_ x 1^ - - 2 , ^ - 1 x 1 - x is equal to - Vidyadip

MCQ

$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt \pi - \sqrt {2\,{{\sin }^{ - 1}}x} }}{{\sqrt {1 - x} }}$ is equal to

A
$\frac{1}{{\sqrt {2\pi } }}$
✓
$\sqrt {\frac{2}{\pi }} $
C
$\sqrt {\frac{\pi }{2}} $
D
$\sqrt \pi $

✓

Answer

Correct option: B.

$\sqrt {\frac{2}{\pi }} $

b
$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} }}{{\sqrt {1 - x} }} \times \frac{{\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} }}{{\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} }}$

$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{2\left( {\frac{\pi }{2} - {{\sin }^{ - 1}}x} \right)}}{{\sqrt {1 - x} \left( {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} } \right)}}$

$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{2{{\cos }^{ - 1}}x}}{{\sqrt {1 - x} }}.\frac{1}{{2\sqrt \pi }}$

Assuming $x = \cos \theta $

$\mathop {\lim }\limits_{\theta \to {0^ + }} \frac{{2\theta }}{{\sqrt 2 \sin \left( {\frac{\theta }{2}} \right)}}.\frac{1}{{2\sqrt \pi }} = \sqrt {\frac{2}{\pi }} $

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