Strategies to solve a limit $\lim_{x\to \pi/4} \frac{\cos(2x)}{2\cos(x)-\sqrt 2}$ 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?

If you set $x=u+\frac{\pi}4$ with $u\to 0$ then you get
$$f(x)=\dfrac{\cos(2x)}{2\cos(x)-\sqrt{2}}=\dfrac{-2\sin(u)\overbrace{\cos(u)}^{\to 1}}{\sqrt{2}(\underbrace{\cos(u)-1}_{\ll\ \sin(u)}-\sin(u))}\sim\dfrac{-2\sin(u)}{-\sqrt{2}\sin(u)}\to\sqrt{2}$$
The fact that $\, \cos(u)-1\ll \sin(u)$ can be inferred from:
$\left|\dfrac{\cos(u)-1}{\sin(u)}\right|=\left|\dfrac{2\sin(u/2)^2}{\sin(u)}\right|=\underbrace{\left|\dfrac{\sin(u/2)^2}{u^2/4}\right|}_{\to 1}\times\underbrace{\left|\dfrac{u}{\sin(u)}\right|}_{\to 1}\times\underbrace{|8u|}_{\to 0}\to 0$
