David Young

2021-12-30

Prove that:
$\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(\mathrm{cos}x\right)}{\mathrm{ln}\left(1-\frac{{x}^{2}}{2}\right)}=1$
without LHopitals rule.

Using $\mathrm{ln}\left(1-y\right)=-\sum _{n=1}^{\mathrm{\infty }}\frac{{y}^{n}}{n}$ we get immediately
$\mathrm{ln}\left(1-\frac{{x}^{2}}{2}\right)=-\frac{{x}^{2}}{2}+O\left({x}^{4}\right)$,and also, after observing that
$\mathrm{cos}x=1-\frac{{x}^{2}}{2}+O\left({x}^{4}\right),\mathrm{ln}\mathrm{cos}x=-\frac{{x}^{2}}{2}+O\left({x}^{4}\right)$
Therefore
$\frac{\mathrm{ln}\left(\mathrm{cos}x\right)}{\mathrm{ln}\left(1-\frac{{x}^{2}}{2}\right)}=1+O\left({x}^{2}\right)$

Gerald Lopez

$\frac{\mathrm{ln}\left(\mathrm{cos}x\right)}{\mathrm{ln}\left(1-\frac{{x}^{2}}{2}\right)}=\frac{\mathrm{ln}\left(1+\mathrm{cos}x-1\right)}{\mathrm{cos}x}\cdot \frac{-\frac{{x}^{2}}{2}}{\mathrm{ln}\left(1-\frac{{x}^{2}}{2}\right)}\cdot \frac{{\mathrm{sin}}^{2}\frac{x}{2}}{\frac{{x}^{2}}{4}}\to 1$

Vasquez

On OP's request I am converting my comment into an answer. The limit in question is easily solved if one rewrites the given expression as
$\frac{\mathrm{log}\left(1+\mathrm{cos}x-1\right)}{\mathrm{cos}x-1}\cdot \frac{\mathrm{cos}x-1}{{x}^{2}}\cdot \frac{{x}^{2}}{\left(\frac{-{x}^{2}}{2}\right)}\cdot \frac{\left(-\frac{{x}^{2}}{2}\right)}{\mathrm{log}\left(1-\frac{{x}^{2}}{2}\right)}$
And then the limit is easily seen to be $1\cdot \left(-\frac{1}{2}\right)\cdot \left(-2\right)\cdot 1=1.$
Other limits in your question (after the fold) don't have any problem as we can just plug x=0 to evaluate them (the functions concerned are continuous at 0).

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