STD 11 - 12. limits — JEE कक्षा 11 साइन्स — Question

$\therefore \,\,\,\mathop {\lim }\limits_{x \to - 1} \,\frac{{\sqrt \pi - \sqrt {{{\cos }^{ - 1}}x} }}{{\sqrt {x + 1} }} = \mathop {\lim }\limits_{y \to \pi } \,\frac{{\sqrt \pi - \sqrt y }}{{\sqrt {1 + \cos y} }}$

$ = \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 } }}.$