Evaluate the limit of $\ln(\cos 2x)/\ln (\cos 3x)$ as $x\to 0$

Question

Evaluate Limits

$$\lim_{x\to 0}\frac{\ln(\cos(2x))}{\ln(\cos(3x))}$$

Method 1 :Using L'Hopital's Rule to Evaluate Limits (indicated by $\stackrel{LHR}{=}$. LHR stands for L'Hôpital Rule)

\begin{align*} \lim _{x\to \:0}\left(\frac{\ln \left(\cos \left(2x\right)\right)}{\ln \left(\cos \left(3x\right)\right)}\right)&\stackrel{LHR}{=}\\ &= \lim _{x\to \:0}\left(\frac{\ln \left(\cos \left(2x\right)\right)}{\ln \left(\cos \left(3x\right)\right)}\right)&\\ &=\lim _{x\to \:0}\frac{\left(\ln \left(\cos \left(2x\right)\right)\right)'}{\left(\ln \left(\cos \left(3x\right)\right)\right)'}\\ &=\lim _{x\to \:0}\left(\frac{-\frac{2\sin \left(2x\right)}{\cos \left(2x\right)}}{-\frac{3\sin \left(3x\right)}{\cos \left(3x\right)}}\right)\\ &=\lim _{x\to \:0}\left(\frac{2\sin \left(2x\right)\cos \left(3x\right)}{3\cos \left(2x\right)\sin \left(3x\right)}\right)\\ &\stackrel{LHR}{=}\lim _{x\to \:0}\frac{\left(2\sin \left(2x\right)\cos \left(3x\right)\right)'}{\left(3\cos \left(2x\right)\sin \left(3x\right)\right)'}\\ &=\lim _{x\to \:0}\left(\frac{4\cos \left(2x\right)\cos \left(3x\right)-6\sin \left(3x\right)\sin \left(2x\right)}{9\cos \left(3x\right)\cos \left(2x\right)-6\sin \left(2x\right)\sin \left(3x\right)}\right)\\ &=\lim _{x\to \:0}\left(\frac{2\left(2\cos \left(2x\right)\cos \left(3x\right)-3\sin \left(3x\right)\sin \left(2x\right)\right)}{3\left(3\cos \left(2x\right)\cos \left(3x\right)-2\sin \left(3x\right)\sin \left(2x\right)\right)}\right)\\ &=\frac{\lim _{x\to \:0}\left(2\left(2\cos \left(2x\right)\cos \left(3x\right)-3\sin \left(3x\right)\sin \left(2x\right)\right)\right)}{\lim _{x\to \:0}\left(3\left(3\cos \left(2x\right)\cos \left(3x\right)-2\sin \left(3x\right)\sin \left(2x\right)\right)\right)}=\dfrac{4}{9} \end{align*}

• Could we do it in others ways?

Answer 1

If you know $\lim_{x\to0}\frac{\ln(1+ x)}{x} = 1$, then

\begin{align} \lim_{x\to 0} \frac{\ln(\cos 2x)}{\ln(\cos 3x)} = \lim_{x\to 0} \frac{\ln(1 + \cos 2x - 1)}{\cos 2x -1} \frac{\cos 3x -1} {\ln(1 + \cos 3x - 1)} \frac{\cos 2x -1}{\cos 3x -1} \end{align}

We have $\lim_{x\to 0} \frac{\ln(1 + \cos 2x - 1)}{\cos 2x -1} = 1$, $\lim_{x\to 0} \frac{\ln(1 + \cos 3x - 1)}{\cos 3x -1} = 1$

and $\lim_{x\to 0}\frac{\cos 2x -1}{\cos 3x -1} = \frac{4}{9}$, since $\cos x = 1 - \frac{x^2}{2} + o(x^3)$ when $x \to 0$. We can also use L'Hopital's rule here

Answer 2

Once you apply L'Hopital's Rule once, and simplify you get $\displaystyle \lim_{x \to 0}\frac{-\frac{2\sin \left(2x\right)}{\cos \left(2x\right)}}{-\frac{3\sin \left(3x\right)}{\cos \left(3x\right)}} = \lim_{x \to 0}\dfrac{2\tan 2x}{3 \tan 3x}$.

Now, applying L'Hopital's Rule one more time gives you $\displaystyle\lim_{x \to 0}\dfrac{4\sec^2 2x}{9 \sec^2 3x} = \dfrac{4}{9}$.

Alternatively, $\displaystyle \lim_{x \to 0}\frac{-\frac{2\sin \left(2x\right)}{\cos \left(2x\right)}}{-\frac{3\sin \left(3x\right)}{\cos \left(3x\right)}} = \lim_{x \to 0}\dfrac{2}{3} \cdot \dfrac{\cos 3x}{\cos 2x} \cdot \dfrac{\frac{\sin 2x}{x}}{\frac{\sin 3x}{x}} = \lim_{x \to 0}\dfrac{4}{9} \cdot \dfrac{\cos 3x}{\cos 2x} \cdot \dfrac{\frac{\sin 2x}{2x}}{\frac{\sin 3x}{3x}}$,

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.

Answer 3

As told in answers, if you are just concerned by $$\lim_{x\to 0}\frac{\ln\Big(\cos(ax)\Big)}{\ln\Big(\cos(bx)\Big)}$$ the shortest way is L'Hopital's rule using the hints given by RecklessReckoner and JimmyK4542. You will then arrive to $$\lim_{x\to 0}\frac{\ln\Big(\cos(ax)\Big)}{\ln\Big(\cos(bx)\Big)}=\frac{a^2}{b^2}$$

But, if you want to know the path to this limit, Taylor expansion is really simple. Start with the series expansion $$\cos(t)= 1-\frac{t^2}{2}+\frac{t^4}{24}+O\left(t^5\right)$$ and the series expansion of $$\log(1+y)=y-\frac{y^2}{2}+O\left(y^3\right)$$ in which you replace now $y$ by $(-\frac{t^2}{2}+\frac{t^4}{24})$ to arrive to $$\log\Big(cos(t)\Big)=-\frac{t^2}{2}-\frac{t^4}{12}+O\left(t^6\right)$$ Now, replace $t$ by $ax$ for getting the numerator and $t$ by $bx$ for getting the denominator and perform the polynomial division. You should then get $$\frac{\ln\Big(\cos(ax)\Big)}{\ln\Big(\cos(bx)\Big)}= \frac{a^2}{b^2}+\frac{1}{6} a^2 \left(\frac{a^2}{b^2}-1\right)x^2+O\left(x^3\right)$$ which shows you that the limit is reached from above if $a \gt b$ and from below if $a \lt b$.