Given, $f(x)=\left\{\begin{array}{c}x \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, \quad x=0\end{array}\right.$
For continuity at $x=0$,
$\mathrm{LHL}=f(0-0)=\lim _{h \rightarrow 0} f(0-h)$
$\begin{aligned} &=\lim _{h \rightarrow 0}(-h) \sin \left(-\frac{1}{h}\right) \\ &=0 \times(\text { finite quantity }) \\ &=0 \\ \text { RHL } &=f(0+0)=\lim _{h \rightarrow 0} f(0+h) \\ &=\lim _{h \rightarrow 0}(h) \sin \left(\frac{1}{h}\right) \end{aligned}$
$=0 \times($ finite quantity $)$
$=0$
and $f(0)=0$
$\therefore \quad f(0)=\mathrm{LHL}=\mathrm{RHL}$
$\therefore f(x)$ is continuous at $x=0 .$
For differentiability at $x=0$,
$R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h \sin \left(\frac{1}{h}\right)-0}{h} \\ &=\lim _{h \rightarrow 0} \sin \left(\frac{1}{h}\right) \\ &=(\text { a finite quantity persist between }-1\\ & &\text { to }+1) \\ &=\text { does not exist } \\ L f^{\prime}(0) &=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{(-h) \sin \left(-\frac{1}{h}\right)-0}{-h} \\=& \lim _{h \rightarrow 0}\left\{-\sin \left(\frac{1}{h}\right)\right\} \\=&(\text { finite quantity persist between }\\ & &-1 \text { to }+1) \\ &=\text { does not exist } \\ \because R f^{\prime}(0) \neq L^{\prime}(0) & \end{aligned}$
$\therefore f(x)$ is not differentiable at $x=0$.