Official solution
\[\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }} \times {{\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}} } \over {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} }}\]
\[\mathop {\lim }\limits_{x \to {1^ - }} {{2\left( {{\pi \over 2} - {{\sin }^{ - 1}}x} \right)} \over {\sqrt {1 - x} \left( {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} } \right)}}\]
\[\mathop {\lim }\limits_{x \to {1^ - }} {{2{{\cos }^{ - 1}}x} \over {\sqrt {1 - x} }}.{1 \over {2\sqrt \pi }}\]
Put \[x = \cos \theta \]
\[\mathop {\lim }\limits_{x \to {0^ + }} {{2\theta } \over {\sqrt 2 \sin \left( {{\theta \over 2}} \right)}}.{1 \over {2\sqrt \pi }}\]
\[ = \sqrt {{2 \over \pi }} \]