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

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

$$ \Rightarrow x = \cos t$$

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

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

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

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

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

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

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