L'Hôpital's Rule

Let

$$L= \lim_{x\to0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}-x}{x^3}$$

Using binomial expansion,

$$\sqrt{1+t} = 1+\frac t2-\frac{t^2}{8}+\frac{t^3}{16}+o(t^3)$$

and

$$\sqrt{1-t} = 1-\frac t2-\frac{t^2}{8}-\frac{t^3}{16}+o(t^3)$$

Subtracting,

$$\sqrt{1+t}-\sqrt{1-t} = t+\frac{t^3}{8}+o(t^3)$$

Put

$$t=\sin x$$

Hence,

$$\sqrt{1+\sin x}-\sqrt{1-\sin x} = \sin x+\frac{\sin^3x}{8}+o(x^3)$$

Using

$$\sin x=x-\frac{x^3}{6}+o(x^3),$$

we get

$$\sin^3x=x^3+o(x^3)$$

Therefore,

$$\sqrt{1+\sin x}-\sqrt{1-\sin x}-x$$

$$= \left(x-\frac{x^3}{6}\right) +\frac{x^3}{8} -x+o(x^3)$$

$$= x^3\left(-\frac16+\frac18\right)+o(x^3)$$

$$= -\frac{x^3}{24}+o(x^3)$$

Hence,

$$L=-\frac1{24}$$

Therefore,

$$48\lambda = 48\left(-\frac1{24}\right)$$

$$=-2$$

Step 1: Verify the Indeterminate Form
Let's substitute $$x = 0$$ into the numerator and denominator to check the initial state of the limit:
* Numerator: $$e^{0 \cdot \cos 0} - 1 - 0 = e^0 - 1 = 1 - 1 = 0$$
* Denominator: $$\ln(1 + 0^2 - 0^3) = \ln(1) = 0$$

Since the expression yields the indeterminate form $$\frac{0}{0}$$, we can apply L'Hôpital's Rule.

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Step 2: Apply L'Hôpital's Rule (First Differentiation)
L'Hôpital's Rule states that if the limit is indeterminate of the form $$\frac{0}{0}$$, then:
$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

Let's find the derivative of the numerator using the chain rule and product rule:
$$\frac{d}{dx}\left(e^{x \cos x} - 1 - x\right) = e^{x \cos x} \cdot \frac{d}{dx}(x \cos x) - 1$$
$$= e^{x \cos x} \cdot (\cos x - x \sin x) - 1$$

Now let's find the derivative of the denominator:
$$\frac{d}{dx}\left(\ln(1 + x^2 - x^3)\right) = \frac{1}{1 + x^2 - x^3} \cdot (2x - 3x^2)$$

Substituting these derivatives back into the limit:
$$L = \lim_{x \to 0} \frac{e^{x \cos x}(\cos x - x \sin x) - 1}{\frac{2x - 3x^2}{1 + x^2 - x^3}}$$

To prevent the denominator algebra from becoming too messy, we can split out the term that does not approach zero:
$$\lim_{x \to 0} (1 + x^2 - x^3) = 1$$

Thus, we can rewrite the limit as:
$$L = \lim_{x \to 0} \frac{e^{x \cos x}(\cos x - x \sin x) - 1}{2x - 3x^2}$$

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Step 3: Verify the Indeterminate Form Again
Let's test $$x = 0$$ once more on our simplified expression:
* New Numerator: $$e^0(\cos 0 - 0) - 1 = 1(1) - 1 = 0$$
* New Denominator: $$2(0) - 3(0)^2 = 0$$

The limit is still in the indeterminate form $$\frac{0}{0}$$, so we apply L'Hôpital's Rule a second time.

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Step 4: Apply L'Hôpital's Rule (Second Differentiation)
Differentiate the numerator using the product rule:
$$\frac{d}{dx}\left[e^{x \cos x}(\cos x - x \sin x) - 1\right]$$
$$= \left[\frac{d}{dx}(e^{x \cos x})\right](\cos x - x \sin x) + e^{x \cos x}\left[\frac{d}{dx}(\cos x - x \sin x)\right]$$
$$= \left[e^{x \cos x}(\cos x - x \sin x)\right](\cos x - x \sin x) + e^{x \cos x}(-\sin x - \sin x - x \cos x)$$
$$= e^{x \cos x}(\cos x - x \sin x)^2 + e^{x \cos x}(-2\sin x - x \cos x)$$

Now differentiate the denominator:
$$\frac{d}{dx}(2x - 3x^2) = 2 - 6x$$

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Step 5: Substitute $$x = 0$$ to Find the Final Limit
Assemble the second derivatives back into the limit expression:
$$L = \lim_{x \to 0} \frac{e^{x \cos x}(\cos x - x \sin x)^2 + e^{x \cos x}(-2\sin x - x \cos x)}{2 - 6x}$$

Now substitute $$x = 0$$ directly:
$$L = \frac{e^0(\cos 0 - 0)^2 + e^0(-2\sin 0 - 0 \cdot \cos 0)}{2 - 6(0)}$$
$$L = \frac{1(1)^2 + 1(0)}{2}$$
$$L = \frac{1}{2}$$

$$\ln(1 + x^2) = x^2 - \frac{(x^2)^2}{2} + \frac{(x^2)^3}{3} + \dots = x^2 - \frac{x^4}{2} + \frac{x^6}{3} + O(x^8)$$

$$\sin(x) = x - \frac{x^3}{6} + O(x^5)$$

$$\sin^2(x) = \left(x - \frac{x^3}{6}\right)^2 = x^2 - 2(x)\left(\frac{x^3}{6}\right) + O(x^6) = x^2 - \frac{x^4}{3} + O(x^6)$$

$$\text{Denominator} = \left( x^2 - \frac{x^4}{2} \right) - \left( x^2 - \frac{x^4}{3} \right) + O(x^6)$$

$$\text{Denominator} = \left(-\frac{1}{2} + \frac{1}{3}\right)x^4 + O(x^6) = -\frac{1}{6}x^4 + O(x^6)$$

Since the denominator's leading term is of order $$x^4$$, we must expand the numerator up to the $$x^4$$ term as well.

The numerator has two parts: $$\cos(x)e^{x^2/2}$$ and $$\sqrt{1 - x^4}$$.

• Part A: $$\cos(x) \cdot e^{x^2/2}$$

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

$$e^{x^2/2} = 1 + \left(\frac{x^2}{2}\right) + \frac{1}{2}\left(\frac{x^2}{2}\right)^2 + O(x^6) = 1 + \frac{x^2}{2} + \frac{x^4}{8} + O(x^6)$$

$$\cos(x) \cdot e^{x^2/2} = \left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right)\left(1 + \frac{x^2}{2} + \frac{x^4}{8}\right)$$

$$= 1 + \frac{x^2}{2} + \frac{x^4}{8} - \frac{x^2}{2} - \frac{x^4}{4} + \frac{x^4}{24} + O(x^6)$$

$$\left(\frac{1}{8} - \frac{1}{4} + \frac{1}{24}\right)x^4 = \left(\frac{3}{24} - \frac{6}{24} + \frac{1}{24}\right)x^4 = -\frac{2}{24}x^4 = -\frac{1}{12}x^4$$

So, $$\cos(x)e^{x^2/2} = 1 - \frac{1}{12}x^4 + O(x^6)$$.

$$\sqrt{1 - x^4} = 1 - \frac{1}{2}x^4 + O(x^8)$$

$$\text{Numerator} = \left(1 - \frac{1}{12}x^4\right) - \left(1 - \frac{1}{2}x^4\right) + O(x^6)$$

$$\text{Numerator} = \left(-\frac{1}{12} + \frac{1}{2}\right)x^4 + O(x^6) = \left(-\frac{1}{12} + \frac{6}{12}\right)x^4 + O(x^6) = \frac{5}{12}x^4 + O(x^6)$$

$$\lim_{x \to 0} \frac{\frac{5}{12}x^4 + O(x^6)}{-\frac{1}{6}x^4 + O(x^6)}$$

$$\lim_{x \to 0} \frac{\frac{5}{12} + O(x^2)}{-\frac{1}{6} + O(x^2)} = \frac{\frac{5}{12}}{-\frac{1}{6}}$$

$$\frac{5}{12} \cdot (-6) = -\frac{30}{12} = -\frac{5}{2}$$