<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] Set $f_n(x)=\cos x\cos2x\cdots\cos nx$ for simplicity. We want to prove by induction that
$$
\lim_{x\to0}\frac{1-f_n(x)}{x^2}=\frac{1}{12}n(n+1)(2n+1)
$$
The statement is true for $n=1$. Assume it is true for $n$. Then
$$
f_{n+1}(x)=f_n(x)(\cos nx\cos x-\sin nx\sin x)
$$
Thus
\begin{align}
\frac{1-f_{n+1}(x)}{x^2}
&=\frac{1-f_n(x)\cos nx\cos x}{x^2}+\frac{\sin nx\sin x}{x^2}\\[6px]
&=\frac{1-f_n(x)}{x^2}\cos nx\cos x+\frac{1-\cos nx\cos x}{x^2}+\frac{\sin nx\sin x}{x^2}
\end{align}
The limit of the second summand is, by standard computations, $(n^2+1)/2$; the limit of the third summand is $n$.
Thus, by the induction hypothesis, the limit is
\begin{align}
\frac{1}{12}n(n+1)(2n+1)+\frac{n^2+1}{2}+n
&=\frac{1}{12}n(n+1)(2n+1)+\frac{(n+1)^2}{2}\\[6px]
&=\frac{n+1}{12}(2n^2+n+6n+6)\\[6px]
&=\frac{1}{12}(n+1)(n+2)(2n+3)
\end{align}
as required.
</s>