Solving $\lim\limits_{x \to 0}\frac{\ln \cos 3x}{\ln \cos (-x)}$ without L'Hospital rule

Question

How can I solve this limit not using L'Hospital rule?

$\lim\limits_{x \to 0}\frac{\ln \cos 3x}{\ln \cos (-x)}$

Thank you very much.

Answer 1

Using standard tools only, I showed in THIS ANSWER that the log function satisfies the inequalities

$$\frac{x-1}{x}\le \log x \le x-1 \tag 1$$

Using $(1)$, we find that $f(x)=\dfrac{\log \cos (3x)}{\log \cos(-x)}$ is bounded below and above as

$$\begin{align} \frac{1}{\cos(3x)}\frac{\sin^2(3x/2)}{\sin^2(x/2)}&=\frac{\cos (3x)-1}{\cos(3x)(\cos(x)-1)}\\\\ &\le \frac{\log \cos(3x)}{\log \cos(x)}\\\\ &\le \frac{\cos (x)(\cos(3x)-1)}{\cos(x)-1}=\cos(x)\frac{\sin^2(3x/2)}{\sin^2(x/2)} \tag 2 \end{align}$$

The right-hand and left-hand sides of $(2)$ both approach $9$ since $\frac{\sin x}{x}\to 1$.

By the squeeze theorem, we find that the limit is $9$. Therefore, we have

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\left(\frac{\log \cos (3x)}{\log \cos(-x)}\right)=9}$$

Answer 2

$$\cos(3x)=\cos(x)\cdot\left(4\cos(x)^2-3\right), $$ hence: $$ \frac{\log\cos(3x)}{\log\cos(-x)}=\frac{\log(\cos x)+\log\left(4\cos(x)^2-3\right)}{\log(\cos x)}=1+2\cdot\frac{\log(1-4\sin^2 x)}{\log(1-\sin^2 x)} $$ and since for $t$ close to zero we have $\log(1-t)\sim -t$, $$ \lim_{x\to 0}\frac{\log(\cos(3x))}{\log\cos(-x)} = 1+2\cdot 4 = \color{red}{9}.$$

Answer 3

Taylor's formula applied twice yields, $$\log \cos (3x)=\log\left(1-\frac{9}{2}x^2+O(x^4)\right)=-\frac{9}{2}x^2+O(x^4)$$ as $x\to 0$, similarly by Taylor's formula, $$\log\cos(-x)=\log\cos(x)=\log\left(1-\frac{1}{2}x^2+O(x^4)\right)=-\frac{1}{2}x^2+O(x^4)$$ as $x\to 0$, note that $\cos(-x)=\cos(x)$. Hence the limit is, $$\lim_{x\to0}\frac{-\dfrac{9}{2}x^2+O(x^4)}{-\dfrac{1}{2}x^2+O(x^4)}=9.$$

Answer 4

A variant:

$\cos 3x=1-\dfrac{9x^2}2+o(x^2)$, $\cos(-x)=1-\dfrac{x^2}2+o(x^2)$ and $\ln(1-u)\sim_0 -u$, hence $$\frac{\ln \cos 3x}{\ln \cos (-x)}\sim_0 \frac{-\frac{9x^2}2}{-\frac{x^2}2}=9.$$