<s>[INST] Evaluate ${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$ 
Evaluate $${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$$

It should be $$\frac{1}{12}n(n+1)(2n+1)$$ but I don't know how to prove that. I am also aware that $\lim\limits_{\theta \to 0} \dfrac{1-\cos \theta}{\theta ^2}=\dfrac{1}{2}$, but I don't know how to use that.
 [/INST] This is handled in the same manner as in this answer.
Split the numerator like $$1-\cos  x+\cos x(1-\cos 2x\cos 3x\dots\cos nx) $$ and the desired limit is equal to $$\lim_{x\to 0}\frac {1-\cos x} {x^2}+\lim_{x\to 0}\cos x\cdot \frac{1-\cos 2x\dots\cos nx} {x^2}$$ which is same as $$\frac {1} {2}+\lim_{x\to 0} \frac{1-\cos 2x\dots\cos nx} {x^2}$$ Applying same technique we see that the above is equal to $$\frac{1}{2}+\frac {2^2} {2}+\lim_{x\to 0} \frac{1-\cos 3x\dots\cos nx} {x^2}$$ and continuing in same fashion we see that the desired limit is equal to $$\frac{1^2+2^2+\dots+n^2}{2}=\frac{n(n+1)(2n+1)}{12}$$
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