lim_(x→π) ((x−π)/(sinx))=?


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Question Number 184891 by mathlove last updated on 13/Jan/23

$\lim_{x \to π} \frac{x - π}{s i n x} = ?$
	Answered by Mathspace last updated on 13/Jan/23

	$c h a n g . x - π = t g i v e$ ${l i m}_{x \to π} \frac{x - π}{s i n x} = {l i m}_{t \to 0} \frac{t}{s i n (π + t)}$ $= - {l i m}_{t \to 0} \frac{t}{s i n t} = - 1$
	Answered by alephzero last updated on 13/Jan/23

	$\lim_{x \to π} \frac{x - π}{\sin x} = \lim_{x \to π} \frac{1}{\cos x} = \frac{1}{- 1} = - 1$
	Answered by TUN last updated on 14/Jan/23

	$= \lim_{x \to π} \frac{- (π - x)}{s i n (π - x)}$ $= - \lim_{x \to π} \frac{π - x}{s i n (π - x)}$ $= - 1$
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