Mathematics - Continuity - Differentiability Question with Solution | TestHub

We have; $f(x)=\{\begin{array}{ll}(x-1)\sin \left(\frac{1}{x-1}\right)& \text{ if }x\ne 1\\ 0& \text{ if }x=1\end{array}$ $R{f}^{\prime }(1)={\lim }_{h\to 0}\frac{f(1+h)-f(1)}{h}$ $={\lim }_{h\to 0}\frac{h\sin \frac{1}{h}-0}{h}={\lim }_{h\to 0}\sin \frac{1}{h}$ which does not exist. $\therefore f$ is not differentiable at $x=1$ Also ${f^{\prime }(0)=\sin \frac{1}{(x-1)}-\frac{x-1}{(x-1{)}^2}\cos \left(\frac{1}{x-1}\right)]}_{x=0}$ $=-\sin 1+\cos 1$ $\therefore f$ is differentiable at $x=0$