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!

The known limits you might wanna use are, for $x\to 0$
$$\frac{\log(1+x)}x\to 1$$
$$\frac{\sin x }x\to 1$$
With them, you get
$$\begin{align}\lim\limits_{x\to 0}\frac{\log(\cos 2x)}{x\sin x}&=\lim\limits_{x\to 0}\frac{\log(1-2\sin ^2 x)}{-2\sin ^2 x}\frac{-2\sin ^2 x}{x\sin x}\\&=-2\lim\limits_{x\to 0}\frac{\log(1-2\sin ^2 x)}{-2\sin ^2 x}\lim\limits_{x\to 0}\frac{\sin  x}{x}\\&=-2\lim\limits_{u\to 0}\frac{\log(1+u)}{u}\lim\limits_{x\to 0}\frac{\sin  x}{x}\\&=-2\cdot 1 \cdot 1 \\&=-2\end{align}$$
