Answer
$0$
Work Step by Step
Since $-1\leq \sin \frac{\pi}{x-1}\leq 1$, then we have
$$-(x-1)\leq (x-1)\sin\frac{\pi}{x-1}\leq (x-1).$$
Moreover, $\lim\limits_{x \to 1}(x-1)=\lim\limits_{x \to 1}-(x-1)=0$. Then by the Squeeze Theorem, we have
$$\lim\limits_{x \to 1}(x-1)\sin\frac{\pi}{x-1}=0.$$
