<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] $$\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}\sqrt[3] {\cos 3x}}{x^2}=$$
$$=\lim_{x\rightarrow0}\frac{\cos^6x-\cos^32x\cos^23x}{6x^2}=$$
$$=\lim_{x\rightarrow0}\frac{\cos^2x(1-\cos^2x)(128\cos^8x-256\cos^6x+200\cos^4x-69\cos^2x+9)}{6x^2}=$$
$$=\frac{128-256+200-69+9}{6}=2.$$
I used $$x^6-y^6=(x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^5+y^5),$$
$$\lim_{x\rightarrow0}\frac{\sin{x}}{x}=1$$ and
$$\lim_{x\rightarrow0}\cos{x}=1.$$
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