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Question Number 99505 by bemath last updated on 21/Jun/20

Commented by Dwaipayan Shikari last updated on 21/Jun/20

$A n s : 1$
Commented by Dwaipayan Shikari last updated on 21/Jun/20

$= e^{l i \underset{x \to \frac{π}{2}}{m} (s i n x) (1 - t a n (\frac{x}{2}))}$ $= e^{l i \underset{x \to \frac{π}{2}}{m} (s i n x - 2 {s i n}^{2} (\frac{x}{2}))}$ $= e^{(1 - 1)} = e^{0} = 1$
	Answered by john santu last updated on 22/Jun/20

	$let \tan (\frac{x}{2}) = t$ $\lim_{t \to 1} (\frac{1 - 2 t + t^{2}}{1 + t^{2}}) \overset{t - 1}{} = e^{\lim_{t \to 1} \ln {(\frac{1 - 2 t + t^{2}}{1 + t^{2}})}^{t - 1}}$ $e^{\lim_{t \to 1} \frac{\ln (\frac{1 - 2 t + t^{2}}{1 + t^{2}})}{t - 1}} = e^{\lim_{t \to 1} \frac{(\frac{1 + t^{2}}{{(t - 1)}^{2}}) (\frac{(2 t - 2) (1 + t^{2}) - 2 t {(t - 1)}^{2}}{{(1 + t^{2})}^{2}})}{1}}$ $= e^{0} = 1$
	Answered by mathmax by abdo last updated on 21/Jun/20

	$let f (x) = {(1 - sinx)}^{\tan (\frac{x}{2} - 1)} \Rightarrow f (x) = e^{(\tan (\frac{x}{2}) - 1) \ln (1 - sinx)}$ $changement x = \frac{π}{2} - t give f (x) = e^{(\tan (\frac{π}{4} - \frac{t}{2}) - 1) \ln (1 - cost)} = g (t)$ $(t \to 0) we have \tan (\frac{π}{4} - \frac{t}{2}) - 1 = \frac{1 - \tan (\frac{t}{2})}{1 + \tan (\frac{t}{2})} - 1 = \frac{- 2 \tan (\frac{t}{2})}{1 + \tan (\frac{t}{2})} \sim \frac{- t}{1 + \frac{t}{2}} \sim - t$ $1 - cost \sim \frac{t^{2}}{2} \Rightarrow \ln (1 - cost) \sim 2 lnt - \ln (2) \Rightarrow$ $g (t) \sim e^{- t (2 lnt - \ln 2)} \to 1 (t \to 0) \Rightarrow \lim_{x \to \frac{π}{2}} f (x) = 1$
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