Find $\lim_{x\to0}\frac{1-\cos x\cos2x\cos3x}{x^2}$ 
Find $$\lim_{x\to0}\dfrac{1-\cos x\cos2x\cos3x}{x^2}$$

My attempt is as follows:-
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(2\cos x\cos2x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos3x+\cos x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos^23x+\cos x\cos3x\right)}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\dfrac{1+\cos6x}{2}+\dfrac{1}{2}\left(\cos4x+\cos2x\right)\right)}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{4}\left(1+\cos6x+\cos4x+\cos2x\right)}{x^2}$$
Applying L'Hospital as we have $\dfrac{0}{0}$ form
$$\lim_{x\to0}\dfrac{-\dfrac{1}{4}\left(-6\sin6x-4\sin4x-2\sin2x\right)}{2x}$$
Again applying L'Hospital as we have $\dfrac{0}{0}$ form
$$\lim_{x\to0}\dfrac{-\dfrac{1}{4}\left(-36\cos6x-16\cos4x-4\cos2x\right)}{2}=\dfrac{36+16+4}{8}=\dfrac{56}{8}=7$$
But actual answer is $3$, what am I messing up here? 
 A: Consider that
$$
1-\cos x\cos2x=1-\cos x+\cos x-\cos x\cos2x=(1-\cos x)+\cos x(1-\cos 2x)
$$
Going a further step on,
$$
1-\cos x\cos2x\cos3x=(1-\cos x)+\cos x(1-\cos2x)+\cos2x(1-\cos3x)
$$
and, by induction,
$$
1-\prod_{k=1}^n\cos(kx)=\sum_{k=1}^{n}\cos((k-1)x)(1-\cos(kx))
$$
Now
$$
\lim_{x\to0}\frac{1-\cos(kx)}{x^2}=\lim_{x\to0}k^2\,\frac{1-\cos(kx)}{(kx)^2}=\frac{k^2}{2}
$$
Therefore
$$
\lim_{x\to0}\frac{1-\prod_{k=1}^n\cos(kx)}{x^2}=\sum_{k=1}^n\frac{k^2}{2}=\frac{n(n+1)(2n+1)}{12}
$$
For $n=3$ we get
$$
\frac{3\cdot4\cdot7}{12}=7
$$

Your method in the now deleted answer is good as well, but you computed wrongly the derivative: if $f(x)=\cos x\cos2x\cos3x$, then
$$
\frac{f'(x)}{f(x)}=\frac{-\sin x}{\cos x}+\frac{-2\sin2x}{\cos2x}+\frac{-3\sin3x}{\cos3x}
$$
so
$$
f'(x)=-\cos x\cos2x\cos3x(\tan x+2\tan2x+3\tan3x)
$$
and so applying l'Hôpital yields
$$
\lim_{x\to0}\frac{\cos x\cos2x\cos3x(\tan x+2\tan2x+3\tan3x)}{2x}
$$
and you can evaluate this as
$$
\frac{1}{2}(1+2\cdot2+3\cdot3)=\frac{14}{2}=7
$$
A: Using $\cos x=1-x^2/2+O(x^4)$ gives
$\cos 2x=1-2x^2+O(x^4)$ and
$\cos 3x=1-9x^2/2+O(x^4)$. Therefore
$$\cos x\cos 2x\cos 3x=1-7x^2+O(x^4)$$
and so
$$\lim_{x\to0}\frac{1-\cos x\cos 2x\cos 3x}{x^2}=7.$$
A: My preferred way is using Taylor expansion of cosine
$$\cos{x}=1-{x^2\over 2}+o(x^3)$$
This yields
$$\begin{align}
\cos{x}\cos{2x}\cos{3x}=&\left(1-{x^2\over 2}+o(x^3)\right)\cdot\\
& \left(1-{4x^2\over 2}+o(x^3)\right)\cdot\\
&\left(1-{9x^2\over 2}+o(x^3)\right)=1-7x^2+o(x^2)
\end{align}$$
And so
$${1-\cos{x}\cos{2x}\cos{3x}\over x^2}=7+o(x)$$
This method generalises to
$$F_n(x)={1-\cos{x}\cdot\cos{2x}\cdots\cos{nx}\over x^2}$$
And we can prove that
$$\lim_{x\to 0}F_n(x)={n(n+1)(2n+1)\over 12}$$
A: Hint:
Method$\#1:$
For $x\to0,$
$$1-\cos2ax\cos2bx\cos2cx\cdots=1-(1-2\sin^2ax)(1-2\sin^2bx)(1-2\sin^2cx)\cdots$$
$$\approx2(ax)^2+2(bx)^2++2(cx)^2+\cdots+O(x^4)$$  as $\lim_{x\to0}\dfrac{\sin px}{px}=1$
Method$\#2:$
$$1-\cos2ax\cos2bx\cos2cx\cdots=\dfrac{1-(1-\sin^22ax)(1-\sin^22bx)(1-\sin^22cx)}{1+\cos2ax\cos2bx\cos2cx\cdots}$$
For $x\to0,$ the numerator $\approx(2ax)^2+(2bx)^2+(2cx)^2+\cdots+O(x^4)$
A: $$P_n=\prod_{k=1}^n\cos(kx)\implies \log(P_n)=\sum_{k=1}^n\log(\cos(kx))$$ Now, by Taylor expansion
$$\log(\cos(kx))=-\frac{k^2}{2}x^2-\frac{k^4 }{12}x^4+O\left(x^6\right)$$
$$\log(P_n)=-\frac{x^2}{2}\sum_{k=1}^n k^2-\frac{x^4 }{12}\sum_{k=1}^n k^4+\cdots$$ $$\log(P_n)=-\frac{n (n+1) (2 n+1)}{12}  x^2-\frac{n (n+1) (2 n+1) \left(3 n^2+3 n-1\right)}{360}  x^4+O\left(x^6\right)$$
$$P_n=e^{\log(P_n)}=1-\frac{n (n+1) (2 n+1) }{12} x^2+\frac{n (n+1) (2 n+1) (10 n^3+3 n^2-7 n+4)
   }{1440}x^4+O\left(x^6\right)$$
$$1-P_n=\frac{n (n+1) (2 n+1) }{12} x^2-\frac{n (n+1) (2 n+1) (10 n^3+3 n^2-7 n+4)
   }{1440}x^4+O\left(x^6\right)$$
$$\frac{1-P_n}{x^2}=\frac{n (n+1) (2 n+1) }{12} \left(1+\frac{ 10 n^3+3 n^2-7 n+4
   }{120}x^2+O\left(x^4\right) \right)$$ shows the limit and how it is approached.
