Evaluating $\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}$ I want to evaluate:

$$\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}.$$

Here's what I did. We know that as soon as ${x\to 0}$ 
$$1 - \cos x = \frac{x^2}{2} + O(x^4)$$
Therefore
$$\cos x = 1 - \frac{x^2}{2} + O(x^4)$$
Now we apply this to cosines in limit
$$\cos 2x = 1 - 2x^2 + O(x^4)$$
$$\cos 3x = 1 - \frac{9x^2}{2} + O(x^4)$$
Then use equality
$$ (1 + x)^n = 1 + xn + o(x)$$
It yields us
$$\lim_{x\to 0} \frac{1 - (1 - \frac{x^2}{2})(1 - x^2)(1 - \frac{3x^2}{2})}{x^2}$$
After simplification 1s cancel and we get $-(-1/2 - 1 - 3/2) = 3$.

The $x^2$ at the denominator suggests we can use the fact that
$$
\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2}
$$
Rewrite the numerator as
$$
(1-\sqrt{\cos2x}\sqrt[3]{\cos3x})+\sqrt{\cos2x}\sqrt[3]{\cos3x}(1-\cos x)
$$
Since
$$
\lim_{x\to0}\frac{\sqrt{\cos2x}\sqrt[3]{\cos3x}(1-\cos x)}{x^2}=
\frac{1}{2}
$$
we are reduced to compute
$$
\lim_{x\to0}\frac{1-\sqrt{\cos2x}\sqrt[3]{\cos3x}}{x^2}
$$
Let's retry the trick, rewriting the numerator as
$$
(1-\sqrt[3]{\cos3x})+\sqrt[3]{\cos3x}(1-\sqrt{\cos2x}\,)
$$
Now
$$
\lim_{x\to0}\frac{1-\sqrt{\cos2x}}{x^2}=
\lim_{x\to0}4\frac{1-\cos2x}{(2x)^2}\frac{1}{1+\sqrt{\cos2x}}=
4\cdot\frac{1}{2}\cdot\frac{1}{2}=1
$$
This leaves us with
$$
\lim_{x\to0}\frac{1-\sqrt[3]{\cos3x}}{x^2}=
\lim_{x\to0}9\frac{1-\cos3x}{(3x)^2}\frac{1}{1+\sqrt[3]{\cos3x}+\sqrt[3]{\cos^23x}}=9\cdot\frac{1}{2}\cdot\frac{1}{3}=\frac{3}{2}
$$
Thus our limit is
$$
\frac{3}{2}+1+\frac{1}{2}=3
$$
Alternative method. Define
$$
g_n(t)=\log\sqrt[n]{\cos(n\sqrt{t})}=\frac{1}{n}\log\cos(n\sqrt{t})
$$
defined in a suitable neighborhood of $0$.
For $t\ne0$, the derivative is
$$
g_n'(t)=
\frac{1}{n}
\frac{1}{\cos(n\sqrt{t})}
(-\sin(n\sqrt{t}))\frac{n}{2\sqrt{t}}
$$
and the limit at $0$ is easily seen to be $-n/2$, so by l'Hôpital the function $g_n$ is also differentiable at $0$.
The function you have to compute the limit of is even, so we can as well compute the limit from the right and do the substitution $x=\sqrt{t}$, so we have
$$
\lim_{t\to0^+}\frac{1-\cos\sqrt{t}\sqrt{\cos(2\sqrt{t})}\sqrt[3]{\cos(3\sqrt{t})}}{t}
$$
which is the negative of the derivative at $0$ of the function
$$
f(t)=\cos\sqrt{t}\sqrt{\cos(2\sqrt{t})}\sqrt[3]{\cos(3\sqrt{t})}
$$
and we can do the logarithmic derivative, so
$$
\log f(t)=g_1(t)+g_2(t)+g_3(t)
$$
and so
$$
\frac{f'(t)}{f(t)}=g_1'(t)+g_2'(t)+g_3'(t)
$$
Since $f(0)=1$, our limit is
$$
-f'(0)=-(g_1'(t)+g_2'(t)+g_3'(t))=\frac{1}{2}+\frac{2}{2}+\frac{3}{2}=3
$$
