$\text{ Let }f\left(x\right)=\left\{\begin{array}{ll}{x}^p\sin \frac{1}{x},x\ne 0& \\ 0& ,x=0\end{array}\right.\text{ then }$$\mathrm{f}\left(\mathrm{x}\right)\text{ is continuous but }\mathrm{not}$$\text{ differentiable at }\mathrm{x}=0\text{ if }$

$\begin{array}{l}f\left(x\right)=x^p\sin \frac{1}{x},x\ne 0\text{ and }f\left(x\right)=0,x=0\\ \text{ Since at x}=0,\text{f}\left(\text{x}\right)\text{ is a continuous function }\\ \therefore \underset{x\to 0}{\lim }\text{f}\left(\text{x}\right)=\text{f}\left(0\right)=0\\ \Rightarrow \underset{x\to 0}{\lim }{x}^p\sin \frac{1}{x}=0\Rightarrow p>0\\ \text{f}\left(\text{x}\right)\text{ is differentiable at x}=0,\text{ if }\underset{\text{x}\to 0}{\lim }\frac{\text{f}\left(\text{x}\right)-\text{f}\left(0\right)}{\text{x}-0}\text{ exists }\Rightarrow \underset{x\to 0}{\lim }\frac{x^p\sin \frac{1}{x}-0}{x-0}\text{ exists }\\ \Rightarrow \underset{x\to 0}{\lim }{x}^{p-1}\sin \frac{1}{x}\text{ exists }\Rightarrow \text{p}-1>0\text{ or p}>1\\ \therefore \text{ for }0<\text{p}\le 1,\text{f}\left(\text{x}\right)\text{ is a continuous function at x}=0\text{ but not differentiable. }\end{array}$