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$\lim _{x \rightarrow \pi} \frac{\sin x}{x-\pi}$ is equal to
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1274 Upvotes
Verified Answer
The correct answer is:
$-1$
$-1$
Since, $\lim _{x \rightarrow \pi} \frac{\sin x}{x-\pi}=\lim _{x \rightarrow \pi}\left[\frac{\sin (\pi-x)}{-(\pi-x)}\right]$
$$
[\because \sin x=\sin (\pi-x)]
$$
$$
=-\left[\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{(\pi-x)}\right]=-1\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]
$$
$$
[\because \sin x=\sin (\pi-x)]
$$
$$
=-\left[\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{(\pi-x)}\right]=-1\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]
$$
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