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

Question

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?

Answer 1

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.

Answer 2

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:

$\left|\dfrac{\cos(u)-1}{\sin(u)}\right|=\left|\dfrac{2\sin(u/2)^2}{\sin(u)}\right|=\underbrace{\left|\dfrac{\sin(u/2)^2}{u^2/4}\right|}_{\to 1}\times\underbrace{\left|\dfrac{u}{\sin(u)}\right|}_{\to 1}\times\underbrace{|8u|}_{\to 0}\to 0$