<s>[INST] How can I calculate $\lim_{x \to 0} \log(\cos(x))/\log(\cos(3x))$ without l'Hopital? How can I calculate the following limit without using, as Wolfram Alpha does, without using l'Hôpital?
$$
\lim_{x\to 0}\frac{\log\cos x}{\log\cos 3x}
$$
 [/INST] Note that since $x \to 0$ both $\cos x$ and $\cos 3x$ tend to $1$ and then using $\lim\limits_{t \to 1}\dfrac{\log t}{t - 1} = 1$ we get $$\lim_{x \to 0}\frac{\log\cos x}{\cos x - 1} = 1 = \lim_{x \to 0}\frac{\log \cos 3x}{\cos 3x - 1}$$ We can now easily see that $$\begin{aligned}L &= \lim_{x \to 0}\frac{\log \cos x}{\log \cos 3x}\\
&= \lim_{x \to 0}\frac{\log \cos x}{\cos x - 1}\cdot\frac{\cos x - 1}{\cos 3x - 1}\cdot\frac{\cos 3x - 1}{\log \cos 3x}\\
&= \lim_{x \to 0}\frac{\cos x - 1}{\cos 3x - 1}\\
&= \lim_{x \to 0}\frac{(\cos x - 1)(\cos x + 1)}{\cos x + 1}\cdot\frac{\cos 3x + 1}{(\cos 3x - 1)(\cos 3x + 1)}\\
&= \lim_{x \to 0}\frac{1 + \cos 3x}{1 + \cos x}\cdot\frac{\sin^{2}x}{\sin^{2}3x}\\
&= \lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot \frac{x^{2}}{(3x)^{2}}\cdot\frac{(3x)^{2}}{\sin^{2}3x}\\
&= \frac{1}{9}\end{aligned}$$
</s>