We are given,

$\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\sin x – \cos x}}{{x – \frac{\pi }{4}}}$

$\sin x – \cos x = \sqrt 2 \left( {\frac{{\sin x}}{{\sqrt 2 }} – \frac{{\cos x}}{{\sqrt 2 }}} \right)$

$ = \sqrt 2 \left( {\sin x\cos \left( {\frac{\pi }{4}} \right) – \operatorname{cosx} \sin \left( {\frac{\pi }{4}} \right)} \right)$

Using the formula,

$\sqrt 2 \left( {\sin x\cos \left( {\frac{\pi }{4}} \right) – \cos x\sin \left( {\frac{\pi }{4}} \right)} \right) = \sqrt 2 \sin \left( {x – \frac{\pi }{4}} \right)$

So,

$\sin x – \cos x = \sqrt 2 \sin \left( {x – \frac{\pi }{4}} \right)$

Now evaluating the limit to get,

$\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\sin x – \cos x}}{{x – \frac{\pi }{4}}} = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\sqrt 2 \sin \left( {x – \frac{\pi }{4}} \right)}}{{x – \frac{\pi }{4}}}$

$ = \sqrt 2 \mathop {\lim }\limits_{x \to \frac{7}{4}} \frac{{\sin \left( {x – \frac{\pi }{4}} \right)}}{{x – \frac{\pi }{4}}}$

Applying, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$ and substituting the limit value to get,

$\sqrt 2 \mathop {\lim }\limits_{x \to \frac{ – }{4}} \frac{{\sin \left( {x – \frac{\pi }{4}} \right)}}{{x – \frac{\pi }{4}}} = \sqrt 2  \cdot 1$

$ = \sqrt 2 $

Therefore, $\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\sin x – \operatorname{cosx} }}{{x – \frac{\pi }{4}}} = \sqrt 2 $.