Evaluating $\lim_{x \to 0}\frac{1-\cos (1- \cos x)}{x^4}$ $$\lim_{x \to 0}\frac{1-\cos (1- \cos x)}{x^4}$$
I don't think L'Hôpital's rule is a good idea here. 
I will not finish this until the evening and it's easy to make mistake. Maybe I can expand $\cos$ in a series? 
But I don't know how to use this trick...
 A: Let's approach it elementarily:
$$\displaystyle\lim_{x \to 0}\frac{1-\cos (1- \cos x)}{x^4}=\displaystyle\lim_{x \to 0}\frac{\sin^2 (1- \cos x)}{x^4(1+\cos (1- \cos x))}=\lim_{x \to 0}\frac{\sin^2 (1- \cos x)}{2x^4}= $$
$$\lim_{x \to 0}\left( \frac{\sin (1- \cos x)}{(1-\cos x)}\right )^2 \cdot \frac{1}{2}\lim_{x \to 0}\left(\frac{1-\cos x}{x^2}\right)^{2} = \frac{1}{8}.$$  
NOTE: for the above limit i resorted to the auxiliary limit:
 $$\lim_{x\to 0} \frac{1-\cos x}{x^2} = \lim_{x\to 0} \frac{\sin^2 x}{x^2(1+\cos x)}=\frac{1}{2}.$$
The proof is complete.
A: You could try and use the fact that 
$$ \lim_{x \to 0}\frac{1-\cos x}{x^2}=\frac{1}{2} $$
This can be proved easy, using l'Hospital, or just writing $1-\cos x=2\sin^2\frac{x}{2}$.
So returning to your problem you can write your limit as
$$\lim_{x \to 0}\frac{1-\cos (1-\cos x)}{(1-\cos x)^2}\cdot \frac{(1-\cos x)^2}{x^4} $$
and use two times the limit described at the beginning of the answer.
l'Hospital also works but you'd probably have to differentiate four times until you get the result.
A: You can expand cos in a series, like you said:
$$1 - \cos\left(1 - \cos x\right) = 1 - \cos\left(1 - \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\right)\right) $$
$$= 1 - \cos\left(\frac{x^2}{2!} - \frac{x^4}{4!} + \cdots\right) $$
$$= 1 - \left[1 - \frac{1}{2!}\left(\frac{x^2}{2!} - \frac{x^4}{4!} + \cdots\right)^2 + \frac{1}{4!}\left(\frac{x^2}{2!} - \frac{x^4}{4!} + \cdots\right)^4 \cdots \right]$$
$$= \frac{1}{2!}\left(\frac{x^2}{2!} - \frac{x^4}{4!} + \cdots\right)^2 - \frac{1}{4!}\left(\frac{x^2}{2!} - \frac{x^4}{4!} + \cdots\right)^4 $$
Now since we are taking the limit as $x \to 0$ of that over $x^4$, all terms of fifth degree or higher go to $0$.  So the limit is just $\frac{1}{x^4}\frac{1}{2!}\left(\frac{x^2}{2!}\right)^2 = \frac{1}{8}$.
