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 $2$.

Hint:
$$\cos x-(\cos2x)^{1/2}(\cos3x)^{1/3}=1-\cos x-[1-(\cos2x)^{1/2}(\cos3x)^{1/3}]$$
Now $\lim_{x\to0}\dfrac{1-\cos x}{x^2}=\cdots=\dfrac12$
On rationalization using $a^6-b^6=(a-b)(\cdots),$
$$1-(\cos2x)^{1/2}(\cos3x)^{1/3}=\dfrac{1-\cos^32x\cos^23x}{\sum_{r=0}^5[(\cos2x)^{1/2}(\cos3x)^{1/3}]^r}$$
$\lim_{x\to0}\sum_{r=0}^5[(\cos2x)^{1/2}(\cos3x)^{1/3}]^r=\sum_{r=0}^51=?$
Finally,
$1-\cos^32x\cos^23x=1-(1-2\sin^2x)^3(1-\sin^23x)$
$\approx1-(1-2x^2)^3(1-9x^2)=x^2(6+9)+O(x^4)$
as $\lim_{x\to0}\dfrac{\sin mx}{mx}=1$
