<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] I don't think there is a way to evaluate the limit without using the l'Hospital rule in some form. (Using Taylor series is basically the same, because they are obtained by calculating the derivatives.)  
I would calculate the limit by applying l'Hospital once and then using known series expansions:
\begin{align}
&~~~~\lim_{x \to 0}\frac{\sin^2{x}+2\ln\left(\cos{x}\right)}{x^4}
\\
&=\lim_{x \to 0} \frac{ \sin(2x)- 2 \tan(x)}{4x^3}
\\\
&=\lim_{x \to 0} \frac{2x-\frac 8 6 x^3 + \mathcal O(x^5) - 2(x + \frac 1 3 x^3 +\mathcal O(x^5)) } {4 x^3} 
\\
&= \lim_{x \to 0}\frac{-2 x^3 + \mathcal O(x^5)}{4x^3} = -\frac 1 2
\end{align}
</s>