Let f(x)=cases(x-1)1x-1, & x10, & x=1cases.

Question

Let $f(x)=\begin{cases}(x-1)\sin\dfrac{1}{x-1}, & x
e1\0, & x=1\end{cases}$. Then which one of the following is true?

Accepted Answer

At $x=1$: RHD $=\displaystyle\lim_{h	o0}\sin\frac1h$ (does not exist, oscillates), LHD $=-\displaystyle\lim_{h	o0}\sin\frac1h$ (also oscillates) --- different behavior, so $f$ is not differentiable at $x=1$.

At $x=0$: $f'(0)=\sin\left(\dfrac{1}{x-1}ight)+\dfrac{-(x-1)}{(x-1)^2}\cos\left(\dfrac{1}{x-1}ight)\Big|_{x=0}=-\sin1+\cos1$ --- a finite value, so $f$ is differentiable at $x=0$.

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