Differentiability at a Point

For

$$x\ne0$$

given function is

$$f(x)=x^p\sin\left(\frac1x\right)$$

Differentiate using product rule.

We know,

$$\frac d{dx}\left(\sin\frac1x\right)=\cos\left(\frac1x\right)\left(-\frac1{x^2}\right)$$

Therefore,

$$f'(x)=\frac d{dx}\left(x^p\right)\sin\left(\frac1x\right)+x^p\frac d{dx}{\left(\sin\frac1x\right)}$$

$$=px^{p-1}\sin\left(\frac1x\right)-x^{p-2}\cos\left(\frac1x\right)$$

Now differentiate again.

Differentiate first term :

$$\frac d{dx}\left[px^{p-1}\sin\left(\frac1x\right)\right]$$

Using product rule,

$$=p(p-1)x^{p-2}\sin\left(\frac1x\right)+px^{p-1}\cos\left(\frac1x\right)\left(-\frac1{x^2}\right)$$

$$=p(p-1)x^{p-2}\sin\left(\frac1x\right)-px^{p-3}\cos\left(\frac1x\right)$$

Now differentiate second term :

$$\frac d{dx}\left[-x^{p-2}\cos\left(\frac1x\right)\right]$$

Again using product rule,

$$=-(p-2)x^{p-3}\cos\left(\frac1x\right)-x^{p-2}\left[-\sin\left(\frac1x\right)\left(-\frac1{x^2}\right)\right]$$

$$=-(p-2)x^{p-3}\cos\left(\frac1x\right)-x^{p-4}\sin\left(\frac1x\right)$$

Now combine both parts.

$$f''(x)=p(p-1)x^{p-2}\sin\left(\frac1x\right)-px^{p-3}\cos\left(\frac1x\right)$$

$$-(p-2)x^{p-3}\cos\left(\frac1x\right)-x^{p-4}\sin\left(\frac1x\right)$$

Group similar terms :

$$f''(x)=\left[p(p-1)x^{p-2}-x^{p-4}\right]\sin\left(\frac1x\right)$$

$$-(2p-2)x^{p-3}\cos\left(\frac1x\right)$$

Now check existence of $$f''(0)$$.

First,

$$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}$$

$$=\lim_{h\to0}h^{p-1}\sin\left(\frac1h\right)$$

Since $$\sin\left(\frac1h\right)$$ is bounded,

$$f'(0)=0 \qquad \text{for } p>1$$

Now,

$$f''(0)=\lim_{h\to0}\frac{f'(h)-f'(0)}{h}$$

Since $$f'(0)=0$$,

$$f''(0)=\lim_{h\to0}\left[p\,h^{p-2}\sin\left(\frac1h\right)-h^{p-3}\cos\left(\frac1h\right)\right]$$

For this limit to exist, both terms must approach zero.

Therefore,

$$p-2>0,\qquad p-3>0$$

which gives

$$p>3$$

and hence

$$f''(0)=0$$

Now for continuity of $$f''(x)$$ at $$x=0$$

we need

$$\lim_{x\to0}f''(x)=f''(0)$$

Since $$\sin\left(\frac1x\right)$$ and $$\cos\left(\frac1x\right)$$ are bounded between $$-1$$ and $$1$$,

the powers of $$x$$ must force every term to go to zero.

The powers present are

$$p-2,\quad p-3,\quad p-4$$

The smallest power is

$$p-4$$

For the limit to exist and equal zero,

$$p-4>0$$

$$p>4$$

Therefore,

$$\lim_{x\to0}f''(x)=0=f''(0)$$

Hence,

$$f''(x)$$

is continuous at

$$x=0$$

when

$$\boxed{p>4}$$

To find the value of the limit $$L$$, we first determine the derivative function $$f'(x)$$ using the first principles of differentiation and the given functional equation:

1. Find the general expression for the derivative $$f'(x)$$:
By the definition of the derivative, we have:
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

Using the given functional equation to substitute for $$f(x + h)$$, we get:
$$f'(x) = \lim_{h \to 0} \frac{f(x)e^{h} + f(h)e^{x} + 4xh - f(x)}{h}$$

Rearrange and group the terms to isolate the limits:
$$f'(x) = \lim_{h \to 0} \frac{f(x)(e^{h} - 1) + f(h)e^{x} + 4xh}{h}$$
$$f'(x) = f(x) \cdot \lim_{h \to 0} \left(\frac{e^{h} - 1}{h}\right) + e^{x} \cdot \lim_{h \to 0} \left(\frac{f(h) - f(0)}{h}\right) + 4x$$

Since $$\lim_{h \to 0} \frac{e^h - 1}{h} = 1$$ and $$\lim_{h \to 0} \frac{f(h) - f(0)}{h} = f'(0) = 2$$, substituting these standard limits yields the first-order differential equation:
$$f'(x) = f(x) + 2e^{x} + 4x$$

2. Find the second derivative function $$f''(x)$$:
Differentiate both sides of our newly found expression with respect to $$x$$:
$$f''(x) = f'(x) + 2e^{x} + 4$$

3. Evaluate the given limit expression $$L$$:
We are asked to evaluate the limit:
$$L = \lim_{x \to 1} \frac{f'(x) - f'(1)}{x^2 - 1}$$

We can factor the denominator as $$(x - 1)(x + 1)$$ and rewrite the expression to reveal a standard derivative structure:
$$L = \lim_{x \to 1} \left[ \frac{f'(x) - f'(1)}{x - 1} \cdot \frac{1}{x + 1} \right]$$

By the definition of the second derivative, the first part of the limit is exactly $$f''(1)$$:
$$L = f''(1) \cdot \frac{1}{1 + 1} = \frac{f''(1)}{2}$$

4. Compute the value of $$f''(1)$$:
Using the differential equations established in the earlier steps, we substitute $$x = 1$$:
$$f''(1) = f'(1) + 2e^{1} + 4$$

Now substitute the expression for $$f'(1)$$ into the equation above:
$$f'(1) = f(1) + 2e^{1} + 4(1)$$
$$f''(1) = (f(1) + 2e + 4) + 2e + 4 = f(1) + 4e + 8$$

To find $$f(1)$$, we solve the linear differential equation $$f'(x) - f(x) = 2e^x + 4x$$. The integrating factor is $$e^{-x}$$.
Multiplying both sides by the integrating factor gives:
$$\frac{d}{dx}[f(x)e^{-x}] = 2 + 4xe^{-x}$$

Integrating both sides from 0 to 1:
$$\left[f(x)e^{-x}\right]_0^1 = \int_0^1 (2 + 4xe^{-x}) dx$$
$$f(1)e^{-1} - f(0)e^0 = [2x]_0^1 + 4 \cdot \left([-xe^{-x}]_0^1 + \int_0^1 e^{-x} dx\right)$$
$$f(1)e^{-1} = 2 + 4 \cdot \left(-e^{-1} + [-e^{-x}]_0^1\right)$$
$$f(1)e^{-1} = 2 + 4 \cdot (-e^{-1} - e^{-1} + 1)$$
$$f(1)e^{-1} = 2 - 8e^{-1} + 4 = 6 - 8e^{-1}$$

Multiplying the entire equation by $$e$$ gives:
$$f(1) = 6e - 8$$

Now, substitute $$f(1)$$ back into the expression for $$f''(1)$$:
$$f''(1) = (6e - 8) + 4e + 8 = 10e$$

$$ L = 5e $$

$$ \frac{L}{e} = 5$$

We evaluate the left and right behaviors around the origin.
For $$x > 0$$, $$\text{sgn}(x) = 1 \implies f(x) = \frac{\ln(1 + \sin^2 x)}{x}$$.
Using Taylor expansions as $$x \to 0^+$$: $$\ln(1 + \sin^2 x) \approx \sin^2 x \approx x^2$$.
$$RHL = \lim_{x \to 0^+} \frac{x^2}{x} = 0$$.

For $$x < 0$$, $$\text{sgn}(x) = -1 \implies f(x) = \frac{\ln(1 - \sin^2 x)}{x} = \frac{\ln(\cos^2 x)}{x} = \frac{2 \ln(\cos x)}{x}$$.
Using expansions near zero: $$\ln(\cos x) \approx \ln(1 - \frac{x^2}{2}) \approx -\frac{x^2}{2}$$.
$$LHL = \lim_{x \to 0^-} \frac{2(-\frac{x^2}{2})}{x} = \lim_{x \to 0^-} (-x) = 0$$.
Since $$LHL = RHL = f(0) = 0$$, the function is continuous at $$x = 0$$.

Now evaluate differentiability using first principles:
$$RHD = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\ln(1 + \sin^2 h)}{h^2} = 1$$.
$$LHD = \lim_{h \to 0^+} \frac{f(-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{\frac{2\ln(\cos(-h))}{-h} - 0}{-h} = \lim_{h \to 0^+} \frac{2\ln(\cos h)}{h^2}$$.
Using our expansion: $$\lim_{h \to 0^+} \frac{2(-\frac{h^2}{2})}{h^2} = -1$$.
Since $$RHD = 1$$ and $$LHD = -1$$, the derivatives do not match. The function is continuous but not differentiable.