$\displaystyle \lim_{x\rightarrow 0}\left ( \frac{1-\textup{cos}(1-\textup{cosx})}{\textup{x}^4} \right )$
Of course, series expansion would be easier than hospital's rule many times!Indeed some complicated limits which are impossible to be solved with hospital's rule could be solved using series expansion!
$\displaystyle 1/8-x^2/48-x^4/960+O(x^6)$
See:
lim ((1-cos(1-cosx))/x^4 as x->0 - Wolfram|Alpha
Try using this $\displaystyle \sin^2\theta=\dfrac{1-\cos2\theta}{2}$
$\displaystyle \lim\limits_{x\to0}\frac{1-\cos(1-\cos{x})}{x^4}=\lim\limits_{x\to0}\frac{1-\cos\left(2\sin^2\frac{x}{2}\right)}{x^4}=2\lim\li mits_{x\to0}\frac{\sin^2\sin^2\frac{x}{2}}{x^4}=$
$\displaystyle =\frac{1}{8}\lim\limits_{x\to0}\frac{\sin^4\frac{x }{2}}{\frac{x^4}{16}}\frac{\sin^2\sin^2\frac{x}{2} }{\sin^4\frac{x}{2}}=\frac{1}{8}\cdot1^4\cdot1^2=\ frac{1}{8}.$