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..
$$\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{1-\cos x \cos 2x}{x^2}-\frac {5}{2}\right)\cos2x}{x^2} $$

First of all note that $$\lim_{x\to 0} \frac{1-\cos x\cos 2x}{x^2}=\frac{5}{2}\tag{1}$$ To see why the above equation is true you can observe that the numerator in expression under limit is of the form $1-ab$ where $a, b$ tend to $1$. And we can write $1-ab$ as $$1-a+a(1-b)$$ so that limit in equation $(1)$ is equal to $$\lim_{x\to 0}\frac{1-\cos x} {x^2}+\lim_{x\to 0}\frac{1-\cos 2x}{x^2}$$ and thus the desired limit in $(1)$ is $1/2+2=5/2$.
Next we can use the same technique for the limit in question as the numerator is of the form $1-ab$ with $a,b$ tending to $1$. Thus the limit in question is equal to $$\lim_{x\to 0}\frac{1-\cos t} {x^2}+\lim_{x\to 0}\frac{1-\cos 2x}{x^2}\tag{2}$$ where $$t=\frac{1-\cos x\cos 2x}{x^2}-\frac{5}{2}\tag{3}$$ The second term in equation $(2)$ evaluates to $2$ and the first term can be written as $$\lim_{x\to 0}\frac{1-\cos t} {t^2}\cdot\left(\frac{t}{x}\right)^2$$ which is same as $$\frac{1}{2}\lim_{x\to 0}\left(\frac{t}{x}\right) ^2$$ Now we have $$\frac {t} {x} =\frac{2-2\cos x\cos 2x-5x^2}{2x^3}\tag{4}$$ and using $\cos 2x=1-2\sin^2x$ and similarly for $\cos x$ we get the numerator in $(4)$ as $$4\sin^2x+4\sin^2(x/2)-5x^2-8\sin^2x\sin^2(x/2)$$ Clearly the last term upon division with $2x^3$ tends to $0$ and hence the limit of fraction in $(4)$ is equal to the limit $$\frac{1}{4}\lim_{u\to 0}\frac{\sin^22u+\sin^2u-5u^2}{u^3}$$ via substitution $x=2u$ and this is the same as $$2\lim_{u\to 0}\frac{\sin 2u-2u}{(2u)^2}\cdot\frac{\sin 2u+2u}{2u}+\frac{1}{4}\lim_{u\to 0}\frac{\sin u-u} {u^2}\cdot\frac{\sin u+u} {u} $$ Since $(\sin u-u) /u^2\to 0$ it follows that the above limit is $0$ and hence the expression in $(4)$ namely $t/x\to 0$. The desired limit is thus $2$.
Note that the above solution is a bit lengthy as it avoids use of L'Hospital's Rule and Taylor series. 
