<s>[INST] Evaluating $\lim_{x\to0}{\frac{x^2+2\ln(\cos x)}{x^4}}$ without l'Hopital's rule or Taylor series Can anyone please help me find this limit without l'Hopital's rule, I already used it to evaluate the limit, but I didn't know how to calculate it without l'Hopital's rule.
$$\lim_{x\to0}{\frac{x^2+2\ln(\cos x)}{x^4}}$$
Any tips will be helpful.
Sorry, but I don't want to use the Taylor series as well.
 [/INST] Hint:
Use Taylor expansion at order $4$: as
$$\cos x=1-\frac{x^2}2+\frac{x^4}{24}+o(x^4),$$
setting $u=-\dfrac{x^2}2+\dfrac{x^4}{24}+o(x^4)$, we have to expand $\ln (1+u)$ at order $2$ in $u$ and truncate the result at order $4$ (in $x$):
\begin{align}
\ln(\cos x)&=\ln(1+u)=u-\frac{u^2}2+o(u^2)=-\dfrac{x^2}2+\dfrac{x^4}{24}-\frac12\biggl(-\dfrac{x^2}2+\dfrac{x^4}{24}\biggr)^2+o(x^4)\\
&=-\dfrac{x^2}2+\dfrac{x^4}{24}-\dfrac{x^4}{8}+o(x^4)=-\dfrac{x^2}2-\dfrac{x^4}{12}+o(x^4)
\end{align}
so that the numerator is
$$x^2+2\ln(\cos x) =-\dfrac{x^4}{6}+o(x^4)\sim_0 -\dfrac{x^4}{6}.$$
</s>