4
$\begingroup$

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

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...

3 Answers 3

16

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.

  • 0
    Very nice solution2012-03-11
2

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.

1

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

  • 0
    Checked Wolfram Alpha http://tinyurl.com/7cqjgv3 $\frac{1}{8}$ seems to be the answer you get there as well.2012-03-14