<s>[INST] How can I solve this crazy limit? $\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{1-\cos x \cos 2x}{x^2}-\frac {5}{2}\right)\cos2x}{x^2} $ Firstly, I think this can be done with equivalent infinitesimal, but it seems so much complicated. I'm not very brave to do L'Hospital's rule on this question. And I don't think trig formulas can simplify this..
$$\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{1-\cos x \cos 2x}{x^2}-\frac {5}{2}\right)\cos2x}{x^2} $$
 [/INST] Note that $\cos(x)\cos(2x)= \frac 1 2 (\cos(x) + \cos(3x))$. Hence 
$$\frac{1-\cos(x)\cos(2x)}{x^2}-5/2= \frac{1-\cos(x)}{2x^2}+ \frac {1-\cos(3x)}{x^2}-5/2\\
=\frac {x^2/2+O(x^4)}{2x^2}+ \frac {9x^2/2+O(x^4)}{2x^2}-\frac 5 2=\frac 5 2 +O(x^2)-\frac 5 2=O(x^2). $$
Also observe that 
$$ \cos(x^2)\cos(2x)= (1-\frac{x^4}{2}+O(x^6))(1-2x^2+O(x^3))=1-2x^2+O(x^3).$$
Therefore we know  the nominator  up to second order. But this is enough to calculate the limit:
$$\frac{1-\cos(\frac{1-\cos(x)\cos(2x)}{x^2}-5/2)\cos(2x)}{x^2} \\
=\frac{1-\cos(O(x^4))\cos(2x)}{x^2}\\ = \frac{1-[1-2x^2+O(x^3)]}{x^2}=2 $$
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