Find a limit of a function W/OUT l'Hopital's rule.

I've got an expression: $\lim_{x\to 0}$ $\frac {log(6-\frac 5{cosx})}{\sin^2 x}$

The question is: how to find limit without l'Hopital's rule?

Answer

Hint:

$$\dfrac{\ln\left(6-\dfrac5{\cos x}\right)}{\sin^2x}=\dfrac{\ln(6\cos x-5)}{\sin^2x}+\dfrac{\ln(1-\sin^2x)}{-2\sin^2x}$$

Now the second limit can be managed by $\lim_{h\to}\dfrac{\ln(1+h)}h=1$

For the first limit $6\cos x-5=6\left(1-2\sin^2\dfrac x2\right)-5=1-12\sin^2\dfrac x2$

and $\sin^2x=4\sin^2\dfrac x2\cos^2\dfrac x2$