<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] Or we could've approached it differently from the beginning, if we observe that
$$2\ln(\cos x)=\ln(\cos^2 x)=\ln(1-\sin^2 x)$$
and use the series for
$$\ln(1-t)=-\sum_{n=1}^{\infty}\frac{t^n}{n}=-t-\frac{t^2}{2}-\frac{t^3}{3}-\cdots.$$
Then for the original limit:
$$\lim_{x\to0}\frac{\sin^2 x+2\ln(\cos x)}{x^4}=
\lim_{x\to0}\frac{\sin^2 x+\ln(1-\sin^2 x)}{x^4}=
\lim_{x\to0}\frac{\sin^2 x-\sin^2 x-\frac{\sin^4 x}{2}-O(\sin^6 x)}{x^4}=
\lim_{x\to0}\frac{-\frac{\sin^4 x}{2}-O(x^6)}{x^4}=
-\frac{1}{2}.$$
</s>