Gardiolo0j

## Answered question

2022-09-30

Evaluate the limit of $\mathrm{ln}\left(\mathrm{cos}2x\right)/\mathrm{ln}\left(\mathrm{cos}3x\right)$ as $x\to 0$
Evaluate Limits
$\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(\mathrm{cos}\left(2x\right)\right)}{\mathrm{ln}\left(\mathrm{cos}\left(3x\right)\right)}$
Method 1 :Using L'Hopital's Rule to Evaluate Limits (indicated by $\stackrel{LHR}{=}$. LHR stands for L'Hôpital Rule)
$\begin{array}{rl}\underset{x\to \phantom{\rule{mediummathspace}{0ex}}0}{lim}\left(\frac{\mathrm{ln}\left(\mathrm{cos}\left(2x\right)\right)}{\mathrm{ln}\left(\mathrm{cos}\left(3x\right)\right)}\right)& \stackrel{LHR}{=}\\ & =\underset{x\to \phantom{\rule{mediummathspace}{0ex}}0}{lim}\left(\frac{\mathrm{ln}\left(\mathrm{cos}\left(2x\right)\right)}{\mathrm{ln}\left(\mathrm{cos}\left(3x\right)\right)}\right)& \\ & =\underset{x\to \phantom{\rule{mediummathspace}{0ex}}0}{lim}\frac{{\left(\mathrm{ln}\left(\mathrm{cos}\left(2x\right)\right)\right)}^{\prime }}{{\left(\mathrm{ln}\left(\mathrm{cos}\left(3x\right)\right)\right)}^{\prime }}\\ & =\underset{x\to \phantom{\rule{mediummathspace}{0ex}}0}{lim}\left(\frac{-\frac{2\mathrm{sin}\left(2x\right)}{\mathrm{cos}\left(2x\right)}}{-\frac{3\mathrm{sin}\left(3x\right)}{\mathrm{cos}\left(3x\right)}}\right)\\ & =\underset{x\to \phantom{\rule{mediummathspace}{0ex}}0}{lim}\left(\frac{2\mathrm{sin}\left(2x\right)\mathrm{cos}\left(3x\right)}{3\mathrm{cos}\left(2x\right)\mathrm{sin}\left(3x\right)}\right)\\ & \stackrel{LHR}{=}\underset{x\to \phantom{\rule{mediummathspace}{0ex}}0}{lim}\frac{{\left(2\mathrm{sin}\left(2x\right)\mathrm{cos}\left(3x\right)\right)}^{\prime }}{{\left(3\mathrm{cos}\left(2x\right)\mathrm{sin}\left(3x\right)\right)}^{\prime }}\\ & =\underset{x\to \phantom{\rule{mediummathspace}{0ex}}0}{lim}\left(\frac{4\mathrm{cos}\left(2x\right)\mathrm{cos}\left(3x\right)-6\mathrm{sin}\left(3x\right)\mathrm{sin}\left(2x\right)}{9\mathrm{cos}\left(3x\right)\mathrm{cos}\left(2x\right)-6\mathrm{sin}\left(2x\right)\mathrm{sin}\left(3x\right)}\right)\\ & =\underset{x\to \phantom{\rule{mediummathspace}{0ex}}0}{lim}\left(\frac{2\left(2\mathrm{cos}\left(2x\right)\mathrm{cos}\left(3x\right)-3\mathrm{sin}\left(3x\right)\mathrm{sin}\left(2x\right)\right)}{3\left(3\mathrm{cos}\left(2x\right)\mathrm{cos}\left(3x\right)-2\mathrm{sin}\left(3x\right)\mathrm{sin}\left(2x\right)\right)}\right)\\ & =\frac{\underset{x\to \phantom{\rule{mediummathspace}{0ex}}0}{lim}\left(2\left(2\mathrm{cos}\left(2x\right)\mathrm{cos}\left(3x\right)-3\mathrm{sin}\left(3x\right)\mathrm{sin}\left(2x\right)\right)\right)}{\underset{x\to \phantom{\rule{mediummathspace}{0ex}}0}{lim}\left(3\left(3\mathrm{cos}\left(2x\right)\mathrm{cos}\left(3x\right)-2\mathrm{sin}\left(3x\right)\mathrm{sin}\left(2x\right)\right)\right)}=\frac{4}{9}\end{array}$
Could we do it in others ways?

### Answer & Explanation

Rachael Singh

Beginner2022-10-01Added 4 answers

If you know $\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+x\right)}{x}=1$, then
$\begin{array}{r}\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(\mathrm{cos}2x\right)}{\mathrm{ln}\left(\mathrm{cos}3x\right)}=\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+\mathrm{cos}2x-1\right)}{\mathrm{cos}2x-1}\frac{\mathrm{cos}3x-1}{\mathrm{ln}\left(1+\mathrm{cos}3x-1\right)}\frac{\mathrm{cos}2x-1}{\mathrm{cos}3x-1}\end{array}$
We have $\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+\mathrm{cos}2x-1\right)}{\mathrm{cos}2x-1}=1$, $\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+\mathrm{cos}3x-1\right)}{\mathrm{cos}3x-1}=1$
and $\underset{x\to 0}{lim}\frac{\mathrm{cos}2x-1}{\mathrm{cos}3x-1}=\frac{4}{9}$, since $\mathrm{cos}x=1-\frac{{x}^{2}}{2}+o\left({x}^{3}\right)$ when $x\to 0$. We can also use L'Hopital's rule here

overrated3245w

Beginner2022-10-02Added 3 answers

Once you apply L'Hopital's Rule once, and simplify you get $\underset{x\to 0}{lim}\frac{-\frac{2\mathrm{sin}\left(2x\right)}{\mathrm{cos}\left(2x\right)}}{-\frac{3\mathrm{sin}\left(3x\right)}{\mathrm{cos}\left(3x\right)}}=\underset{x\to 0}{lim}\frac{2\mathrm{tan}2x}{3\mathrm{tan}3x}$
Now, applying L'Hopital's Rule one more time gives you $\underset{x\to 0}{lim}\frac{4{\mathrm{sec}}^{2}2x}{9{\mathrm{sec}}^{2}3x}=\frac{4}{9}$
Alternatively, $\underset{x\to 0}{lim}\frac{-\frac{2\mathrm{sin}\left(2x\right)}{\mathrm{cos}\left(2x\right)}}{-\frac{3\mathrm{sin}\left(3x\right)}{\mathrm{cos}\left(3x\right)}}=\underset{x\to 0}{lim}\frac{2}{3}\cdot \frac{\mathrm{cos}3x}{\mathrm{cos}2x}\cdot \frac{\frac{\mathrm{sin}2x}{x}}{\frac{\mathrm{sin}3x}{x}}=\underset{x\to 0}{lim}\frac{4}{9}\cdot \frac{\mathrm{cos}3x}{\mathrm{cos}2x}\cdot \frac{\frac{\mathrm{sin}2x}{2x}}{\frac{\mathrm{sin}3x}{3x}}$
which is easily broken up into well known limits.
Note: More often than not, it is better to simplify a limit before applying L'Hopital's Rule.

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