Find this limit. Compute the value of the limit :
$$
\lim_{x\to\infty}{\frac{1-\cos x\cos2x\cos3x}{\sin^2x}}
$$
I've tried simplifying the expression to 
$$
\lim_{x\to\infty}\frac{-8\cos^6x+10\cos^4x-3\cos^2x+1}{\sin^2x}
$$
But I don't know what to do after this. 

Using the identities $\displaystyle \sin^2(x)=\frac{1-\cos(2x)}{2}$, $\displaystyle \cos(4x)=2\cos^2(2x)-1$, and $\displaystyle \cos(x)\cos(3x)=\frac12(\cos(2x)+\cos(4x))$, we obtain
$$\begin{align}
\frac{1-\cos(x)\cos(2x)\cos(3x)}{\sin^2(x)}&=\frac{1-\frac{\cos(2x)\left(\cos(2x)+\overbrace{(2\cos^2(2x)-1)}^{=\cos(4x)}\right)}{2}}{\frac{1-\cos(2x)}{2}}\\\\
&=\frac{-2\cos^3(2x)-\cos^2(2x)+\cos(2x)+2}{1-\cos(2x)}\\\\
&=2\cos^2(2x)+3\cos(2x)+2
\end{align}$$
whence taking the limit as $x\to 0$ yields the coveted result
$$\lim_{x\to 0}\left(\frac{1-\cos(x)\cos(2x)\cos(3x)}{\sin^2(x)}\right)=7$$
