<s>[INST] Evaluating limits Evaluate the limit without L’Hôpital rule:
$$
\lim_{x \to 0}\frac{\sin^2{x}+2\ln\left(\cos{x}\right)}{x^4}
$$
My work is:
\begin{align}
L&=\lim_{x \to 0}\frac{\sin^2{x}-x^2}{x^4}+\lim_{x \to 0} \frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\lim_{x \to 0}\frac{\sin{x}-x}{x^3}
\lim_{x \to 0}\frac{\sin{x}+x}{x}+
\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\frac{-1}{6}\left[\lim_{x \to 0}\frac{\sin x}{x}+1\right]
+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=\frac{-1}{6}\left(2\right)+\lim \limits_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\frac{-1}{3}+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}
\end{align}
I could not evaluate the second limit
 [/INST] With Taylor expansion
$$\sin (x)=x-\frac {x^3}{6}+x^4\epsilon (x) $$
$$\sin^2 (x)=x^2-\frac {x^4}{3}+x^5\epsilon (x) $$
$$\cos (x)-1=-\frac {x^2}{2}+\frac {x^4}{24}+x^5\epsilon (x) $$
$$\ln (X+1)=X-\frac {X^2}{2}+X^2\epsilon (X) $$
$$\ln (\cos (x))=\ln \Bigl(\cos (x)-1+1\Bigr) $$
$$=-\frac {x^2}{2}+\frac {x^4}{24}-\frac {x^4}{8}+x^5\epsilon (x) $$
Replacing that, you should find

$$L=-\frac {1}{3}-\frac {1}{4}+\frac {1}{12}=-\frac {1}{2} $$

</s>