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. 

For $x\rightarrow\infty$ your $\lim$ does not exist.
For $x\rightarrow0$ it's $$\lim_{x\rightarrow0}\frac{-8\cos^6x+8\cos^4x+2\cos^4x-2\cos^2x-\cos^2x+1}{1-\cos^2x}=$$
$$=\lim\limits_{x\rightarrow0}(8\cos^4x-2\cos^2x+1)=7$$
