lim-x-pi-2-1-tan-x-2-1-sin-x-1-tan-x-2-pi-2x-3-

Question Number 137820 by bramlexs22 last updated on 07/Apr/21

\lim_{x \to π / 2} \frac{(1 - \tan \frac{x}{2}) (1 - \sin x)}{(1 + \tan \frac{x}{2}) {(π - 2 x)}^{3}} ?

Answered by SLVR last updated on 07/Apr/21

Commented by SLVR last updated on 07/Apr/21

i s i t o k a y

Commented by bramlexs22 last updated on 07/Apr/21

a n s w e r \frac{1}{32}

Commented by SLVR last updated on 07/Apr/21

i a m s o r r y . . c a l i c u l a t i o n \dots e r r o r

s e e t h i s . . b u t 1 / 32 i h o p e

Commented by SLVR last updated on 07/Apr/21

Answered by liberty last updated on 07/Apr/21

\lim_{x \to π / 2} \frac{\tan (\frac{π}{4} - \frac{x}{2}) (1 - \sin x)}{{(π - 2 x)}^{3}}

l e t x = \frac{π}{2} + z & z \to 0

{\begin{cases} \frac{π}{4} - \frac{x}{2} = - \frac{z}{2} \\ 1 - \sin x = 1 - \cos z \\ π - 2 x = - 8 z^{3} \end{cases}

s o \lim_{z \to 0} \frac{- \tan (\frac{z}{2}) (1 - \cos z)}{- 8 z^{3}} =

\lim_{z \to 0} \frac{\tan (\frac{z}{2}) (2 \sin^{2} (\frac{z}{2})}{8 z^{3}} = \frac{1}{32}

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