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?

Consider rewriting the numerator as $\cos(2x) - \cos(\frac{\pi}2).$ Then we can use the same manipulation to get
$$\cos(2x) - \cos(\frac{\pi}2) = -2\sin\left(\frac{2x + \frac{\pi}{2}}{2}\right)\sin\left(\frac{2x - \frac{\pi}{2}}{2}\right) = -2\sin\left(x + \frac{\pi}{4}\right)\sin\left(x - \frac{\pi}{4}\right)$$
Substituting this in gives us
$$\lim_{x \to \frac{\pi}4} \frac{-2\sin\left(x + \frac{\pi}{4}\right)\sin\left(x - \frac{\pi}{4}\right)}{2\left(-2\sin\left(\frac{x + \frac{\pi}{4}}2\right)\sin\left(\frac{x - \frac{\pi}{4}}2\right)\right)} = \frac12\lim_{x \to \frac{\pi}4}\frac{\sin\left(x + \frac{\pi}{4}\right)}{\sin\left(\frac{x + \frac{\pi}{4}}2\right)} \cdot \lim_{x \to \frac{\pi}4}\frac{\sin\left(x - \frac{\pi}{4}\right)}{\sin\left(\frac{x - \frac{\pi}{4}}2\right)}$$
supposing for the moment that those two limits exist. (also note the additional factor of $2$ in the denominator which seems to have been dropped in your manipulation)
For the first limit we can simply evaluate at $x = \frac{\pi}4$ to get $\frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}.$
For the second limit, consider rewriting $\sin(x - \frac{\pi}4)$ in the numerator as $\sin\left(2\frac{x - \frac{\pi}{4}}{2}\right) = 2\sin\left(\frac{x - \frac{\pi}{4}}{2}\right)\cos\left(\frac{x - \frac{\pi}{4}}{2}\right).$ This gives us that
$$\lim_{x \to \frac{\pi}4}\frac{\sin\left(x - \frac{\pi}{4}\right)}{\sin\left(\frac{x - \frac{\pi}{4}}2\right)} = \lim_{x \to \frac{\pi}4} 2\cos\left(\frac{x - \frac{\pi}{4}}{2}\right) = 2$$
So our total limit is $\frac12 \cdot \sqrt{2} \cdot 2 = \sqrt{2},$ which agrees with your initial solution.
