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$$\displaystyle\lim_{x \to 0}\frac{1-\cos (1- \cos x)}{x^4}$$

I don't think L'Hôpital's rule is a good idea.. I will not finish this till the evening and it's easy to make mistake.. Maybe expand cos in a series? But I don't know how to use this trick..

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up vote 14 down vote accepted

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.

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thank you very much :-) – xan Mar 11 '12 at 12:27
Very nice solution – Geoff Robinson Mar 11 '12 at 14:06

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.

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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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Checked Wolfram Alpha $\frac{1}{8}$ seems to be the answer you get there as well. – Kirthi Raman Mar 14 '12 at 15:52

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