_ x 1 ( 4 ^ - 1 x )^ 1 ( x^2 - 1) is equal to - - Vidyadip

MCQ

$\mathop {\lim }\limits_{x \to 1} {\left( {\frac{4}{\pi }{{\tan }^{ - 1}}x} \right)^{\frac{1}{{({x^2} - 1)}}}}$ is equal to -

A
$-\frac{1}{\pi}$
B
$\frac{1}{\pi}$
C
$e^{-\frac{1}{\pi}}$
✓
$e^{\frac{1}{\pi}}$

✓

Answer

Correct option: D.

$e^{\frac{1}{\pi}}$

d
$\mathop {\lim }\limits_{x \to 1} {\left( {\frac{4}{\pi }{{\tan }^{ - 1}}x} \right)^{\frac{1}{{\left( {{x^2} - 1} \right)}}}} = {e^{\mathop {\lim }\limits_{x \to 1} \frac{1}{{{x^2} - 1}}\left( {\frac{4}{\pi }{{\tan }^{ - 1}}x - 1} \right)}}$

${e^{\mathop {\lim }\limits_{x \to 1} \frac{4}{\pi }\frac{{\left( {{{\tan }^{ - 1}}x - {{\tan }^{ - 1}}1} \right)}}{{\left( {{x^2} - 1} \right)}}}} = {e^{\mathop {\lim }\limits_{x \to 1} \frac{4}{\pi }\frac{{{{\tan }^{ - 1}}\left( {\frac{{x - 1}}{{1 + x}}} \right)}}{{\left( {{x^2} - 1} \right)}}}} = {e^{\frac{1}{\pi }}}$

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