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
 A: 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

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

A: 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:
$$\lim_{x\to0}\frac{\sin^2 x+2\ln(\cos x)}{x^4}=
\lim_{x\to0}\frac{\sin^2 x+\ln(1-\sin^2 x)}{x^4}=
\lim_{x\to0}\frac{\sin^2 x-\sin^2 x-\frac{\sin^4 x}{2}-O(\sin^6 x)}{x^4}=
\lim_{x\to0}\frac{-\frac{\sin^4 x}{2}-O(x^6)}{x^4}=
-\frac{1}{2}.$$
A: 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}
A: Your last line is $$L=-\frac{1}{3}+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}$$ So, as other answers used, by Taylor around $x=0$, $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$\log(\cos(x))=-\frac{x^2}{2}-\frac{x^4}{12}+O\left(x^6\right)$$ $$x^2+2\ln\left(\cos{x}\right)=-\frac{x^4}{6}+O\left(x^6\right)$$ $$\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}=-\frac{1}{6}+O\left(x^2\right)$$ making $$L=-\frac{1}{3}-\frac{1}{6}=-\frac{1}{2}$$
A: 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). 
A: 
This is my solution after alot of tries 
Note:The special limit which used in the solution can be proved without niether Lospital rule nor series
