mathop {lim }limits_{x to {1^ - }} frac{{sqrt | vedclass

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Question

(B) Let $L = \mathop {\lim }\limits_{x 	o {1^ - }} \frac{{\sqrt \pi   - \sqrt {2{{\sin }^{ - 1}}x} }}{{\sqrt {1 - x} }}$.Rationalizing the numerator,

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $L = \mathop {\lim }\limits_{x 	o {1^ - }} \frac{{\sqrt \pi   - \sqrt {2{{\sin }^{ - 1}}x} }}{{\sqrt {1 - x} }}$.Rationalizing the numerator,we get:$L = \mathop {\lim }\limits_{x 	o {1^ - }} \frac{{(\sqrt \pi   - \sqrt {2{{\sin }^{ - 1}}x} )(\sqrt \pi   + \sqrt {2{{\sin }^{ - 1}}x} )}}{{\sqrt {1 - x} (\sqrt \pi   + \sqrt {2{{\sin }^{ - 1}}x} )}}$$L = \mathop {\lim }\limits_{x 	o {1^ - }} \frac{{\pi  - 2{{\sin }^{ - 1}}x}}{{\sqrt {1 - x} (\sqrt \pi   + \sqrt {2{{\sin }^{ - 1}}x} )}}$$L = \mathop {\lim }\limits_{x 	o {1^ - }} \frac{{2(\frac{\pi }{2} - {{\sin }^{ - 1}}x)}}{{\sqrt {1 - x} (\sqrt \pi   + \sqrt {2{{\sin }^{ - 1}}x} )}}$Using the identity $\frac{\pi }{2} - {{\sin }^{ - 1}}x = {{\cos }^{ - 1}}x$,we have:$L = \mathop {\lim }\limits_{x 	o {1^ - }} \frac{{2{{\cos }^{ - 1}}x}}{{\sqrt {1 - x} }} \cdot \frac{1}{{2\sqrt \pi  }}$Let $x = \cos 	heta$,then as $x 	o 1^-$,$	heta 	o 0^+$. Also,$\sqrt{1-x} = \sqrt{1-\cos 	heta} = \sqrt{2\sin^2(	heta/2)} = \sqrt{2}\sin(	heta/2)$.$L = \mathop {\lim }\limits_{	heta 	o 0^+} \frac{{2	heta }}{{\sqrt{2}\sin(	heta/2)}} \cdot \frac{1}{{2\sqrt \pi  }}$Since $\mathop {\lim }\limits_{	heta 	o 0^+} \frac{	heta}{\sin(	heta/2)} = 2$,we get:$L = \frac{2 \cdot 2}{\sqrt{2} \cdot 2\sqrt{\pi}} = \frac{2}{\sqrt{2\pi}} = \sqrt{\frac{2}{\pi}}$.

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

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