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 $2$.
 A: $$\cos x=1-\frac{x^2}2+O(x^4),$$
$$\cos nx=1-\frac{n^2x^2}2+O(x^4),$$
$$(\cos nx)^{1/n}=1-\frac{nx^2}2+O(x^4),$$
$$(\cos 2x)^{1/2}(\cos 3x)^{1/3}
=\left(1-\frac{2x^2}{2}+O(x^4)\right)\left(1-\frac{3x^2}{2}+O(x^4)\right)
=1-\frac{5x^2}{2}+O(x^4)$$
and so
$$\cos x-(\cos 2x)^{1/2}(\cos 3x)^{1/3}=2x^2+O(x^4)$$
so the limit you seek is $2$.
A: 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 $2$.
A: $$\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}\sqrt[3] {\cos 3x}}{x^2}=$$
$$=\lim_{x\rightarrow0}\frac{\cos^6x-\cos^32x\cos^23x}{6x^2}=$$
$$=\lim_{x\rightarrow0}\frac{\cos^2x(1-\cos^2x)(128\cos^8x-256\cos^6x+200\cos^4x-69\cos^2x+9)}{6x^2}=$$
$$=\frac{128-256+200-69+9}{6}=2.$$
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
$$\lim_{x\rightarrow0}\cos{x}=1.$$
A: 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:
$$\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}= \frac{1-\frac{x^2}{2}-1+\frac{5x^2}{2}+o(x^2)}{x^2}=\frac{2x^2+o(x^2)}{x^2}=2+o(1)\to2$$
A: 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 $\lim_{x\to0}\dfrac{\sin mx}{mx}=1$
A: 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
$$\begin{align}{\cos x-\sqrt{\cos2x}\sqrt[3]{\cos3x}\over x^2}
&={\sqrt{\cos2x}(1-\sqrt[3]{\cos3x})+(1-\sqrt{\cos2x})-(1-\cos x)\over x^2}\\\\
&=\sqrt{\cos2x}\left(1-\sqrt[3]{\cos3x}\over x^2 \right)+\left(1-\sqrt{\cos2x}\over x^2\right)-\left(1-\cos x\over x^2\right)\\\\
&\to(1)\left(3\over2\right)+\left(2\over2\right)-\left(1\over2\right)=2
\end{align}$$
