Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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..

share|improve this question

3 Answers 3

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.

share|improve this answer
    
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.

share|improve this answer

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 $\frac{1}{x^4}\frac{1}{2!}\left(\frac{x^2}{2!}\right)^2 = \frac{1}{8}$.

share|improve this answer
    
Checked Wolfram Alpha tinyurl.com/7cqjgv3 $\frac{1}{8}$ seems to be the answer you get there as well. –  Kirthi Raman Mar 14 '12 at 15:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.