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 . 

An answer not using trig identities: rewrite the function as
$$
\begin{align}
&\frac{1}{x^2}\left[1-\left(1-x^2\frac{1-\cos x}{x^2}\right)\left(1-4x^2\frac{1-\cos 2x}{4x^2}\right)\left(1-9x^2\frac{1-\cos 3x}{9x^2}\right)\right]=\\
&\quad=\frac{1}{x^2}[1-(1-x^2f(x))(1-4x^2f(2x))(1-9x^2f(3x))]=\\
&\quad=f(x)+4f(2x)+9f(3x)-x^2[4f(x)f(2x)+9f(x)f(3x)+36f(2x)f(3x)]+36x^4f(x)f(2x)f(3x)
\end{align}
$$
with 
$$
f(x)=\frac{1-\cos x}{x^2}
$$ 
In the limit $x\to0$, only the first three terms survive and they give
$$
\frac{1}{2}(1+4+9)=7
$$
