Let f, g: R arrow R be two functions defined by f(x)= cases x...

$$\begin{aligned}&f(x)= \begin{cases} x \sin \left(\frac{1}{x}\right),& x ≠ 0 \\ 0, & x=0\end{cases} \\ &\text{\text{ and } } g(x)=x f(x) \\ &\text{\text{ For } } f(x) \\ &\mathrm{LHL}=\lim _{h \rightarrow 0^{-}} \left\{-h \sin \left(-\frac{1}{h}\right)\right\} \end{aligned}$$=0 $\times a finite quantity between$-1 and $1 =0 \mathrm{RHL}=\lim _{h \rightarrow 0^{+}}h \sin \frac{1}{h}=0 Also,$ f(0)=0$Thus$\mathrm{LHL}=\mathrm{RHL}$=$ f(0)\therefore f(x)$is continuous at$x=0$g(x)=$\begin{cases}x^2 \sin \frac{1}{x},& x ≠ 0 \\ 0, & x=0\end{cases}$$For$g(x)\mathrm{LHL}=\lim _{h \rightarrow 0^{-}} \left\{-h^2 \sin \left(\frac{1}{h}\right)\right\}$=0^2$\times a finite quantity between $-1 and$1=0\text{ RHL }=\lim _{h \rightarrow 0^{+}}h^2 \sin \left(\frac{1}{h}\right)=0 Also$g(0)=0\therefore g(x)$ is continuous at $x=0