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

Question

$\sqrt {{2 \over \pi }} $

${1 \over {\sqrt {2\pi } }}$

$\sqrt {{\pi \over 2}} $

$\sqrt \pi $

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