pouzdrotf

2022-07-03

This morning a colleague asked me how I would solve quickly this limit,
$\underset{x\to \frac{\pi }{4}}{lim}\frac{\mathrm{cos}\left(2x\right)}{2\mathrm{cos}\left(x\right)-\sqrt{2}}$

Sophia Mcdowell

Expert

Consider rewriting the numerator as $\mathrm{cos}\left(2x\right)-\mathrm{cos}\left(\frac{\pi }{2}\right).$ Then we can use the same manipulation to get
$\mathrm{cos}\left(2x\right)-\mathrm{cos}\left(\frac{\pi }{2}\right)=-2\mathrm{sin}\left(\frac{2x+\frac{\pi }{2}}{2}\right)\mathrm{sin}\left(\frac{2x-\frac{\pi }{2}}{2}\right)=-2\mathrm{sin}\left(x+\frac{\pi }{4}\right)\mathrm{sin}\left(x-\frac{\pi }{4}\right)$
Substituting this in gives us
$\underset{x\to \frac{\pi }{4}}{lim}\frac{-2\mathrm{sin}\left(x+\frac{\pi }{4}\right)\mathrm{sin}\left(x-\frac{\pi }{4}\right)}{2\left(-2\mathrm{sin}\left(\frac{x+\frac{\pi }{4}}{2}\right)\mathrm{sin}\left(\frac{x-\frac{\pi }{4}}{2}\right)\right)}=\frac{1}{2}\underset{x\to \frac{\pi }{4}}{lim}\frac{\mathrm{sin}\left(x+\frac{\pi }{4}\right)}{\mathrm{sin}\left(\frac{x+\frac{\pi }{4}}{2}\right)}\cdot \underset{x\to \frac{\pi }{4}}{lim}\frac{\mathrm{sin}\left(x-\frac{\pi }{4}\right)}{\mathrm{sin}\left(\frac{x-\frac{\pi }{4}}{2}\right)}$
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 $\mathrm{sin}\left(x-\frac{\pi }{4}\right)$ in the numerator as $\mathrm{sin}\left(2\frac{x-\frac{\pi }{4}}{2}\right)=2\mathrm{sin}\left(\frac{x-\frac{\pi }{4}}{2}\right)\mathrm{cos}\left(\frac{x-\frac{\pi }{4}}{2}\right).$ This gives us that
$\underset{x\to \frac{\pi }{4}}{lim}\frac{\mathrm{sin}\left(x-\frac{\pi }{4}\right)}{\mathrm{sin}\left(\frac{x-\frac{\pi }{4}}{2}\right)}=\underset{x\to \frac{\pi }{4}}{lim}2\mathrm{cos}\left(\frac{x-\frac{\pi }{4}}{2}\right)=2$
So our total limit is $\frac{1}{2}\cdot \sqrt{2}\cdot 2=\sqrt{2},$ which agrees with your initial solution.

Expert

If you set $x=u+\frac{\pi }{4}$ with $u\to 0$ then you get

The fact that $\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(u\right)-1\ll \mathrm{sin}\left(u\right)$ can be inferred from:
$|\frac{\mathrm{cos}\left(u\right)-1}{\mathrm{sin}\left(u\right)}|=|\frac{2\mathrm{sin}\left(u/2{\right)}^{2}}{\mathrm{sin}\left(u\right)}|=\underset{\to 1}{\underset{⏟}{|\frac{\mathrm{sin}\left(u/2{\right)}^{2}}{{u}^{2}/4}|}}×\underset{\to 1}{\underset{⏟}{|\frac{u}{\mathrm{sin}\left(u\right)}|}}×\underset{\to 0}{\underset{⏟}{|8u|}}\to 0$

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