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[tex]\displaystyle\lim_{x \to{0}}{\frac{1-\cos(1-\cos x)}{3x^4}}[/tex]

In case we have to apply L'Hospital, appart from it, how could I solve this without it?

Thanks!

- Thread starter Hernaner28
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- #1

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[tex]\displaystyle\lim_{x \to{0}}{\frac{1-\cos(1-\cos x)}{3x^4}}[/tex]

In case we have to apply L'Hospital, appart from it, how could I solve this without it?

Thanks!

- #2

Dick

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You could use the Taylor series expansion of cos(x) around x=0.

- #3

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Thanks for the reply :)

- #4

Dick

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If you don't have Taylor series yet, then you'll probably want to stick with l'Hopital. But the idea is to use cos(x)=1-x^2/2!+ terms of higher order in x. It does make things easier.

Thanks for the reply :)

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- #6

Dick

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l'Hopital will end at the fourth derivative. It has to. Then the denominator becomes a constant. It is a little hard to keep track of the numerator, I will admit.

- #7

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http://www.wolframalpha.com/input/?i=fourth+derivative+of+1-cos(1-cosx)

I agree, it's a nasty numerator, but you can just plug in x = 0 now. Looking at it real quick, and it's looks like the numerator at 0 equals 3, so the limit is 3.

Edit: Oh wait, the limit wouldn't be 3, it would 3/(3*4*3*2*1) = 1/24

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