lim_(x→(π/2)) (1+sin (π−2x))^(5/(sin (x−(π/2)))) .ln (4.((cos (π−2x)−1)/(((π/2)−x)^2 )))=?


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Question Number 170018 by cortano1 last updated on 14/May/22

$\lim_{x \to \frac{π}{2}} {(1 + \sin (π - 2 x))}^{\frac{5}{\sin (x - \frac{π}{2})}} . \ln (4 . \frac{\cos (π - 2 x) - 1}{{(\frac{π}{2} - x)}^{2}}) = ?$
	Answered by Mathspace last updated on 14/May/22

	$u (x) = {(1 + s i n (π - 2 x))}^{\frac{5}{s i n (x - \frac{π}{2})}}$ $a n d v (x) = l n (4 . \frac{c o s (π - 2 x) - 1}{{(\frac{π}{2} - x)}^{2}}$ $\Rightarrow u (x) = {(1 + s i n (2 x))}^{\frac{5}{- c o s x}}$ $= e^{- \frac{5}{c o s x} l n (1 + s i n (2 x)}$ $c h . x - \frac{π}{2} = t g i v e$ $u (x) = u (\frac{π}{2} + t) = e^{- \frac{5}{- s i n t} l n (1 - s i n (2 t))}$ $s i n t \sim t a n d s i n (2 t) \sim 2 t \Rightarrow$ $l n (1 - s i n (2 t)) \sim l n (1 - 2 t) \sim - 2 t \Rightarrow$ $\frac{5}{s i n t} l n (1 - s i n (2 t) \sim \frac{5}{t} (- 2 t) = - 10 \Rightarrow$ $l i m u (x) = e^{- 10}$ $v (x) = l n (4 . \frac{- c o s (2 x) - 1}{{(\frac{π}{2} - x)}^{2}}) (x - \frac{π}{2} = t)$ $= l n (4 \times \frac{- c o s (2 (\frac{π}{2} + t) - 1}{t^{2}})$ $= l n (- 4 . \frac{- c o s (2 t) + 1}{t^{2}})$ $= l n (\frac{4}{t^{2}} (c o s (2 t) - 1))$ $b u t c o s (2 t) - 1 < 0 e r r o r a t t h e q u e s t i o n! . . .$ $c o s (2 t) \sim 1 - 2 t^{2} \Rightarrow 1 - c o s (2 t)$
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