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Cosine IND limit

  1. Mar 17, 2012 #1
    It's the following one:

    [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. jcsd
  3. Mar 17, 2012 #2

    Dick

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    You could use the Taylor series expansion of cos(x) around x=0.
     
  4. Mar 17, 2012 #3
    Mmm I haven't learnt Taylor series expansions yet. So anyway, could you tell me how to apply L'Hospital here? There are a lot of steps! I keep getting indeterminations. I cannot figure it out yet...
    Thanks for the reply :)
     
  5. Mar 17, 2012 #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.
     
  6. Mar 17, 2012 #5
    Sorry but I didn't understand the idea.. could you explain the first steps of the resolution? I keep getting IND 0/0 -- I know I've got to apply l'Hopitale every time I get the indtermination but there're just too many.. it never ends.
     
  7. Mar 17, 2012 #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.
     
  8. Mar 17, 2012 #7
    Yo just plug that mofo numerator equation into WolframAlpha:
    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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