a=\sqrt{2} \quad\text{and}\quad b=\sqrt{1+\cos x}.$$ Multiplying numerator and denominator by $$a+b=\sqrt{2}+\sqrt{1+\cos x}$$ leaves the value of the fraction unchanged but simplifies the denominator.

So we write

$$\frac{\sin^{2}x}{\sqrt{2}-\sqrt{1+\cos x}} \;=\; \frac{\sin^{2}x\;(\sqrt{2}+\sqrt{1+\cos x})} {(\sqrt{2}-\sqrt{1+\cos x})(\sqrt{2}+\sqrt{1+\cos x})}.$$

Applying the identity just quoted to the denominator:

$$ (\sqrt{2})^{2}-(\sqrt{1+\cos x})^{2} \;=\; 2-(1+\cos x) \;=\; 1-\cos x.$$

Hence the limit becomes

$$\lim_{x\to0}\; \dfrac{\sin^{2}x\;(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}.$$

Now we recall the Pythagorean identity

$$1-\cos x = 2\sin^{2}\left(\dfrac{x}{2}\right),$$

and also the double-angle relation

$$\sin x = 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right).$$

Squaring the double-angle formula gives

$$\sin^{2}x = 4\sin^{2}\left(\dfrac{x}{2}\right)\cos^{2}\left(\dfrac{x}{2}\right).$$

Substituting these two expressions into our fraction we get

$$\frac{\sin^{2}x}{1-\cos x} \;=\; \frac{4\sin^{2}\!\left(\dfrac{x}{2}\right)\cos^{2}\!\left(\dfrac{x}{2}\right)} {2\sin^{2}\!\left(\dfrac{x}{2}\right)} \;=\; 2\cos^{2}\!\left(\dfrac{x}{2}\right).$$

Therefore the entire expression simplifies to

$$\left[\,2\cos^{2}\!\left(\dfrac{x}{2}\right)\right] \; \bigl(\sqrt{2}+\sqrt{1+\cos x}\bigr).$$

Now we take the limit as $$x \to 0.$$ We know the standard limits

$$\cos\left(\dfrac{x}{2}\right) \xrightarrow[x\to0]{} 1,$$ $$\cos x \xrightarrow[x\to0]{} 1.$$

Using these values we have

$$2\cos^{2}\!\left(\dfrac{0}{2}\right) \;=\; 2\cdot 1^{2} \;=\; 2,$$

and

$$\sqrt{1+\cos 0} \;=\; \sqrt{1+1} \;=\; \sqrt{2}.$$

Thus the bracket evaluates to

$$\sqrt{2}+\sqrt{2}=2\sqrt{2}.$$

Multiplying the two limiting factors gives

$$2 \times 2\sqrt{2} \;=\; 4\sqrt{2}.$$

So the limit is

$$\boxed{4\sqrt{2}}.$$

Hence, the correct answer is Option A.

We have to evaluate the limit

$$$\lim_{x \to 1^-}\dfrac{\sqrt{\pi}-\sqrt{2\sin^{-1}x}}{\sqrt{1-x}}.$$$

First observe what happens to the numerator and the denominator separately as $$x$$ approaches $$1$$ from the left.

When $$x \to 1^{-}$$, we know that $$\sin^{-1}x$$ (that is, $$\arcsin x$$) approaches $$\dfrac{\pi}{2}$$. Therefore

$$$\sqrt{\pi}-\sqrt{2\sin^{-1}x}\; \longrightarrow\; \sqrt{\pi}-\sqrt{2\cdot\frac{\pi}{2}} = \sqrt{\pi}-\sqrt{\pi}=0.$$$

Simultaneously, the denominator also approaches zero, because

$$\sqrt{1-x}\; \longrightarrow\; \sqrt{1-1}=0.$$

So the limit has the indeterminate form $$\dfrac{0}{0}$$. Whenever we face a $$\dfrac{0}{0}$$ form, L’Hospital’s Rule is applicable. The rule states:

If $$\displaystyle\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0$$ and the derivatives exist near $$a$$, then $$$\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$$$.

So we differentiate the numerator and the denominator with respect to $$x$$.

Derivative of the numerator

The numerator is $$f(x)=\sqrt{\pi}-\sqrt{2\sin^{-1}x}$$. The constant $$\sqrt{\pi}$$ differentiates to $$0$$, so we focus on the second term.

First recall that for any differentiable function $$u(x)$$,

$$$\dfrac{d}{dx}\sqrt{u(x)}=\dfrac{1}{2\sqrt{u(x)}}\,u'(x).$$$

Here $$u(x)=2\sin^{-1}x$$. Thus,

$$$f'(x)=-\dfrac{1}{2\sqrt{2\sin^{-1}x}}\;\dfrac{d}{dx}\bigl(2\sin^{-1}x\bigr).$$$

Now $$$\dfrac{d}{dx}\bigl(2\sin^{-1}x\bigr)=2\cdot\dfrac{1}{\sqrt{1-x^{2}}}$$$ because the derivative of $$\sin^{-1}x$$ is $$\dfrac{1}{\sqrt{1-x^{2}}}$$.

Substituting this derivative back, we obtain

$$$f'(x) =-\dfrac{1}{2\sqrt{2\sin^{-1}x}}\;\Bigl(2\cdot\dfrac{1}{\sqrt{1-x^{2}}}\Bigr) =-\dfrac{1}{\sqrt{2\sin^{-1}x}\,\sqrt{1-x^{2}}}.$$$

Derivative of the denominator

The denominator is $$g(x)=\sqrt{1-x}$$. Using the same square-root derivative formula with $$v(x)=1-x$$, we get

$$$g'(x)=\dfrac{1}{2\sqrt{1-x}}\;\dfrac{d}{dx}(1-x) =\dfrac{1}{2\sqrt{1-x}}\;(-1) =-\dfrac{1}{2\sqrt{1-x}}.$$$

Applying L’Hospital’s Rule

Therefore,

$$$\lim_{x \to 1^-}\dfrac{\sqrt{\pi}-\sqrt{2\sin^{-1}x}}{\sqrt{1-x}} =\lim_{x \to 1^-}\dfrac{f'(x)}{g'(x)} =\lim_{x \to 1^-} \dfrac{-\dfrac{1}{\sqrt{2\sin^{-1}x}\,\sqrt{1-x^{2}}}} {-\dfrac{1}{2\sqrt{1-x}}}. $$$

The negatives cancel, giving

$$$\lim_{x \to 1^-} \dfrac{1}{\sqrt{2\sin^{-1}x}\,\sqrt{1-x^{2}}} \;\bigg/\; \dfrac{1}{2\sqrt{1-x}} =\lim_{x \to 1^-} \dfrac{2\sqrt{1-x}}{\sqrt{2\sin^{-1}x}\,\sqrt{1-x^{2}}}. $$$

We now simplify the square roots in the denominator. Notice that

$$1-x^{2}=(1-x)(1+x),$$

so

$$$\sqrt{1-x^{2}}=\sqrt{(1-x)(1+x)}=\sqrt{1-x}\,\sqrt{1+x}.$$$

Substituting this factorisation, we obtain

$$$\dfrac{2\sqrt{1-x}} {\sqrt{2\sin^{-1}x}\,\bigl(\sqrt{1-x}\,\sqrt{1+x}\bigr)} =\dfrac{2}{\sqrt{2\sin^{-1}x}\,\sqrt{1+x}}.$$$

At this stage the factor $$\sqrt{1-x}$$ cancels completely, so the limit is no longer indeterminate. We can now let $$x$$ approach $$1$$:

When $$x \to 1^{-}$$,

$$\sin^{-1}x \to \dfrac{\pi}{2}$$  and  $$1+x \to 2.$$

Hence

$$$\sqrt{2\sin^{-1}x}\; \longrightarrow\; \sqrt{2\cdot\dfrac{\pi}{2}}=\sqrt{\pi},$$$

and

$$\sqrt{1+x}\;\longrightarrow\;\sqrt{2}.$$

Therefore the limit equals

$$$\dfrac{2}{\sqrt{\pi}\,\sqrt{2}} =\dfrac{2}{\sqrt{2}\,\sqrt{\pi}} =\left(\dfrac{2}{\sqrt{2}}\right)\,\dfrac{1}{\sqrt{\pi}} =\sqrt{2}\,\dfrac{1}{\sqrt{\pi}} =\sqrt{\dfrac{2}{\pi}}.$$$

This value corresponds to Option B.

Hence, the correct answer is Option B.

We have to evaluate the one-sided limit

$$\lim_{x\to 1^{+}}\; \dfrac{ \bigl(1-|x|+\sin|1-x|\bigr)\; \sin\!\left(\dfrac{\pi}{2}\,[\,1-x\,]\right)} {|1-x|\,[\,1-x\,]},$$ where $$[t]$$ denotes the greatest integer less than or equal to $$t$$ (the “greatest-integer” or “floor” function).

Because $$x\to 1^{+}$$ we are interested only in values of $$x$$ that are just a little larger than $$1$$. Write

$$ x = 1+h,\qquad\text{where }h>0\text{ and }h\to 0^{+}. $$

All the pieces of the given expression can now be rewritten in terms of this small positive quantity $$h$$.

First observe that

$$ 1-x \;=\; 1-(1+h)= -h, \qquad\text{so}\qquad |1-x| = |-h| = h. $$

Next we need the value of the floor function $$[\,1-x\,] = [\, -h\,]$$ when $$0<h<1$$. Since $$-1 < -h \le 0,$$ the greatest integer that does not exceed $$-h$$ is $$-1$$. Therefore

$$ [\,1-x\,]= -1\quad\text{for all }x\text{ with }1<x<2. $$

With these elementary facts we can rewrite every factor of the original expression.

(i) The first bracket.

$$ 1-|x|+\sin|1-x| = 1-(1+h)+\sin(h) = -h+\sin h. $$

(ii) The sine containing the floor function.

$$\sin\!\left(\dfrac{\pi}{2}\,[\,1-x\,]\right) =\sin\!\Bigl(\dfrac{\pi}{2}\cdot(-1)\Bigr) =\sin\!\bigl(-\tfrac{\pi}{2}\bigr) =-1.$$

(iii) The entire denominator.

$$ |1-x|\,[\,1-x\,] =h\cdot(-1) =-h. $$

Putting (i), (ii) and (iii) together gives

$$\frac{\bigl(1-|x|+\sin|1-x|\bigr) \sin\!\left(\dfrac{\pi}{2}[1-x]\right)} {|1-x|[1-x]} = \frac{(-h+\sin h)\,(-1)}{-h}.$$

Multiplying out the two minus-signs in the numerator we get

$$ \frac{(-h+\sin h)\,(-1)}{-h} = \frac{h-\sin h}{-h} = -\frac{h-\sin h}{h}. $$

Now recall the standard Maclaurin expansion $$\sin h = h-\dfrac{h^{3}}{6}+O(h^{5})$$ for small $$h$$. Subtracting this from $$h$$ gives

$$h-\sin h = h-\left(h-\dfrac{h^{3}}{6}+O(h^{5})\right) = \dfrac{h^{3}}{6}+O(h^{5}).$$

Therefore

$$-\frac{h-\sin h}{h} = -\frac{\dfrac{h^{3}}{6}+O(h^{5})}{h} = -\left(\dfrac{h^{2}}{6}+O(h^{4})\right).$$

Because the factor in parentheses contains the square of the small quantity $$h$$, it approaches $$0$$ as $$h\to 0^{+}$$. Hence the whole expression tends to $$0$$.

Formally, using the well-known limit $$\displaystyle\lim_{h\to 0}\frac{\sin h}{h}=1,$$ we may write more compactly

$$-\frac{h-\sin h}{h} = -\Bigl(1-\frac{\sin h}{h}\Bigr) \;\xrightarrow[h\to 0]{}\; -(1-1)=0.$$

Thus

$$\boxed{\,\displaystyle\lim_{x\to 1^{+}} \frac{(1-|x|+\sin|1-x|)\sin\!\left(\dfrac{\pi}{2}[1-x]\right)} {|1-x|[1-x]}\;=\;0\,}.$$

Hence, the correct answer is Option A.

We are given the function $$f(x)= [x]-\left[\dfrac{x}{4}\right]$$ where $$[\,\cdot\,]$$ denotes the greatest-integer (floor) function. Our task is to study the behaviour of this function when $$x$$ approaches $$4$$ from the left and from the right, and then decide whether the function is continuous at $$x=4$$.

First recall the definition of the greatest-integer function: for any real number $$t$$, $$[t]$$ is the unique integer satisfying $$[t]\le t<[t]+1$$. We shall repeatedly use this basic property.

We begin with the left-hand neighbourhood of $$4$$. Take $$x$$ such that $$3<x<4$$. Because $$x<4$$, the greatest integer not exceeding $$x$$ is $$3$$, i.e. $$[x]=3$$. Now consider the term $$\dfrac{x}{4}$$. For the same $$x$$ we have

$$

\frac{3}{4}<\frac{x}{4}<1.

$$

Each number lying strictly between $$\dfrac{3}{4}$$ and $$1$$ has greatest integer $$0$$, so

$$

\left[\frac{x}{4}\right]=0\qquad\text{whenever }3<x<4.

$$

Putting these two pieces together, for $$x$$ in the interval $$(3,4)$$ we get

$$

f(x)= [x]-\left[\frac{x}{4}\right]=3-0=3.

$$

Hence the left-hand expression of the function is the constant $$3$$, and therefore the left-hand limit exists and equals

$$

\lim_{x\to4^-}f(x)=3.

$$

Now we turn to the right-hand neighbourhood of $$4$$. Take $$x$$ such that $$4<x<5$$. In this case we clearly have $$[x]=4$$ because $$4<x<5$$. Likewise

$$

1<\frac{x}{4}<\frac{5}{4}=1.25,

$$

so every value of $$\dfrac{x}{4}$$ in this range lies strictly between $$1$$ and $$2$$. The greatest integer not exceeding such a number is $$1$$, and therefore

$$

\left[\frac{x}{4}\right]=1\qquad\text{whenever }4<x<5.

$$

Substituting into the definition of $$f$$ again gives, for $$x$$ in $$(4,5)$$,

$$

f(x)= [x]-\left[\frac{x}{4}\right]=4-1=3.

$$

Thus the value of $$f(x)$$ to the right of $$4$$ is also the constant $$3$$, which means the right-hand limit exists and equals

$$

\lim_{x\to4^+}f(x)=3.

$$

Because both one-sided limits exist and are equal, the two-sided limit exists and equals $$3$$.

Finally we evaluate the actual value of the function at $$x=4$$:

$$

f(4)= [4]-\left[\frac{4}{4}\right]=4-1=3.

$$

We now invoke the continuity criterion: a function is continuous at a point $$c$$ if

$$

\lim_{x\to c}f(x)=f(c).

$$

We have already shown

$$

\lim_{x\to4}f(x)=3\quad\text{and}\quad f(4)=3,

$$

so the equality holds. Hence the function is continuous at $$x=4$$. Both one-sided limits exist, they are equal to each other, and they coincide with the function’s value at the point.

Among the given statements, Option B is exactly the assertion that $$f$$ is continuous at $$x=4$$. None of the other options correctly matches our findings.

Hence, the correct answer is Option B.

We have to evaluate the limit

$$\lim_{x \to 1}\dfrac{x^{2}-ax+b}{x-1}=5.$$

Because the denominator $$x-1$$ approaches $$0$$ as $$x \to 1$$, the limit can stay finite only if the numerator also becomes $$0$$ at $$x=1$$. Therefore we must have

$$1^{2}-a(1)+b=0 \quad\Longrightarrow\quad 1-a+b=0 \quad\Longrightarrow\quad b=a-1.$$

Knowing that $$x^{2}-ax+b$$ vanishes at $$x=1$$, the Factor Theorem tells us that $$x-1$$ is a factor of the quadratic. Hence we can write

$$x^{2}-ax+b=(x-1)(x-c)$$

for some real number $$c$$. Multiplying out the right-hand side gives

$$ (x-1)(x-c)=x^{2}-cx-x+ c=\;x^{2}-(c+1)x+c. $$

Comparing coefficients with the original quadratic $$x^{2}-ax+b,$$ we obtain

$$a=c+1 \quad\text{and}\quad b=c.$$

After canceling the common factor $$x-1,$$ the expression inside the limit simplifies to

$$\dfrac{x^{2}-ax+b}{x-1}=x-c.$$

Now we let $$x \to 1$$:

$$\lim_{x \to 1}(x-c)=1-c.$$

But the given limit equals $$5,$$ so we set

$$1-c=5 \quad\Longrightarrow\quad c=-4.$$

Substituting $$c=-4$$ into the relations $$a=c+1$$ and $$b=c,$$ we get

$$a=-4+1=-3 \quad\text{and}\quad b=-4.$$

Finally,

$$a+b=(-3)+(-4)=-7.$$

Hence, the correct answer is Option D.

We have to evaluate the limit

$$L=\lim_{y\to 0}\dfrac{\sqrt{\,1+\sqrt{\,1+y^{4}\,}\,}-\sqrt2}{y^{4}}.$$

Because the denominator contains the very small power $$y^{4}$$, it is natural to expand every radical in the numerator by using the binomial (Taylor) series of the square-root function near $$1$$.

First, recall the standard expansion valid for $$|t|\ll 1$$:

$$\sqrt{1+t}=1+\dfrac{t}{2}-\dfrac{t^{2}}{8}+O(t^{3}).$$

We apply this formula step by step. Put $$t=y^{4}$$ in it. Then

$$\sqrt{\,1+y^{4}\,}=1+\dfrac{y^{4}}{2}-\dfrac{y^{8}}{8}+O(y^{12}).$$

Now add $$1$$ to the above expression because it appears inside the outer radical:

$$1+\sqrt{\,1+y^{4}\,}=1+\left(1+\dfrac{y^{4}}{2}-\dfrac{y^{8}}{8}+O(y^{12})\right) =2+\dfrac{y^{4}}{2}-\dfrac{y^{8}}{8}+O(y^{12}).$$

The next task is to take the square root of this new quantity. For convenience we factor out the constant $$2$$:

$$2+\dfrac{y^{4}}{2}-\dfrac{y^{8}}{8}=2\Bigl(1+\underbrace{\dfrac{y^{4}}{4}-\dfrac{y^{8}}{16}+O(y^{12})}_{\displaystyle v}\Bigr).$$

Thus

$$\sqrt{\,1+\sqrt{\,1+y^{4}\,}\,} =\sqrt{2}\;\sqrt{1+v}, \qquad v=\dfrac{y^{4}}{4}-\dfrac{y^{8}}{16}+O(y^{12}).$$

We again employ the same series $$\sqrt{1+v}=1+\dfrac{v}{2}-\dfrac{v^{2}}{8}+O(v^{3})$$. Substituting the expression of $$v$$ and keeping only terms up to order $$y^{4}$$ (higher powers will vanish when divided by $$y^{4}$$ in the limit) we get

$$\sqrt{1+v}=1+\dfrac{1}{2}\left(\dfrac{y^{4}}{4}\right)+O(y^{8}) =1+\dfrac{y^{4}}{8}+O(y^{8}).$$

Therefore

$$\sqrt{\,1+\sqrt{\,1+y^{4}\,}\,} =\sqrt{2}\Bigl(1+\dfrac{y^{4}}{8}+O(y^{8})\Bigr) =\sqrt{2}+\dfrac{\sqrt{2}\,y^{4}}{8}+O(y^{8}).$$

Now form the needed numerator:

$$\sqrt{\,1+\sqrt{\,1+y^{4}\,}\,}-\sqrt{2} =\left(\sqrt{2}+\dfrac{\sqrt{2}\,y^{4}}{8}+O(y^{8})\right)-\sqrt{2} =\dfrac{\sqrt{2}\,y^{4}}{8}+O(y^{8}).$$

Divide this by $$y^{4}$$:

$$\dfrac{\sqrt{\,1+\sqrt{\,1+y^{4}\,}\,}-\sqrt{2}}{y^{4}} =\dfrac{\sqrt{2}}{8}+O(y^{4}).$$

As $$y\to 0$$ the term $$O(y^{4})$$ tends to $$0$$, hence the limit equals the constant coefficient just obtained:

$$L=\dfrac{\sqrt{2}}{8}=\dfrac{1}{4\sqrt{2}}.$$

Hence, the correct answer is Option B.

We have to find the value of

$$L=\lim_{x\to 0}\frac{\tan\!\bigl(\pi\sin^{2}x\bigr)\;+\;\bigl(|x|-\sin(x[x])\bigr)^{2}}{x^{2}}.$$

The limit can exist only if the left-hand limit (LHL) and the right-hand limit (RHL) are equal. Because the greatest-integer function $$[x]$$ behaves differently on $$(-1,0)$$ and on $$(0,1)$$, we examine the two sides separately.

1. Preliminary expansions valid for both sides

The standard Maclaurin expansions are

$$\sin t=t-\dfrac{t^{3}}{6}+O(t^{5}),\qquad \tan y=y+\dfrac{y^{3}}{3}+O(y^{5}).$$

For small $$x$$ we first expand $$\sin^{2}x$$:

$$\sin^{2}x=(x-\tfrac{x^{3}}{6}+O(x^{5}))^{2}=x^{2}-\dfrac{x^{4}}{3}+O(x^{6}).$$

Hence

$$\pi\sin^{2}x=\pi x^{2}-\dfrac{\pi x^{4}}{3}+O(x^{6}).$$

Putting this into the expansion of $$\tan$$ we obtain

$$\tan(\pi\sin^{2}x)=\Bigl(\pi x^{2}-\dfrac{\pi x^{4}}{3}+O(x^{6})\Bigr) +\dfrac{1}{3}\Bigl(\pi x^{2}\Bigr)^{3}+O(x^{8}) =\pi x^{2}+O(x^{4}).$$

Thus up to the order required for division by $$x^{2}$$,

$$\tan(\pi\sin^{2}x)=\pi x^{2}+o(x^{2}). \quad -(1)$$

2. Right-hand limit ($$x\to 0^{+}$$)

When $$0\lt x\lt 1$$ we have $$[x]=0.$$ Therefore

$$\sin(x[x])=\sin(x\cdot 0)=\sin 0=0,$$

and, since $$x\gt 0$$, $$|x|=x.$$ So the second part of the numerator becomes

$$\bigl(|x|-\sin(x[x])\bigr)^{2}=(x-0)^{2}=x^{2}. \quad -(2)$$

Adding (1) and (2) we get the entire numerator:

$$\tan(\pi\sin^{2}x)+\bigl(|x|-\sin(x[x])\bigr)^{2} =\bigl(\pi x^{2}+o(x^{2})\bigr)+x^{2} =(\pi+1)x^{2}+o(x^{2}).$$

Dividing by $$x^{2}$$ gives the right-hand limit

$$\text{RHL}=\pi+1.$$

3. Left-hand limit ($$x\to 0^{-}$$)

For $$-1\lt x\lt 0$$ we have $$[x]=-1.$$ Because $$x$$ is negative, $$|x|=-x.$$ Also,

$$\sin(x[x])=\sin\bigl(x\cdot(-1)\bigr)=\sin(-x)=-\sin x.$$

Hence

$$|x|-\sin(x[x])=-x-(-\sin x)=-x+\sin x. \quad -(3)$$

Using $$\sin x=x-\dfrac{x^{3}}{6}+O(x^{5})$$ we obtain

$$-x+\sin x=-x+\Bigl(x-\dfrac{x^{3}}{6}+O(x^{5})\Bigr) =-\dfrac{x^{3}}{6}+O(x^{5}).$$

Squaring,

$$\bigl(|x|-\sin(x[x])\bigr)^{2}=\Bigl(-\dfrac{x^{3}}{6}+O(x^{5})\Bigr)^{2} =\dfrac{x^{6}}{36}+O(x^{8}). \quad -(4)$$

This is of order $$x^{6}$$ and hence negligible compared with the $$x^{2}$$ term we shall divide by.

From (1) and (4) the whole numerator on the left side is

$$\tan(\pi\sin^{2}x)+\bigl(|x|-\sin(x[x])\bigr)^{2} =\pi x^{2}+o(x^{2}).$$

Therefore the left-hand limit is

$$\text{LHL}=\pi.$$

4. Comparison of the two sides

$$\text{RHL}=\pi+1,\qquad \text{LHL}=\pi.$$

Since $$\text{RHL}\neq\text{LHL},$$ the limit as $$x\to 0$$ does not exist.

Hence, the correct answer is Option A.

We begin with the limit

$$L=\lim_{x\to0}\dfrac{x+2\sin x}{\sqrt{x^{2}+2\sin x+1}-\sqrt{\sin^{2}x-x+1}}.$$

Both the numerator and the denominator approach $$0$$ as $$x\to0$$, so L’Hôpital’s Rule is applicable. L’Hôpital’s Rule states that for functions $$f(x)$$ and $$g(x)$$ which satisfy $$f(0)=g(0)=0$$ and have derivatives continuous near $$0$$, we have

$$\lim_{x\to0}\dfrac{f(x)}{g(x)}=\lim_{x\to0}\dfrac{f'(x)}{g'(x)}$$

provided the limit on the right exists.

We set

$$f(x)=x+2\sin x,\qquad g(x)=\sqrt{x^{2}+2\sin x+1}-\sqrt{\sin^{2}x-x+1}.$$

Derivative of the numerator

$$f'(x)=1+2\cos x.$$

Therefore

$$f'(0)=1+2\cos0=1+2\cdot1=3.$$

Derivative of the denominator

Write $$g(x)=u(x)-v(x)$$ with

$$u(x)=\sqrt{x^{2}+2\sin x+1},\qquad v(x)=\sqrt{\sin^{2}x-x+1}.$$

Using $$\dfrac{d}{dx}\sqrt{h(x)}=\dfrac{h'(x)}{2\sqrt{h(x)}}$$ we find

$$u'(x)=\dfrac{2x+2\cos x}{2\sqrt{x^{2}+2\sin x+1}} =\dfrac{x+\cos x}{\sqrt{x^{2}+2\sin x+1}},$$

$$v'(x)=\dfrac{2\sin x\cos x-1}{2\sqrt{\sin^{2}x-x+1}} =\dfrac{\sin x\cos x-\dfrac12}{\sqrt{\sin^{2}x-x+1}}.$$

Hence

$$g'(x)=u'(x)-v'(x) =\dfrac{x+\cos x}{\sqrt{x^{2}+2\sin x+1}} -\dfrac{\sin x\cos x-\dfrac12}{\sqrt{\sin^{2}x-x+1}}.$$

Now evaluate $$g'(x)$$ at $$x=0$$. We know $$\sin0=0,\;\cos0=1$$, and both radicals become $$\sqrt1=1$$. Substituting,

$$g'(0)=\dfrac{0+1}{1}-\dfrac{0\cdot1-\dfrac12}{1} =1-\Bigl(-\dfrac12\Bigr)=1+\dfrac12=\dfrac32.$$

Applying L’Hôpital’s Rule

$$L=\lim_{x\to0}\dfrac{f(x)}{g(x)} =\lim_{x\to0}\dfrac{f'(x)}{g'(x)} =\dfrac{f'(0)}{g'(0)} =\dfrac{3}{\dfrac32}=2.$$

So, the answer is $$2$$.

We have to evaluate the limit

$$\displaystyle L=\lim_{x\to\frac{\pi}{4}} \frac{\cot^{3}x-\tan x}{\cos\!\left(x+\dfrac{\pi}{4}\right)}.$$

First we note the values of the functions at the point of approach. When $$x=\dfrac{\pi}{4}$$ we get $$\tan\dfrac{\pi}{4}=1,\qquad \cot\dfrac{\pi}{4}=1,$$ hence

$$\cot^{3}\dfrac{\pi}{4}-\tan\dfrac{\pi}{4}=1^{3}-1=0.$$

Also

$$\cos\!\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}\right) =\cos\dfrac{\pi}{2}=0.$$

Because both the numerator and the denominator approach $$0,$$ the limit is of the indeterminate form $$\dfrac{0}{0}.$$ Therefore we can apply L’Hôpital’s Rule, which says:

$$\text{If } \displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}= \frac{0}{0}\text{ or }\frac{\infty}{\infty}, \text{ then } \displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}= \displaystyle\lim_{x\to a}\frac{f'(x)}{g'(x)},$$ provided the latter limit exists.

So we differentiate the numerator and the denominator with respect to $$x.$$

For the numerator $$f(x)=\cot^{3}x-\tan x$$ we write

$$f'(x)=\frac{d}{dx}\bigl(\cot^{3}x\bigr)-\frac{d}{dx}\bigl(\tan x\bigr).$$

We recall the derivatives $$\frac{d}{dx}(\cot x)=-\csc^{2}x \quad\text{and}\quad \frac{d}{dx}(\tan x)=\sec^{2}x.$$

Using the chain rule on the first term we get

$$\frac{d}{dx}\bigl(\cot^{3}x\bigr)=3\cot^{2}x\cdot(\!-\csc^{2}x) =-3\cot^{2}x\,\csc^{2}x.$$

Thus

$$f'(x)=-3\cot^{2}x\,\csc^{2}x-\sec^{2}x.$$

For the denominator $$g(x)=\cos\!\left(x+\dfrac{\pi}{4}\right)$$ we use the derivative $$\dfrac{d}{dx}\cos u=-\sin u\cdot\dfrac{du}{dx}.$$ Here $$u=x+\dfrac{\pi}{4},$$ so $$\dfrac{du}{dx}=1,$$ and consequently

$$g'(x)=-\sin\!\left(x+\dfrac{\pi}{4}\right).$$

Now we evaluate the derivatives at $$x=\dfrac{\pi}{4}.$$

First compute the trigonometric values:

$$\cot\dfrac{\pi}{4}=1,\; \csc\dfrac{\pi}{4}=\frac{1}{\sin\dfrac{\pi}{4}} =\frac{1}{\dfrac{\sqrt2}{2}}=\sqrt2,\; \sec\dfrac{\pi}{4}=\frac{1}{\cos\dfrac{\pi}{4}} =\frac{1}{\dfrac{\sqrt2}{2}}=\sqrt2.$$

Hence

$$\cot^{2}\dfrac{\pi}{4}=1,\qquad \csc^{2}\dfrac{\pi}{4}=2,\qquad \sec^{2}\dfrac{\pi}{4}=2.$$

Substituting in $$f'(x)$$ gives

$$f'\!\left(\dfrac{\pi}{4}\right) =-3(1)(2)-2=-6-2=-8.$$

For the denominator we find

$$g'\!\left(\dfrac{\pi}{4}\right) =-\sin\!\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}\right) =-\sin\dfrac{\pi}{2}=-1.$$

Applying L’Hôpital’s Rule we have

$$L=\lim_{x\to\frac{\pi}{4}}\frac{f(x)}{g(x)} =\lim_{x\to\frac{\pi}{4}}\frac{f'(x)}{g'(x)} =\frac{f'\!\left(\dfrac{\pi}{4}\right)} {g'\!\left(\dfrac{\pi}{4}\right)} =\frac{-8}{-1}=8.$$

Hence, the correct answer is Option D.

We have the functions $$f(x)=5-|x-2|$$ and $$g(x)=|x+1|$$ defined for all real $$x$$.

First, we find the point $$\alpha$$ at which $$f(x)$$ attains its maximum value. For any real number $$t$$, the quantity $$|t|$$ is always non-negative and its minimum possible value is $$0$$. In the expression $$f(x)=5-|x-2|$$ the term $$|x-2|$$ will therefore be smallest, namely $$0$$, when $$x-2=0$$, that is when $$x=2$$. So $$|x-2|_{\min}=0$$ and consequently $$f(x)_{\max}=5-0=5$$ at $$x=2$$. Hence $$\alpha=2$$.

Next, we determine the point $$\beta$$ at which $$g(x)=|x+1|$$ attains its minimum value. Exactly as before, $$|x+1|$$ is minimized when its inside quantity is zero: $$x+1=0 \implies x=-1$$. Thus $$|x+1|_{\min}=0$$, and this occurs at $$x=-1$$. Hence $$\beta=-1$$.

We now need the limit $$ \lim_{x \to -\alpha\beta} \frac{(x-1)(x^2-5x+6)}{x^2-6x+8}. $$ First we compute $$-\alpha\beta$$: $$\alpha\beta = (2)(-1)=-2 \quad\Longrightarrow\quad -\alpha\beta = -(-2)=2.$$ So we shall evaluate the limit as $$x \to 2$$.

Before substituting $$x=2$$, we simplify the rational function algebraically. We factor every quadratic completely:

$$x^2-5x+6 = (x-2)(x-3),$$ $$x^2-6x+8 = (x-2)(x-4).$$

Substituting these factorizations, the original fraction becomes

$$ \frac{(x-1)(x^2-5x+6)}{x^2-6x+8} =\frac{(x-1)\bigl(x-2\bigr)\bigl(x-3\bigr)}{\bigl(x-2\bigr)\bigl(x-4\bigr)}. $$

For $$x \neq 2$$ the common factor $$x-2$$ can be cancelled:

$$ \frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)} =\frac{(x-1)(x-3)}{x-4}. $$

Now the limit as $$x \to 2$$ can be found by straightforward substitution because the denominator is no longer zero at $$x=2$$:

Numerator at $$x=2$$: $$ (2-1)(2-3)=1\cdot(-1)=-1. $$

Denominator at $$x=2$$: $$ 2-4=-2. $$

Therefore the value of the fraction at $$x=2$$ is

$$ \frac{-1}{-2}= \frac12. $$

So $$ \lim_{x \to 2} \frac{(x-1)(x^2-5x+6)}{x^2-6x+8}= \frac12. $$

Hence, the correct answer is Option B.

We want to find the limit

$$\displaystyle \lim_{x \to 0} \frac{(27+x)^{\tfrac13}-3}{9-(27+x)^{\tfrac23}}.$$

First, set

$$A=(27+x)^{\tfrac13}.$$

Because taking the cube root and the square of the cube root will both be required, this single symbol will make every step clearer. Re-writing the original fraction in terms of $$A$$ we obtain

$$\frac{(27+x)^{\tfrac13}-3}{9-(27+x)^{\tfrac23}} =\frac{A-3}{\,9-A^{2}\,}.$$

Now use the algebraic identity for the difference of squares:

$$u^{2}-v^{2}=(u-v)(u+v).$$

With $$u=3$$ and $$v=A$$, this identity gives

$$9-A^{2}=(3)^{2}-A^{2}=(3-A)(3+A).$$

Notice that $$3-A=-(A-3)$$. Substituting this factorisation into the denominator we have

$$9-A^{2}=-(A-3)(3+A).$$

Placing this form in the fraction yields

$$\frac{A-3}{9-A^{2}} =\frac{A-3}{-(A-3)(3+A)} =\frac{A-3}{-1\cdot(A-3)(3+A)}.$$

Because $$A$$ tends to $$3$$ as $$x$$ tends to $$0$$, $$A-3$$ tends to $$0$$ but is not identically zero, so we can safely cancel the common factor $$A-3$$ appearing in the numerator and denominator:

$$\frac{A-3}{-1\cdot(A-3)(3+A)}=-\frac{1}{3+A}.$$

Now recall that $$A=(27+x)^{\tfrac13}$$. As $$x \to 0$$ we have $$A \to (27)^{\tfrac13}=3$$. Substituting $$A \to 3$$ into the simplified expression gives

$$\lim_{x \to 0}\left(-\frac{1}{3+A}\right) =-\frac{1}{3+3} =-\frac{1}{6}.$$

Hence, the correct answer is Option A.

We are asked to evaluate

$$\lim_{x \to 0^{+}} \; x\Bigl(\,[\tfrac1x]+[\tfrac2x]+\dots +[\tfrac{15}{x}]\,\Bigr),$$

where $$[\,t\,]$$ denotes the greatest integer less than or equal to $$t$$ (the floor function).

First, recall (state) the basic relation between any real number $$y$$, its greatest-integer part $$[y]$$ and its fractional part $$\{y\}$$:

$$y \;=\; [y] \;+\; \{y\}, \qquad\text{with } 0\le \{y\}<1.$$

Re-arranging this equality gives the useful identity

$$[y] \;=\; y \;-\; \{y\}.$$

We shall apply this identity to every term $$\bigl[\tfrac{k}{x}\bigr]$$ inside the bracketed sum, where $$k=1,2,\dots,15$$ and $$x>0$$ (because the limit approaches $$0$$ from the right).

For a fixed $$k$$ we therefore write

$$\Bigl[\tfrac{k}{x}\Bigr] \;=\; \frac{k}{x}\;-\;\Bigl\{\tfrac{k}{x}\Bigr\}.$$

Multiplying both sides by $$x$$ (which is positive) gives

$$x\Bigl[\tfrac{k}{x}\Bigr] \;=\; x\!\left(\frac{k}{x}-\Bigl\{\tfrac{k}{x}\Bigr\}\right) = k - x\Bigl\{\tfrac{k}{x}\Bigr\}.$$

Now we add these equalities for all $$k=1$$ to $$15$$:

$$x\sum_{k=1}^{15}\Bigl[\tfrac{k}{x}\Bigr] =\sum_{k=1}^{15}\!\Bigl(k - x\bigl\{\tfrac{k}{x}\bigr\}\Bigr) =\sum_{k=1}^{15}k \;-\; x\sum_{k=1}^{15}\Bigl\{\tfrac{k}{x}\Bigr\}.$$

The first summation is a simple arithmetic series. Using the standard formula

$$1+2+\dots+n=\frac{n(n+1)}{2},$$

with $$n=15$$, we have

$$\sum_{k=1}^{15} k \;=\;\frac{15\cdot16}{2}=120.$$

Hence our overall expression becomes

$$x\sum_{k=1}^{15}\Bigl[\tfrac{k}{x}\Bigr] = 120 \;-\; x\sum_{k=1}^{15}\Bigl\{\tfrac{k}{x}\Bigr\}.$$

Observe next that each fractional part satisfies $$0\le\{\,\cdot\,\}<1$$, so

$$0\;\le\;\sum_{k=1}^{15}\Bigl\{\tfrac{k}{x}\Bigr\}\;\le\;15.$$ Multiplying by the positive number $$x$$ gives

$$0\;\le\;x\sum_{k=1}^{15}\Bigl\{\tfrac{k}{x}\Bigr\}\;\le\;15x.$$

As $$x\to 0^{+}$$ the quantity $$15x$$ tends to $$0$$, and by squeeze (sandwich) reasoning the product $$x\sum_{k=1}^{15}\bigl\{\tfrac{k}{x}\bigr\}$$ also tends to $$0$$.

Therefore, taking the limit of both sides, we get

$$\lim_{x\to0^{+}} x\sum_{k=1}^{15}\Bigl[\tfrac{k}{x}\Bigr] = 120 - 0 = 120.$$

Hence, the correct answer is Option D.

We are asked to evaluate the limit

$$\lim_{x \to 0}\dfrac{x\tan 2x\;-\;2x\tan x}{\left(1-\cos 2x\right)^{2}}.$$

Because both numerator and denominator tend to $$0$$ as $$x\to 0$$, this is an indeterminate form $$\dfrac{0}{0}$$. In such cases the Maclaurin (Taylor) expansions of the standard trigonometric functions about $$x=0$$ are very useful. We first recall the series we will need:

For small $$x$$,

$$\tan x = x + \dfrac{x^{3}}{3} + \dfrac{2x^{5}}{15} + \cdots,$$

$$\cos x = 1 - \dfrac{x^{2}}{2} + \dfrac{x^{4}}{24} - \dfrac{x^{6}}{720} + \cdots.$$

Now we substitute $$x$$ by $$2x$$ wherever required and write every term up to the power that will finally survive after simplification.

1. Expansion of $$\tan 2x$$

Replacing $$x$$ with $$2x$$ in the series of $$\tan x$$ we get

$$\tan 2x = 2x + \dfrac{(2x)^{3}}{3} + \dfrac{2(2x)^{5}}{15} + \cdots = 2x + \dfrac{8x^{3}}{3} + \dfrac{32x^{5}}{15} + \cdots.$$

2. Expansion of $$x\tan 2x$$

Multiplying by the outside factor $$x$$ gives

$$x\tan 2x = x\left(2x + \dfrac{8x^{3}}{3} + \dfrac{32x^{5}}{15} + \cdots\right) = 2x^{2} + \dfrac{8x^{4}}{3} + \dfrac{32x^{6}}{15} + \cdots.$$

3. Expansion of $$2x\tan x$$

First write $$\tan x = x + \dfrac{x^{3}}{3} + \dfrac{2x^{5}}{15} + \cdots,$$ then multiply by $$2x$$:

$$2x\tan x = 2x\left(x + \dfrac{x^{3}}{3} + \dfrac{2x^{5}}{15} + \cdots\right) = 2x^{2} + \dfrac{2x^{4}}{3} + \dfrac{4x^{6}}{15} + \cdots.$$

4. Numerator $$x\tan 2x - 2x\tan x$$

Subtracting series term-by-term we obtain

$$x\tan 2x - 2x\tan x = \bigl(2x^{2} + \tfrac{8x^{4}}{3} + \tfrac{32x^{6}}{15} + \cdots\bigr) \;-\;\bigl(2x^{2} + \tfrac{2x^{4}}{3} + \tfrac{4x^{6}}{15} + \cdots\bigr).$$

The $$2x^{2}$$ terms cancel, giving

$$x\tan 2x - 2x\tan x = \left(\dfrac{8x^{4}}{3}-\dfrac{2x^{4}}{3}\right) + \left(\dfrac{32x^{6}}{15}-\dfrac{4x^{6}}{15}\right) + \cdots = 2x^{4} + \dfrac{28x^{6}}{15} + \cdots.$$

Thus, up to the smallest non-zero power, the numerator is

$$x\tan 2x - 2x\tan x = 2x^{4} + O(x^{6}).$$

5. Expansion of $$1-\cos 2x$$

Substituting $$x$$ by $$2x$$ in the cosine series gives

$$\cos 2x = 1 - \dfrac{(2x)^{2}}{2} + \dfrac{(2x)^{4}}{24} - \dfrac{(2x)^{6}}{720} + \cdots = 1 - 2x^{2} + \dfrac{2x^{4}}{3} - \dfrac{4x^{6}}{45} + \cdots.$$

Therefore

$$1 - \cos 2x = 2x^{2} - \dfrac{2x^{4}}{3} + \dfrac{4x^{6}}{45} + \cdots.$$

6. Denominator $$\left(1-\cos 2x\right)^{2}$$

We square the above expression, keeping terms up to $$x^{6}$$ (because the numerator starts from $$x^{4}$$):

$$\left(1-\cos 2x\right)^{2} = \left(2x^{2} - \dfrac{2x^{4}}{3} + \cdots\right)^{2}.$$

Using the algebraic identity $$(a+b)^{2}=a^{2}+2ab+b^{2},$$ and remembering that any term of power higher than $$x^{6}$$ can be ignored for the limit, we compute:

First term: $$\bigl(2x^{2}\bigr)^{2}=4x^{4}.$$
Cross term: $$2\left(2x^{2}\right)\left(-\dfrac{2x^{4}}{3}\right)=-\dfrac{8x^{6}}{3}.$$
Square of the smaller term $$\left(-\dfrac{2x^{4}}{3}\right)^{2}$$ is of order $$x^{8}$$ and can be dropped.

Thus

$$\left(1-\cos 2x\right)^{2} = 4x^{4} - \dfrac{8x^{6}}{3} + \cdots = 4x^{4} + O(x^{6}).$$

7. Forming the ratio

We now divide the expanded numerator by the expanded denominator:

$$\dfrac{x\tan 2x - 2x\tan x}{\left(1-\cos 2x\right)^{2}} = \dfrac{\,2x^{4} + \dfrac{28x^{6}}{15} + \cdots\,} {\,4x^{4} - \dfrac{8x^{6}}{3} + \cdots\,}.$$

Factor $$x^{4}$$ out of both numerator and denominator and cancel it:

$$= \dfrac{\,2 + \dfrac{28x^{2}}{15} + \cdots\,} {\,4 - \dfrac{8x^{2}}{3} + \cdots\,}.$$

Now let $$x\to 0$$. All the terms containing positive powers of $$x$$ vanish, leaving

$$\lim_{x \to 0}\dfrac{x\tan 2x - 2x\tan x}{(1-\cos 2x)^{2}} = \dfrac{2}{4} = \dfrac{1}{2}.$$

Hence, the correct answer is Option D.