<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] Note that by product to sum formula
$$\cos x \cos 3x = \frac12 (\cos 2x+\cos 4x)\\\implies \cos x \cos 2x \cos 3x = \frac12 (\cos^2 2x+\cos 2x \cos 4x)=\frac12(2\cos^3 2x+\cos^2 2x - \cos 2x )$$
then
$$\frac{1 - \cos x \cos 2x \cos 3x}{x^2}=
\frac{1 - \cos^3 2x-\frac12\cos^2 2x + \frac12\cos 2x }{x^2}=\\\frac{\frac12(1-\cos 2x)(2\cos^2 2x+3\cos 2x+2)}{x^2}=\\2\frac{1-\cos 2x}{4x^2}(2\cos^2 2x+3\cos 2x+2)\to 1 \cdot 7=7$$
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