<s>[INST] How can I calculate $\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$ without L'Hôpital's rule? How can I calculate following limit without L'Hôpital's rule

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

I tried L'Hôpital's rule and I found the result $2$.
 [/INST] $$\cos x=1-\frac{x^2}2+O(x^4),$$
$$\cos nx=1-\frac{n^2x^2}2+O(x^4),$$
$$(\cos nx)^{1/n}=1-\frac{nx^2}2+O(x^4),$$
$$(\cos 2x)^{1/2}(\cos 3x)^{1/3}
=\left(1-\frac{2x^2}{2}+O(x^4)\right)\left(1-\frac{3x^2}{2}+O(x^4)\right)
=1-\frac{5x^2}{2}+O(x^4)$$
and so
$$\cos x-(\cos 2x)^{1/2}(\cos 3x)^{1/3}=2x^2+O(x^4)$$
so the limit you seek is $2$.
</s>