Evaluating $\lim_{x\to0}{\frac{x^2+2\ln(\cos x)}{x^4}}$ without l'Hopital's rule or Taylor series Can anyone please help me find this limit without l'Hopital's rule, I already used it to evaluate the limit, but I didn't know how to calculate it without l'Hopital's rule.
$$\lim_{x\to0}{\frac{x^2+2\ln(\cos x)}{x^4}}$$
Any tips will be helpful.
Sorry, but I don't want to use the Taylor series as well.


Result 1: $\displaystyle\lim_{x\to0}\dfrac{x^2 - \sin^2x}{x^4} = \frac{1}{3}$ 

Proof. Note that $\sin x = x - \frac{x^3}{3!} + o(x^5).$ 
Thus, $\sin^2x = x^2 - 2x\frac{x^3}{3!} + o(x^5).$
This, gives $x^2 - \sin^2 x = \frac{x^4}{3} + o(x^5),$ and the result follows.


Result 2: $\displaystyle\lim_{x\to0} \dfrac{\sin^4x}{x^4} = 1$

Proof. Follows trivially from $\displaystyle \lim_{x\to0} \frac{\sin x}{x} = 1.$


Result 3: $\ln(1 - x) = -x - \dfrac{x^2}{2} - \dfrac{x^3}{3} + o(x^4).$ (Expansion is valid near $0$)

Proof. Standard result. This is the Taylor expansion of $\ln(1-x)$ near $0$.

Solution.
$\displaystyle\lim_{x\to0}\dfrac{x^2 + 2\ln(\cos x)}{x^4}$
$=\displaystyle\lim_{x\to0}\dfrac{x^2 + \ln(\cos^2 x)}{x^4}$
$=\displaystyle\lim_{x\to0}\dfrac{x^2 + \ln(1 - \sin^2 x)}{x^4}$
$=\displaystyle\lim_{x\to0}\dfrac{x^2 + (-\sin^2x - \frac{\sin^4x}{2} + o(x^6))}{x^4}$
$=\displaystyle\lim_{x\to0}\dfrac{x^2 - \sin^2x}{x^4} - \dfrac{1}{2}\displaystyle\lim_{x\to0}\dfrac{\sin^4x}{x^4} + 0$
$=\dfrac{1}{3} - \dfrac{1}{2}$
$=\boxed{-\dfrac{1}{6}}$
