A limit involving $\cos x$ and $x^2$ The question is finding the value of $L = \lim_{x \to 0}\frac{1 - \cos x \cos 2x \cos 3x}{x^2}$ if it exists . I've found the answer using taylor's formula but I'm looking for other solutions maybe using trigonometric identities . 

Observe you have
\begin{align}
\cos x \cos 2x \cos 3x= \cos x(\cos 5x+\sin 2x \sin 3x)
\end{align}
and
\begin{align}
\cos x \cos 5x =\cos 6x+ \sin x\sin 5x
\end{align}
which means
\begin{align}
\frac{1-\cos x \cos 2x \cos 3x}{x^2} =&\ \frac{1-\cos 6x}{x^2}-\frac{\sin x\sin 5x}{x^2}-\frac{\cos x \sin 2x\sin 3x}{x^2}\\
=&\ 18\frac{\sin^2 3x}{(3x)^2}-5\frac{\sin x\sin 5x}{x(5x)}-6\frac{\cos x \sin 2x\sin 3x}{(2x)(3x)}
\end{align}
In the limit, we have
\begin{align}
18-5 - 6= 7
\end{align}
