<s>[INST] 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 . 
 [/INST] Alternatively:
$$\cos 3x=\cos (2x+x)=\cos 2x\cos x-\sin 2x\sin x.$$
$$L = \lim_{x \to 0}\frac{1 - \cos x \cos 2x \cos 3x}{x^2}=\lim_{x \to 0}\frac{1 - \cos x \cos 2x (\cos 2x\cos x-\sin 2x\sin x)}{x^2}=\\
\lim_{x \to 0}\frac{1 - \cos^2 x(1-\sin ^22x)+\cos x\cos 2x\sin 2x\sin x}{x^2}=\\
\lim_{x \to 0}\frac{\sin^2x + \cos^2x\sin ^22x+\cos x\cos 2x\sin 2x\sin x}{x^2}=\\
\lim_{x\to 0} \frac{\sin^2x}{x^2}+\lim_{x\to 0} \frac{4\sin^22x}{(2x)^2}+\lim_{x\to 0} \frac{2\sin 2x\sin x}{2x^2}=1+4+2=7.$$
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