calculus - Find $lim limits_{xto 0}frac{logleft(cos xright)}{x^2}$ without L'Hopital

$$\lim_{x\to 0}\frac{\log\left(\cos x\right)}{x^2}$$

I've been triyng to:

1. show $\displaystyle -\frac{\pi}{2}




2. find a function so that $\displaystyle f(x)<\frac{\log\left(\cos x\right)}{x^2}$ and $\displaystyle \lim\limits_{x\to0}f(x) = -\frac{1}{2}$

And then apply the squeeze principle, but haven't managed any of these.

Answer

HINT:

$$\dfrac{\log(\cos x)}{x^2}=\dfrac{\log(\cos^2x)}{2x^2}=-\dfrac12\cdot\dfrac{\log(1-\sin^2x)}{-\sin^2x}\cdot\left(\dfrac{\sin x}x\right)^2$$