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. 

The limit at $\infty$ does not exist: consider the sequence
$$
a_n=\frac{\pi}{3}+2n\pi
$$
Then the numerator evaluated at $a_n$ is
$$
1-\cos\frac{\pi}{3}\cos\frac{2\pi}{3}\cos\pi=1-\frac{1}{4}=\frac{3}{4}
$$
and the denominator is $\frac{3}{4}$, so the quotient is $1$.
On the other hand, for the sequence
$$
b_n=\frac{\pi}{6}+2n\pi
$$
the numerator is $1$ and the denominator is $1/4$. So
$$
1=\lim_{n\to\infty}\frac{1-\cos a_n\cos2a_n\cos3a_n}{\sin^2a_n}
\ne
\lim_{n\to\infty}\frac{1-\cos b_n\cos2b_n\cos3b_n}{\sin^2b_n}=4
$$
For the limit at $0$, which is probably what you have to compute, the easiest way is to use l’Hôpital: after the first run you have
$$
\lim_{x\to0}\frac{\sin x\cos2x\cos3x+2\cos x\sin 2x\cos3x+3\cos x\cos2x\sin3x}{2\sin x\cos x}
$$
that you can rewrite, using $2\sin x\cos x=\sin2x$ and that
$$
\lim_{x\to0}\frac{\sin3x}{\sin2x}=\frac{3}{2}
$$
as
$$
\lim_{x\to0}
\left(
\frac{\cos2x\cos3x}{2\cos x}
+2\cos x\cos 3x
+3\cos x\cos2x\frac{\sin3x}{\sin2x}
\right)=
\frac{1}{2}+2+3\cdot\frac{3}{2}=7
$$
