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$\lim\limits_{x\to {-1}}\large\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}}$ is given by

$(a)\;\large\frac{1}{\sqrt{\pi}}$$\qquad(b)\;\large\frac{1}{\sqrt{2\pi}}$$\qquad(c)\;1\qquad(d)\;0$

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Put $\cos^{-1}x=y$
$x=\cos y$
$\therefore$ As $ x\to -1,y\to \pi$
Given limit $=\lim\limits_{y\to \pi}\large\frac{\sqrt {\pi}-\sqrt{y}}{\sqrt{1+\cos y}}$
$\Rightarrow \lim\limits_{y\to \pi}\large\frac{\sqrt{\pi}-\sqrt{y}}{\sqrt 2\cos\large\frac{y}{2}}$
$\Rightarrow \lim\limits_{y\to \pi}\large\frac{\sqrt{\pi}-\sqrt{y}}{\sqrt{2}\sin(\pi/2-y/2)}.\frac{\pi/2-y/2}{\pi/2-y/2}$
$\Rightarrow \lim\limits_{y\to \pi}\large\frac{1}{\sqrt{2}\sin(\pi/2-y/2)}.\frac{\pi/2-y/2}{\pi/2-y/2}$
$\Rightarrow \lim\limits_{y\to \pi}\large\frac{1}{\Large\frac{\sqrt 2}{2}(\sqrt{\pi}+\sqrt{y})}.\frac{1}{\Large\frac{\sin(\pi/2-y/2)}{\pi/2-y/2}}$
$\Rightarrow \large\frac{1}{\sqrt{2\pi}}$
Hence (b) is the correct option.
answered Dec 30, 2013 by sreemathi.v
 

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