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 $2$.

By power series we need the following two:

$$\cos x=1-\frac{x^2}{2}+o(x^2)$$
$$(1+x)^n=1+nx+o(x)$$

Thus
$$\sqrt {\cos 2x}=\left( 1-\frac{4x^2}{2}+o(x^2)\right)^\frac12=1-x^2+o(x^2)$$
$$\sqrt[3] {\cos 3x}=\left( 1-\frac{9x^2}{2}+o(x^2)\right)^\frac13=1-\frac{3x^2}{2}+o(x^2)$$
Thus
$$\sqrt {\cos 2x}×\sqrt[3] {\cos 3x}=\left(1-x^2+o(x^2)\right) \left(1-\frac{3x^2}{2}+o(x^2)\right)=1-\frac{5x^2}{2}+o(x^2)$$
And then:
$$\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}= \frac{1-\frac{x^2}{2}-1+\frac{5x^2}{2}+o(x^2)}{x^2}=\frac{2x^2+o(x^2)}{x^2}=2+o(1)\to2$$
