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. 
 A: 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
$$
\lim_{x\to0}
\left(
\frac{\cos2x\cos3x}{2\cos x}
+2\cos x\cos 3x
+3\cos x\cos2x\frac{\sin3x}{\sin2x}
\right)=
\frac{1}{2}+2+3\cdot\frac{3}{2}=7
$$
A: 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
$$\lim_{x\to 0}\left(\frac{1-\cos(x)\cos(2x)\cos(3x)}{\sin^2(x)}\right)=7$$
A: For $x\rightarrow\infty$ your $\lim$ does not exist.
For $x\rightarrow0$ it's $$\lim_{x\rightarrow0}\frac{-8\cos^6x+8\cos^4x+2\cos^4x-2\cos^2x-\cos^2x+1}{1-\cos^2x}=$$
$$=\lim\limits_{x\rightarrow0}(8\cos^4x-2\cos^2x+1)=7$$
A: Short answer:
From 
$$\cos x=1-\frac{x^2}2+o(x^3)$$ you see that the product will have the constant term $1$ and a quadratic term
$$-\frac{1+2^2+3^2}2.$$
Hence after simplifications, $7$.
A: This is not exactly an answer because you are seeking the limit when $x\to\infty $ (that part is trivial as the denominator vanishes infinitely often when $x\to\infty$). I want to highlight that the limit as $x\to 0$ can be evaluated very easily and other answers here are taking unnecessarily complicated approach. 

We can observe that if $x\to 0$ then the denominator $\sin^{2}x$ can be replaced by $x^{2}$ using the limit $(\sin x) /x\to 1$. And we can simplify denominator as $$1-\cos x+\cos x(1-\cos 2x\cos3x)$$ so the desired limit is split into two terms where first term tends to $1/2$ and the second limit is $$\lim_{x\to 0}\frac{1-\cos 2x\cos 3x}{x^{2}}$$ Applying the same technique we can see that the limit of above expression is $$2+\lim_{x\to 0}\frac{1-\cos 3x}{x^{2}}$$ The limit for the original expression is thus $1/2 +2 + 9/2=7$. There is no need to simplify the numerator as a polynomial in $\cos x$ which requires a bit of labor. 
A: Generalization:
For $$\lim_{x\to0}\dfrac{1-\cos ax\cos bx\cos cx}{\sin^2x}=\lim_{x\to0}\dfrac1{1+\cos ax\cos bx\cos cx}\cdot\lim_{x\to0}\dfrac{1-\cos^2ax\cos^2bx\cos^2cx}{\sin^2x}$$
$$=\dfrac12\cdot\lim_{x\to0}\dfrac{1-(1-\sin^2ax)(1-\sin^2bx)(1-\sin^2cx)}{\sin^2x}$$
$$=\dfrac12\cdot\lim_{x\to0}\left(\left(\dfrac{\sin ax}{\sin x}\right)^2+\left(\dfrac{\sin bx}{\sin x}\right)^2+\left(\dfrac{\sin cx}{\sin x}\right)^2+\text{terms containing multiple of }\sin^2\right)$$
$$=\dfrac{a^2+b^2+c^2}2$$
A: 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.
