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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 :

JEE · Math · previous-year question

  1. A.$$\sqrt 2 $$
  2. B.$$ - \sqrt 2 $$
  3. C.$${1 \over {\sqrt 2 }}$$
  4. D.$$ - {1 \over {\sqrt 2 }}$$correct

Answer

D. $$ - {1 \over {\sqrt 2 }}$$

Explanation

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