# I don't know how to solve this limit? Can you do it for me?

Find the limit when $x$ approaches zero of $$\lim\limits_{x \to 0}{\frac{1-\cos(1-\cos x)}{x^4}}$$

My teacher already told us that the result is $1/8$

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Got to applaud the honest approach. – Will Jagy Nov 10 '12 at 23:21

I use that $$\lim\limits_{x\to 0}\frac{1-\cos x}{x^2}=\frac 1 2$$

Now, consider the following manipulation $$\lim\limits_{x\to 0}\frac{1-\cos(1-\cos x )}{x^4}=\\ \lim\limits_{x\to 0}\frac{1-\cos(1-\cos x )}{(1-\cos x )^2}\frac {(1-\cos x )^2}{x^4}=\\ \lim\limits_{x\to 0}\frac{1-\cos(1-\cos x )}{(1-\cos x )^2}\left(\frac {1-\cos x }{x^2}\right)^2=$$

When $x\to 0$, $1-\cos x \to 0$, so

$$\lim\limits_{u\to 0}\frac{1-\cos u}{u^2}\lim\limits_{x\to 0}\left(\frac {1-\cos x }{x^2}\right)^2=\frac 1 2 \frac 1 4=\frac 1 8$$

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Nice solution.. – Berci Nov 11 '12 at 11:30

What about the L'Hospital rule? $$\frac{1-\cos(1-\cos x)}{x^4}$$ Differentiate both the nominator and the denominator: $$\big(1-\cos(1-\cos x)\big)' = \sin(1-\cos x)\cdot(1-\cos x)' = \sin(1-\cos x)\cdot\sin x$$ and so on..

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@Cameron Buie $0\over0$ seems indeterminant to me... – Daryl Nov 10 '12 at 23:39
@CameronBuie, the first person to edit the question got it wrong. Berci's version, a single fraction, is what was originally posted. – Will Jagy Nov 10 '12 at 23:45
@CameronBuie $1-\cos(1-\cos(0))=1-\cos(1-1)=1-\cos(0)=1-1=0$. The denominator is clearly $0$. Have I made an error? – Daryl Nov 10 '12 at 23:52
@Daryl, Cameron was reacting to an incorrect version of the expression that was visible for about ten minutes, until I fixed it to agree with the actual question asked. You are doing fine. – Will Jagy Nov 10 '12 at 23:57