Math Help - calculate the limit

1. calculate the limit

$\lim_{x\rightarrow 0}\left ( \frac{1-\textup{cos}(1-\textup{cosx})}{\textup{x}^4} \right )$

2. I would suggest using L'Hopital's rule, four times.

3. YA i know that....but is there any shorter way

4. L'Hospital's Rule is nearly always the shortest way...

5. 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!
$1/8-x^2/48-x^4/960+O(x^6)$
See:
lim &#40;&#40;1-cos&#40;1-cosx&#41;&#41;&#47;x&#94;4 as x->0 - Wolfram|Alpha

6. Originally Posted by banku12
$\lim_{x\rightarrow 0}\left ( \frac{1-\textup{cos}(1-\textup{cosx})}{\textup{x}^4} \right )$
Try using this $\sin^2\theta=\dfrac{1-\cos2\theta}{2}$

$\lim\limits_{x\to0}\frac{1-\cos(1-\cos{x})}{x^4}=\lim\limits_{x\to0}\frac{1-\cos\left(2\sin^2\frac{x}{2}\right)}{x^4}=2\lim\li mits_{x\to0}\frac{\sin^2\sin^2\frac{x}{2}}{x^4}=$

$=\frac{1}{8}\lim\limits_{x\to0}\frac{\sin^4\frac{x }{2}}{\frac{x^4}{16}}\frac{\sin^2\sin^2\frac{x}{2} }{\sin^4\frac{x}{2}}=\frac{1}{8}\cdot1^4\cdot1^2=\ frac{1}{8}.$

7. thanks