Let $f(x)=\{\begin{array}{ll}(x-1)\sin \left(\frac{1}{x-1}\right)& \text{if }x\ne 1\\ 0& \text{if }x=1\end{array}$. Then which one of the following is true?

At $x=1$:$f^{\prime }(1)={\lim }_{h\to 0}\frac{f(1+h)-f(1)}{h}={\lim }_{h\to 0}\frac{h\sin (1/h)-0}{h}={\lim }_{h\to 0}\sin (1/h)$.The limit ${\lim }_{h\to 0}\sin (1/h)$ does not exist (it oscillates between $-1$ and $1$).So, $f$ is not differentiable at $x=1$.At $x=0$:$f(x)$ is a product of $(x-1)$ and $\sin (1/(x-1))$, both of which are differentiable functions at $x=0$ because $(x-1)$ is never $0$ near $x=0$.$f^{\prime }(0)=\frac{d}{dx}[(x-1)\sin (1/(x-1)){]}_{x=0}$ Using product rule: $f^{\prime }(x)=\sin \left(\frac{1}{x-1}\right)+(x-1)\cos \left(\frac{1}{x-1}\right)\cdot \frac{-1}{(x-1{)}^2}=\sin \left(\frac{1}{x-1}\right)-\frac{1}{x-1}\cos \left(\frac{1}{x-1}\right)$.At $x=0$, $f^{\prime }(0)=\sin (-1)-\frac{1}{-1}\cos (-1)=-\sin (1)+\cos (1)$, which is a finite value.So $f$ is differentiable at $x=0$.