<s>[INST] Solving limit without using L'Hopital $$\lim_{x\to 0} \frac{\ln(\cos x)}{\ln(\cos 3x)}$$
The answer is $1/9$, but I don't want to use L'Hopital because I'm not supposed to.
Any help would be greatly appreciated and thanks in advance.
 [/INST] Since $\ln(x)=\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}(x-1)^n}{n!}}}$ and $\cos x=\displaystyle{\sum_{n=0}^{\infty}{\frac{(-1)^nx^{2n}}{(2n)!}}}=1-\frac{x^2}{2}+\frac{x^4}{24}+\ldots$ we can estimate $\ln(\cos x)$ and $\ln(\cos 3x)$ as follows
\begin{align}
\ln(\cos x)&=-\frac{1}{2}x^2+O(x^4)  &\text{and}  & &\ln(\cos 3x)&=-\frac{1}{2}(3x)^2+O(x^4)=-\frac{9}{2}x^2+O(x^4) \\
\end{align}
where $O(x^4)$ means that all remaining terms are of order $4$ and greater. So
\begin{align}
\frac{\ln(\cos x)}{\ln(\cos 3x)}&=\frac{-\frac{1}{2}x^2+O(x^4)}{-\frac{9}{2}x^2+O(x^4)} \\
&= \frac{-\frac{1}{2}x^2[1+O(x^2)]}{-\frac{1}{2}x^2[9+O(x^2)]}\\
&= \frac{1+O(x^2)}{9+O(x^2)}\\
\end{align}
Hence $\displaystyle{\lim_{x\rightarrow0}{\frac{\ln(\cos x)}{\ln(\cos 3x)}}=\frac{1}{9}}$.
</s>