(a) Verify whether the function f defined by is continuous at x=0 or not.

$\displaystyle \lim _{x \rightarrow 0} \mathrm{f}(x)=\lim _{x \rightarrow 0} x \cdot \sin \frac{1}{x}=0 \times$ Finite value in $\displaystyle [-1,1]=0=\mathrm{f}(0)$

$\displaystyle \therefore \mathbf{f}(\mathbf{x})$ is a continuous function.

Or (b) LHD $\displaystyle =\lim _{x \rightarrow 5^{-}} \frac{|x-5|-0}{x-5}=\lim _{x \rightarrow 5^{-}} \frac{-(x-5)}{x-5}=-1$

$$\begin{aligned} & \text { RHD }=\lim _{x \rightarrow $\displaystyle 5$^{+}} \frac{|x-5|-0}{x-5}=\lim _{x \rightarrow $\displaystyle 5$^{+}} \frac{(x-5)}{x-5}=$\displaystyle 1$ & \text { LHD } \neq \text { RHD, } \therefore \mathrm{f}(x) \text { is not differentiable at } x=$\displaystyle 5$ \end{aligned}