Answer

(b) -1
Explanation: Given, $\lim _{x \rightarrow \pi} \frac{\sin x}{x-\pi}=\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{-(\pi-x)}$
$=-1\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right.$ and $\left.\pi-x \rightarrow 0 \Rightarrow x \rightarrow \pi\right]$

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