Official solution
\[\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}\]
Let \[{\cos ^{ - 1}}x = t\]
\[ \Rightarrow x = \cos t\]
When \[x \to {1 \over {\sqrt 2 }}\], then \[t \to {\cos ^{ - 1}}\left( {{1 \over {\sqrt 2 }}} \right) \to {\pi \over 4}\]
\[\therefore\] \[\mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - \tan (t)}}\]
\[ = \mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - {{\sin t} \over {\cos t}}}}\]
\[ = \mathop {\lim }\limits_{t \to {\pi \over 4}} {{(\sin t - \cos t)(\cos t)} \over {(\cos t - \sin t)}}\]
\[ = \mathop {\lim }\limits_{t \to {\pi \over 4}} - \cos t\]
\[ = - \mathop {\lim }\limits_{t \to {\pi \over 4}} \cos t\]
\[ = - {1 \over {\sqrt 2 }}\]