Mathematics - Continuity - Differentiability Question with Solution | TestHub

We have,

$f\left(x\right)=\left\{\begin{array}{l}\begin{array}{cc}{x}^p\sin \frac{1}{x},& x\ne 0\end{array}\\ \begin{array}{cc}0, & x=0\end{array}\end{array}\right.$

For $f\left(x\right)$ to be continuous at $x=0$

$\underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)$

$\Rightarrow \underset{x\to 0}{\lim }\left(x^p\sin \frac{1}{x}\right)=0$

$\Rightarrow \underset{x\to 0}{\lim }\left(x^p\right)\times \underset{x\to 0}{\lim }\left(\sin \frac{1}{x}\right)=0$

$\Rightarrow \underset{x\to 0}{\lim }\left(x^p\right)\times \left(\text{value between }-1\text{ to} 1\right)=0$

$\therefore p>0$

Now,

RHD at $\text{RHD at }\left(x=0\right)=\underset{h\to 0}{\lim }\left[\frac{h^p\sin {\displaystyle \frac{1}{h}}-0}{h}\right]=\underset{h\to 0}{\lim }\left[h^{p-1}\sin \frac{1}{h}\right]$

$\text{LHD at }\left(x=0\right)=\underset{h\to 0}{\lim }\left[\frac{{\left(-h\right)}^p\sin \left(-{\displaystyle \frac{1}{h}}\right)-0}{-h}\right]=\underset{h\to 0}{\lim }\left[{\left(-1\right)}^p{\left(-h\right)}^{p-1}\sin \left(\frac{1}{h}\right)\right]$

Since, $f\left(x\right)$ is not differentiable at $x=0$, therefore

$p\le 1$

Hence, $0<p\le 1$.