<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] Since in a neighbourhood of the origin $\cos(x)= 1-\frac{x^2}{2}+o(x^2)$, it follows that $\cos(2x)= 1-2x^2+o(x^2)$ and $\cos(3x)= 1-\frac{9x^2}{2}+o(x^2)$. Since $\sqrt{1+z}=1+\frac{z}{2}+o(z)$ and $\sqrt[3]{1+z}=1+\frac{z}{3}+o(z)$ we get that:
$$ \cos(x)= 1-\frac{x^2}{2}+o(x^2),\quad \sqrt{\cos(2x)}= 1-x^2+o(x^2),\quad \sqrt[3]{\cos(3x)}= 1-\frac{3x^2}{2}+o(x^2) $$
hence their product behaves like $1-\left(\frac{1}{2}+1+\frac{3}{2}\right)x^2=1-3x^2$ and the wanted limit is $\color{red}{\large 3}$.
The neglected parts in the above manipulations are $o(x^2)$, hence they do not affect the final outcome.
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