Evaluate the following limits:

$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{1-\sqrt{2} \sin x}$

Let,

$y=x-\frac{\pi}{4}$

$=\lim _{y \rightarrow 0} \frac{2 \tan x}{1-\cos x+\sin x}$

$=\lim _{y \rightarrow 0} \frac{\frac{2 \cos \frac{x}{2}}{\cos x}}{\sin \frac{x}{2}+\cos \frac{x}{2}}$

$=2$

$\therefore \lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{1-\sqrt{2} \sin x}=2$