<s>[INST] 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. 
 [/INST] Letting
$c = \cos(x)$,
$\cos x\cos2x\cos3x
=c(2c^2-1)(4c^3-3c)
=c^2(2c^2-1)(4c^2-3)
$
so
$\dfrac{1-\cos x\cos2x\cos3x}{\sin^2x}
=\dfrac{1-c^2(2c^2-1)(4c^2-3)}{1-c^2}
=8 c^4 - 2 c^2 + 1
$
(according to Wolfy).
Putting
$c^2 = 1-s^2$
(s = sin),
this becomes
$8(1-s^2)^2-2(1-s^2)+1
=8(1-2s^2+s^4)-2+2s^2+1
=8s^4-14s^2+7
$.
As $x \to \infty$,
the limit of this
does not exist
since it oscillates
from 1 to 7.
As $x \to 0$,
the limit is 7.
</s>