"$$\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}$$ is equal to :"

Question

Accepted Answer

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

Let ${\cos ^{ - 1}}x = t$

$\Rightarrow x = \cos t$

When $x \to {1 \over {\sqrt 2 }}$, then $t \to {\cos ^{ - 1}}\left( {{1 \over {\sqrt 2 }}} \right) \to {\pi \over 4}$

$\therefore$ $$\mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - \tan (t)}}$$

$$ = \mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - {{\sin t} \over {\cos t}}}}$$

$$ = \mathop {\lim }\limits_{t \to {\pi \over 4}} {{(\sin t - \cos t)(\cos t)} \over {(\cos t - \sin t)}}$$

$= \mathop {\lim }\limits_{t \to {\pi \over 4}} - \cos t$

$= - \mathop {\lim }\limits_{t \to {\pi \over 4}} \cos t$

$= - {1 \over {\sqrt 2 }}$

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