Finding the limit $ \lim_{x\to 0} \frac{1-\cos x \cos(2x)}{x^2}$ I cannot find the following limit:
$$ \lim_{x\to 0} \frac{1-\cos x \cos(2x)}{x^2} \, . $$
Please, help me.

Here is a proof using Tools 1-3 below and, I think, nothing else.
Tool 1: $2\cos(a)\cos(b)=\cos(a+b)+\cos(a-b)$.
Tool 2: $1-\cos(a)=2\sin^2(\frac{a}2)$.
Tool 3: Let $u(z)=\frac{\sin(z)}z$, then $u(z)\to1$ when $z\to0$. (You said you knew that.)
Using Tools 1 and 2, one gets
$$
\frac{1-\cos x\cos 2x}{x^2}\stackrel{\text{(Tool 1)}}{=}\frac{1-\cos 3x}{2x^2}+\frac{1-\cos x}{2x^2}\stackrel{\text{(Tool 2)}}{=}\frac{\sin^2(x/2)}{x^2}+\frac{\sin^2(3x/2)}{x^2},
$$
hence
$$
\frac{1-\cos x\cos 2x}{x^2}=\frac{u^2(\frac{x}2)+9u^2(\frac{3x}2)}4.
$$
Now, Tool 3 yields $u(\frac{x}2)\to1$ and $u(\frac{3x}2)\to1$, hence the limit is $\frac{1^2+9\cdot1^2}4=\frac52$.
