Mathematics - Continuity - Differentiability Question with Solution | TestHub

$$f^{\prime }(1)=\underset{h\to 0}{\lim }\frac{f(1+h)-f(1)}{h}$$ $$\begin{array}{rl}& \Rightarrow f^{\prime }(1)=\underset{h\to 0}{\lim }\frac{(1+h-1)\sin \left(\frac{1}{1+h-1}\right)-0}{h}=\underset{h\to 0}{\lim }\frac{h}{h}\sin \left(\frac{1}{h}\right)\\ & \Rightarrow f^{\prime }(1)=\underset{h\to 0}{\lim }\sin \left(\frac{1}{h}\right)\end{array}$$ $\therefore \mathrm{f}$ is not differentiable at $x=1$. Similarly, $f^{\prime }(0)={\lim }_{h\to 0}\frac{f(h)-f(0)}{h}$ $\Rightarrow f^{\prime }(0)={\lim }_{h\to 0}\frac{(h-1)\sin \left(\frac{1}{h-1}\right)-\sin (1)}{h}$ $\Rightarrow \mathrm{f}$ is also not differentiable at $\mathrm{x}=0$.