This morning a colleague asked me how I would solve quickly this limit,

$$\lim\_{x\to \frac{\pi}{4}} \frac{\cos(2x)}{2\cos(x)-\sqrt 2}$$

Probably because I am a member of this community but I have suggest to do:

$$\frac{\cos(2x)}{2\cos(x)-\sqrt 2}=\frac{\cos(2x)}{(2\cos(x)-\sqrt 2)}\frac{(2\cos(x)+\sqrt 2)}{(2\cos(x)+\sqrt 2)}$$
Being $\cos(2x)=2\cos^2 x-1$,

$$\frac{\cos(2x)(2\cos(x)+\sqrt 2)}{2(2\cos^2 x-1)}=\frac{2\cos(x)+\sqrt 2}{2}$$

and we can found the limit for $x\to \frac{\pi}4$.

But I have now look the denominator $2\cos(x)-\sqrt 2=2\left(\cos x-\cos \frac{\pi}{4}\right)$. Using the prostapheresis formulas I will have:

$$2\left(\cos x-\cos \frac{\pi}{4}\right)=-2\left(\sin\frac{(x+\pi/4)}{2}\cdot\sin\frac{(x-\pi/4)}{2}\right)$$
If $x\to \frac{\pi}4$ then $$-2\left(\sin\frac{(x+\pi/4)}{2}\cdot\sin\frac{(x-\pi/4)}{2}\right)\to 0$$

Hence the strategy with the prostapheresis formulas is it not good or is there a way to use the notable limits?