MCQ
$\mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt \pi - \sqrt {{{\cos }^{ - 1}}x} }}{{\sqrt {x + 1} }}$ is given by
- A$\frac{1}{{\sqrt \pi }}$
- ✓$\frac{1}{{\sqrt {2\pi } }}$
- C$1$
- D$0$
$\therefore \,\,\,\mathop {\lim }\limits_{x \to - 1} \,\frac{{\sqrt \pi - \sqrt {{{\cos }^{ - 1}}x} }}{{\sqrt {x + 1} }} = \mathop {\lim }\limits_{y \to \pi } \,\frac{{\sqrt \pi - \sqrt y }}{{\sqrt {1 + \cos y} }}$
$ = \mathop {\lim }\limits_{y \to \pi } \,\frac{{\sqrt \pi - \sqrt y }}{{\sqrt 2 \,\cos \,(y/2)}}\, $
$= \mathop {\lim }\limits_{y \to \pi } \,\,\frac{{\sqrt \pi - \sqrt y }}{{\sqrt 2 \,\sin \,\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}\frac{{\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}{{\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}$
$ = \mathop {\lim }\limits_{y \to \pi } \,\frac{1}{{\frac{{\sqrt 2 }}{2}(\sqrt \pi + \sqrt y )}}.\frac{1}{{\frac{{\sin \,\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}{{\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}}} = \frac{1}{{\sqrt {2\pi } }}.$
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