MCQ
$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt \pi - \sqrt {2\,{{\sin }^{ - 1}}x} }}{{\sqrt {1 - x} }}$ =
- A$\frac{1}{{\sqrt {2\pi } }}$
- ✓$\sqrt {\frac{2}{\pi }} $
- C$\sqrt {\frac{\pi }{2}} $
- D$\sqrt \pi $
$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{2\left( {\frac{\pi }{2} - {{\sin }^{ - 1}}x} \right)}}{{\sqrt {1 - x} \left( {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} } \right)}}$
$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{2{{\cos }^{ - 1}}x}}{{\sqrt {1 - x} }}.\frac{1}{{2\sqrt \pi }}$
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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