Mathematics - Continuity - Differentiability Question with Solution | TestHub

$$\begin{array}{rl}& f(x)=\{\begin{array}{ll}x\sin \left(\frac{1}{x}\right),& x\ne 0\\ 0,& x=0\end{array}\\ & \text{ and }g(x)=xf(x)\\ & \text{ For }f(x)\\ & \mathrm{LHL}=\underset{h\to 0^{-}}{\lim }\left\{-h\sin \left(-\frac{1}{h}\right)\right\}\end{array}$$ $=0\times$ a finite quantity between $-1$ and 1 $$=0$$ $$\mathrm{RHL}=\underset{h\to 0^{+}}{\lim }h\sin \frac{1}{h}=0$$ Also, $f(0)=0$ Thus $\mathrm{LHL}=\mathrm{RHL}=f(0)$ $\therefore f(x)$ is continuous at $x=0$ $$g(x)=\{\begin{array}{ll}{x}^2\sin \frac{1}{x},& x\ne 0\\ 0,& x=0\end{array}$$ For $g(x)$ $$\mathrm{LHL}=\underset{h\to 0^{-}}{\lim }\left\{-h^2\sin \left(\frac{1}{h}\right)\right\}$$ $=0^2\times$ a finite quantity between $-1$ and $1=0$ $$\text{ RHL }=\underset{h\to 0^{+}}{\lim }{h}^2\sin \left(\frac{1}{h}\right)=0$$ Also $g(0)=0$ $\therefore g(x)$ is continuous at $x=0$