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