How to evaluate this trigonometric limit?

multiply numerator and denominator by $\sqrt{2}\cos x+1$. you get $$\frac{2\cos^2 x-1}{(\cot x -1)( \sqrt{2}\cos x+1 )}$$ now $2\cos^2 x-1=\cos 2x$ and $\cos 2x=\frac{1-\tan^2x}{1+\tan^2 x}$

also $\cot x-1=\frac{1-\tan x}{\tan x}$ using all these you get $$\frac{\sqrt{2}\cos x-1}{\cot x-1}=\frac{\tan x(1+\tan x)}{(1+\tan^2 x)( \sqrt{2}\cos x+1 )}$$

Alternative method:

Your expression can be written as: $$ \lim_{x\to\pi/4} \dfrac{ \sqrt2 \cos x - 1} {\cos x - \sin x} \cdot\sin x $$ $$ = \lim_{h\to 0} \dfrac{ \sqrt2 \cos (h + \pi/4) - 1} {\cos (h + \pi/4) - \sin (h + \pi/4)} \cdot\sin (h + \pi/4) $$

(taking $ x = \pi/4 + h $. And hey, it just looks complicated here! It isn't. )

$$ = \lim_{h\to 0} \dfrac{ (\cos h - \sin h) - 1} {-2\sin h} \cdot\sin (h + \pi/4) $$ (simplifying using the $\sin(A+B), \cos (A+B)$ formulas)

$$ = \lim_{h\to 0} \dfrac{ 1 - \cos h + \sin h} {2\sin h} \cdot\sin (h + \pi/4) $$ $$ = \lim_{h\to 0} \;\;2\sin\left(\frac h 2 \right)\dfrac{ \sin(h/2) + \cos(h/2)} {2 \cdot 2\sin (h/2)\cos(h/2)} \cdot\sin (h + \pi/4) $$

Now, everything breaks down to $0.5$ nicely.