lim_(x→π/2) (([1−tan x/2][1−sin x])/([1+tan x/2][π−2x]^3 ))=?


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Question Number 63918 by raj last updated on 11/Jul/19

$\lim_{x \to π / 2} \frac{[1 - \tan x / 2] [1 - \sin x]}{[1 + \tan x / 2] {[π - 2 x]}^{3}} = ?$
	Answered by Cmr 237 last updated on 20/Aug/19

	$p o s o n s c e t t e l i m i t e e g a l e A$ $p a r l e d e v e l o p e m e n t l i m i t e w e h a v e :$ $a u x v o i s i n a g e d e \frac{π}{2},$ $t a n (\frac{x}{2}) ⌢ 1 + (x - \frac{π}{2}) + o (x)$ $s i n x ⌢ 1 - \frac{{(x - \frac{π}{2})}^{2}}{2} + o (x)$ $A (x) ⌢ \frac{- (x - \frac{π}{2}) (\frac{{(x - \frac{π}{2})}^{2}}{2})}{(2 + (x - \frac{π}{2})) {(π - 2 x)}^{3}} + o (x)$ $= \frac{\frac{1}{16} {(π - 2 x)}^{3}}{(2 + (x - \frac{π}{2})) {(π - 2 x)}^{3}} + o (x)$ $= \frac{\frac{1}{16}}{(2 + (x - \frac{π}{2}))} + o (x)$ $n o w {i t}^{'} s c l e a r t h a t;$ $l i m \underset{x \to \frac{π}{2}}{} A (x) = \frac{1}{32} = A .$
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