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Limits Continuity and Differentiability question

2022 · 26 Jun · Shift 1 · Q25

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JEE Main
Mathematics
Limits Continuity and Differentiability
2022 · 26 Jun · Shift 1 · Q25

Limits Continuity and Differentiability question

2022 · 26 Jun · Shift 1 · Q25

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}

is equal to :

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

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