Solving $\lim\limits_{x \to 0}\frac{\ln \cos 3x}{\ln \cos (-x)}$ without L'Hospital rule 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.
 A: 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
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\left(\frac{\log \cos (3x)}{\log \cos(-x)}\right)=9}$$
A: $$\cos(3x)=\cos(x)\cdot\left(4\cos(x)^2-3\right), $$
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 $\log(1-t)\sim -t$,
$$ \lim_{x\to 0}\frac{\log(\cos(3x))}{\log\cos(-x)} = 1+2\cdot 4 = \color{red}{9}.$$
A: 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
$$\frac{\ln \cos 3x}{\ln \cos (-x)}\sim_0 \frac{-\frac{9x^2}2}{-\frac{x^2}2}=9.$$
