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

We evaluate the limit in question as follows
\begin{align}
L&=\lim_{x\to 0}\frac{\sin^{2}x+2\log\cos x}{x^{4}}\notag\\
&=\lim_{x\to 0} \frac{\log(1-\sin^{2}) +\sin^{2}x}{\sin^{4}x}\cdot\frac{\sin^{4}x}{x^{4}}\notag\\
&=\lim_{t\to 0^{+}}\frac{\log(1-t)+t}{t^{2}}\text{ (putting } t=\sin^{2}x)\notag\\
&=-\frac{1}{2}\text{ (via Taylor series or L'Hospital's Rule)} \notag
\end{align}
I have left the last step (which is almost routine). 
