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!
 A: $\lim_{x\to0}\cos 2x=1$ not $0$ unlike $\sin x$ so we can not write $\lim_{x\rightarrow0}\frac{\ln(2\cos^{2}(x)-1)}{(2\cos^{2}(x)-1)}$ to utilize $\lim_{y\to 0}\frac{\ln(1+y)}y=1$
Instead, we can try in the following way: $$\frac{\ln(1-2\sin^2x)}{x\sin x}=(-2)\frac{\ln(1-2\sin^2x)}{(-2\sin^2x)}\frac{\sin x}x$$
A: As $x$ tends to $0$, $\cos(2x)$ tends to $1$. Hence, using $\ln(1+u) \sim u$, $\sin u \sim u$ and $1 - \cos u\sim \frac{u^2}{2}$,
$$
\frac{\ln(\cos(2x))}{x\sin x} \sim \frac{\cos(2x)-1}{x\times x} \sim \frac{-\frac{(2x)^2}{2}}{x^2} \sim - 2
$$
A: 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
$$\begin{align}\lim\limits_{x\to 0}\frac{\log(\cos 2x)}{x\sin x}&=\lim\limits_{x\to 0}\frac{\log(1-2\sin ^2 x)}{-2\sin ^2 x}\frac{-2\sin ^2 x}{x\sin x}\\&=-2\lim\limits_{x\to 0}\frac{\log(1-2\sin ^2 x)}{-2\sin ^2 x}\lim\limits_{x\to 0}\frac{\sin  x}{x}\\&=-2\lim\limits_{u\to 0}\frac{\log(1+u)}{u}\lim\limits_{x\to 0}\frac{\sin  x}{x}\\&=-2\cdot 1 \cdot 1 \\&=-2\end{align}$$
A: $$\lim_{x\to\ 0}\frac{\log\cos 2x}{x\sin x}\stackrel{\text{L'Hospital}}=\lim_{x\to 0}\frac{-2\tan 2x}{\sin x+x\cos x}{}\stackrel{\text{L'H}}=\lim_{x\to 0}-\frac{4\sec 2x}{2\cos x-x\sin x}=-\frac{4}{2}=-2$$
A: 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 $-2$.
A: $$\lim_{x\to 0}\frac{\ln(\cos(2x))}{x\sin x}=\lim_{x\to 0}\frac{\ln(\cos(2x))}{1-\cos 2x}\times\lim_{x\to 0}\frac{x}{\sin x}\times\lim_{x\to 0}\frac{1-\cos 2x }{(2x)^2}\times4=-1\times1\times2=-2$$
