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$\lim_{x\rightarrow \frac{\pi}{2}}{( \sin x)}^{\tan x}$
in Calculus by Active (5.1k points)
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3 Answers

+3 votes
Best answer
$\lim_{x \to \frac{\pi}{2}}\sin x^{\tan x} = \lim_{x \to \frac{\pi}{2}} e^{\tan x.\ln(\sin x)} \\ e^{\lim_{x \to \frac{\pi}{2}}\tan x.\ln(\sin x) }
\\=e^{\lim_{x \to \frac{\pi}{2}} \frac{\sin x . \ln(\sin x)}{\cos x}  \left(\frac{0}{0} \text{ form, applying L'Hospital's rule}\right) }
\\=e^{\lim_{x \to \frac{\pi}{2}} \frac{ \sin x \frac{\cos x}{\sin x} + \ln (\sin x). \cos x}{- \sin x} }
\\= e^0 = 1$
by Junior (985 points)
selected by
0
yo​ur Answer is right but your explanation is wrong. Please  explain your first line.​
0 votes

Easy way to answer this question ->

Calculate sin (pi/2) using Scientific calculator => You get 1 !

calculate cos(pi/2) => you get 0

10 = 1 

Answer 1

by Boss (41.6k points)
–1 vote
Put x = pi/4

(Sin pi/4)^ tan pi/4
(1/sqrt 2)^1
1/sqrt 2.
by Veteran (60.6k points)
0
wrong plzz check its answer is 1
0
^^ why  1 ???

How u know my answer is wrong ??
0

i think if x-> pi/2 then the answer will be 1 , for this it seems right ( 1/21/2) .

0

sorry , edited x\rightarrow \prod /2

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