Calculate $\lim_{x\to 0}\frac{\ln(\cos(2x))}{x\sin x}$ Problems with calculating 
$$\lim_{x\rightarrow0}\frac{\ln(\cos(2x))}{x\sin x}$$
$$\lim_{x\rightarrow0}\frac{\ln(\cos(2x))}{x\sin x}=\lim_{x\rightarrow0}\frac{\ln(2\cos^{2}(x)-1)}{(2\cos^{2}(x)-1)}\cdot \left(\frac{\sin x}{x}\right)^{-1}\cdot\frac{(2\cos^{2}(x)-1)}{x^{2}}=0$$
Correct answer is -2. Please show where this time I've error.  Thanks in advance!

I tink if you work as follows, you will get the answer better(I hope so):
When $\alpha(x)$ is very small, then $\ln(1+\alpha(x))\sim\alpha(x)$ so $$\ln\left(\cos(2x)\right)=\ln\left(1+(-2\sin^2(x)\right)\sim -2\sin^2(x)$$
Now take your limit with this fact again. It is $-2$.
