<s>[INST] Evaluating $\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}$ I want to evaluate:

$$\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}.$$

Here's what I did. We know that as soon as ${x\to 0}$ 
$$1 - \cos x = \frac{x^2}{2} + O(x^4)$$
Therefore
$$\cos x = 1 - \frac{x^2}{2} + O(x^4)$$
Now we apply this to cosines in limit
$$\cos 2x = 1 - 2x^2 + O(x^4)$$
$$\cos 3x = 1 - \frac{9x^2}{2} + O(x^4)$$
Then use equality
$$ (1 + x)^n = 1 + xn + o(x)$$
It yields us
$$\lim_{x\to 0} \frac{1 - (1 - \frac{x^2}{2})(1 - x^2)(1 - \frac{3x^2}{2})}{x^2}$$
After simplification 1s cancel and we get $-(-1/2 - 1 - 3/2) = 3$.
 [/INST] More generally, for any integer $N\geq 1$,
$$\lim_{x\to 0}\frac{1-\prod_{n=1}^N \sqrt[n]{\cos(nx)}}{x^2}=\frac{N(N+1)}{4},$$
because
$$\prod_{n=1}^N \sqrt[n]{\cos(nx)}=\prod_{n=1}^N \sqrt[n]{1+\frac{(nx)^2}{2}+o(x^2)}=\prod_{n=1}^N \left(1+\frac{(nx)^2}{n2}+o(x^2)\right)\\
=\prod_{n=1}^N \left(1+\frac{nx^2}{2}+o(x^2)\right)=
1+\frac{x^2}{2}\sum_{n=1}^N n+o(x^2)=1+\frac{N(N+1)}{4}\cdot x^2+o(x^2).$$
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