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Limits Continuity and Differentiability question

2019 · 12 Jan · Shift 2 · Q28

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JEE Main
Mathematics
Limits Continuity and Differentiability
2019 · 12 Jan · Shift 2 · Q28

Limits Continuity and Differentiability question

2019 · 12 Jan · Shift 2 · Q28

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

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

is equal to :

A
$\sqrt {{2 \over \pi }}$
B
${1 \over {\sqrt {2\pi } }}$
C
$\sqrt {{\pi \over 2}}$
D
$\sqrt \pi$

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