Continuity 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

$$\boxed{p>4}$$