$\lim\limits_{x\to \large\frac{\pi}{4}}\large\frac{\sqrt 2\cos x-1}{\cot x-1}\qquad$$(0/0)$ Applying L Hospital's Rule
$\Rightarrow \lim\limits_{x\to\Large\frac{\pi}{4}}\large\frac{-\sqrt 2\sin x}{-cosec^2x}$
$\Rightarrow \sqrt 2\lim\limits_{x\to \large\frac{\pi}{4}}\sin^3x$
$\Rightarrow \sqrt 2\big(\large\frac{1}{\sqrt 2}\big)^3$
$\Rightarrow \large\frac{1}{2}$
Hence (b) is the correct answer.