<s>[INST] 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...
 [/INST] 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 $\frac{1}{x^4}\frac{1}{2!}\left(\frac{x^2}{2!}\right)^2 = \frac{1}{8}$.
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