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VITEEE 2012 Paper

Section: Mathematics

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Question : 115 of 120

Marks: +1, -0

The value of

\lim\limits_{x\rightarrow\infty}

\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\frac{1}{x}}

is

0
1
- 1
e

Solution:

Let y =

\lim\limits_{x\rightarrow\infty}

\left(\frac{\pi}{2}-\tan^{-1}x\right)

Taking log on both sides, we get

log y =

\lim\limits_{x\rightarrow\infty}\frac{1}{x}

log

\left(\frac{\pi}{2}-\tan^{-1}x\right)

form

(\text{form }\infty/\infty)

=

\lim\limits_{x\rightarrow\infty}

\frac{-\frac{1}{1+x^{2}}}{\frac{\pi}{2}-\tan^{-1}x}

(using L ‘Hospitals’ rule)

=

\lim\limits_{x\rightarrow\infty}

\frac{\frac{2x}{(1+x^{2})^{2}}}{-\frac{1}{1-x^{2}}}

(using L' Hospital’s rule)

\lim\limits_{x\rightarrow\infty}

-\frac{2x}{1+x^{2}}

= 0 ⇒ y =

e^{0}

= 1

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