QJEEM19Maba4e: JEE Main 2019 Mathematics - Class 12 JEEM19Maba4e Solution

Question

Verified Solution • 2025–26 Session moderate

\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }}

is equal to :

Accepted Answer

\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }} \times {{\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}} } \over {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} }}

\mathop {\lim }\limits_{x \to {1^ - }} {{2\left( {{\pi \over 2} - {{\sin }^{ - 1}}x} \right)} \over {\sqrt {1 - x} \left( {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} } \right)}}

\mathop {\lim }\limits_{x \to {1^ - }} {{2{{\cos }^{ - 1}}x} \over {\sqrt {1 - x} }}.{1 \over {2\sqrt \pi }}

Put

x = \cos \theta

\mathop {\lim }\limits_{x \to {0^ + }} {{2\theta } \over {\sqrt 2 \sin \left( {{\theta \over 2}} \right)}}.{1 \over {2\sqrt \pi }}

= \sqrt {{2 \over \pi }}

Step-by-Step Verified Solution