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  • Evaluate: <img alt="\lim _{x \rightarrow \pi / 2} \sqrt{\frac{\tan x-\sin \left\{\tan ^{-1}(\tan x)\right\}}{\tan x+\cos ^2(\tan x)}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cl

Evaluate: \lim _{x \rightarrow \pi / 2} \sqrt{\frac{\tan x-\sin \left\{\tan ^{-1}(\tan x)\right\}}{\tan x+\cos ^2(\tan x)}}

 

 

Option: 1

1


Option: 2

2
 


Option: 3

3
 


Option: 4

0


Answers (1)

best_answer

We have, right hand limit at x=\pi / 2

=\lim _{x \rightarrow+\pi / 2} \sqrt{\frac{\tan x-\sin (x-\pi)}{\tan x+\cos ^2(\tan x)}}

                                       \left\{\because \tan ^{-1}(\tan x)=x-\pi, \text { when } x>\pi / 2\right\}

=\lim _{x \rightarrow+\pi / 2} \sqrt{\frac{1+\frac{\sin x}{\tan x}}{1+\frac{\cos ^2(\tan x)}{\tan x}}}

=\sqrt{\frac{1+0}{1+0}}=1

L H L at                         x=\pi / 2

                                         =\lim _{x \rightarrow \frac{\pi}{2}^{-}} \sqrt{\frac{\tan x-\sin (x)}{\tan x+\cos ^2(\tan x)}}

                                                                                    \left\{\because \tan ^{-1}(\tan x)=x \text {, when } x<\pi / 2\right\}

    \begin{aligned} & =\lim _{x \rightarrow \frac{\pi^{-}}{2}} \sqrt{\frac{1-\frac{\sin x}{\tan x}}{1+\frac{\cos ^2(\tan x)}{\tan x}}} \\ \\& =\sqrt{\frac{1+0}{1+0}}=1 \end{aligned}

RHL = LHL, hence the given limit = 1

Posted by

Anam Khan

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