<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] \begin{align}
\dfrac{\log(\cos(x))}{\log(\cos(3x))} & = \dfrac{2\log(\cos(x))}{2\log(\cos(3x))} \\
&= \dfrac{\log(\cos^2(x))}{\log(\cos^2(3x))}\\
& =\dfrac{\log(1-\sin^2(x))}{\log(1-\sin^2(3x))}\\
& =\dfrac{\log(1-\sin^2(x))}{\sin^2(x)} \times \dfrac{\sin^2(3x)}{\log(1-\sin^2(3x))} \times \dfrac{\sin^2(x)}{\sin^2(3x)}\\
& =\dfrac{\log(1-\sin^2(x))}{\sin^2(x)} \times \dfrac{\sin^2(3x)}{\log(1-\sin^2(3x))} \times \dfrac{\sin^2(x)}{x^2} \times \dfrac{(3x)^2}{\sin^2(3x)} \times \dfrac19
\end{align}
Now recall the following limits
$$\lim_{y \to 0} \dfrac{\log(1-y)}{y} = -1$$
$$\lim_{z \to 0} \dfrac{\sin(z)}{z} = 1$$
Also, note that as $x \to 0$, $\sin(kx) \to 0$.
Hence,
$$\lim_{x \to 0} \dfrac{\log(1-\sin^2(x))}{\sin^2(x)} = -1$$
$$\lim_{x \to 0} \dfrac{\sin^2(3x)}{\log(1-\sin^2(3x))} = -1$$
$$\lim_{x \to 0} \dfrac{\sin^2(x)}{x^2} = 1$$
$$\lim_{x \to 0} \dfrac{(3x)^2}{\sin^2(3x)} = 1$$
Hence, $$\lim_{x \to 0} \dfrac{\log(\cos(x))}{\log(\cos(3x))} = (-1) \times (-1) \times 1 \times 1 \times \dfrac19 = \dfrac19$$
Hence, the limit is $\dfrac19$.
</s>