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 $2$.

This is essentially the same as MyGlasses's answer (which I overlooked at first), just presented in a different order.
First, a little preliminary:
$$\begin{align}
{1-\sqrt[n]{\cos nx}\over x^2}
&=\left(1-\sqrt[n]{\cos nx}\over1-\cos nx\right)\left(1-\cos nx\over x^2\right)\\
&=\left(1-\sqrt[n]{\cos nx}\over1-\cos nx\right)\left(1-\cos^2nx\over (nx)^2\right)\left(n^2\over1+\cos nx\right)\\
&=\left(1-\sqrt[n]{\cos nx}\over1-\cos nx\right)\left(\sin nx\over nx\right)^2\left(n^2\over1+\cos nx\right)\\
&=\left(1-u\over1-u^n\right)\left(\sin\theta\over\theta\right)^2\left(n^2\over1+\cos nx\right)\\
&=\left(1\over1+u+\cdots+u^{n-1} \right)\left(\sin\theta\over\theta\right)^2\left(n^2\over1+\cos nx\right)\\
&\to\left(1\over1+1+\cdots+1 \right)(1)^2\left(n^2\over1+1 \right)={n\over2}
\end{align}$$
where we have $u=\sqrt[n]{\cos nx}\to1$ as $\theta=nx\to0$.
Now, as in MyGlasses's answer, we have
$$\begin{align}{\cos x-\sqrt{\cos2x}\sqrt[3]{\cos3x}\over x^2}
&={\sqrt{\cos2x}(1-\sqrt[3]{\cos3x})+(1-\sqrt{\cos2x})-(1-\cos x)\over x^2}\\\\
&=\sqrt{\cos2x}\left(1-\sqrt[3]{\cos3x}\over x^2 \right)+\left(1-\sqrt{\cos2x}\over x^2\right)-\left(1-\cos x\over x^2\right)\\\\
&\to(1)\left(3\over2\right)+\left(2\over2\right)-\left(1\over2\right)=2
\end{align}$$
