<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] Write
$$\lim_{x\to0}\dfrac{\cos x-1+1-\sqrt{\cos 2x}+\sqrt{\cos 2x}-\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}=$$
$$\lim_{x\to0}\dfrac{\cos x-1}{x^2}+\lim_{x\to0}\dfrac{1-\sqrt{\cos 2x}}{x^2}+\lim_{x\to0}\sqrt{\cos 2x}\lim_{x\to0}\dfrac{1-\sqrt[3]{\cos 3x}}{x^2}$$
and we have
\begin{align}
&\lim_{x\to0}\dfrac{\cos x-1}{x^2}=\lim_{x\to0}\dfrac{-2\sin^2\frac{x}{2}}{x^2}=-\dfrac{1}{2} \\
&\lim_{x\to0}\dfrac{1-\sqrt{\cos 2x}}{x^2}=\lim_{x\to0}\dfrac{1-\cos 2x}{x^2(1+\sqrt{\cos 2x})}=\lim_{x\to0}\dfrac{2\sin^2x}{x^2(1+\sqrt{\cos 2x})}=1 \\
&\lim_{x\to0}\sqrt{\cos 2x}=1 \\
&\lim_{x\to0}\dfrac{1-\sqrt[3]{\cos 3x}}{x^2}=\lim_{x\to0}\dfrac{1-\cos3x}{x^2(1+\sqrt[3]{\cos 3x}+\sqrt[3]{\cos^23x})}=\dfrac{3}{2}
\end{align}
the answer is $2$.
</s>