limits - Avoid L'hopital's rule

$$\lim_{x\to 0} {\ln(\cos x)\over \sin^2x} = ?$$

I can solve this by using L'Hopital's rule but how would I do this without this?

Answer

$$\frac{\log\left(\cos\left(x\right)\right)}{\sin^{2}\left(x\right)}=\frac{1}{2}\frac{\log\left(1-\sin^{2}\left(x\right)\right)}{\sin^{2}\left(x\right)}=-\frac{\sin^{2}\left(x\right)+O\left(\sin^{4}\left(x\right)\right) }{2\sin^{2}\left(x\right)}\stackrel{x\rightarrow0}{\rightarrow}-\frac{1}{2}.$$