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An interesting limit

  1. Oct 20, 2008 #1
    I was having a debate with a friend about how to show the following limit.

    [tex] \lim_{n \to \infty} \cos( \frac{2 \pi}{2n - 2} )^n = 1 [/tex]

    I claim that you can just hand-wavingly say that since cosine of 0 is 1, and 1^infinity is 1, the limit is 1. He claims I need to show this using some sort of limit theorem (I don't want to get into delta-epsilons).

    Is there a cool limit theorem I can use?I
     
    Last edited: Oct 20, 2008
  2. jcsd
  3. Oct 20, 2008 #2

    statdad

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    Homework Helper

    Your latex didn't compile - at least I can't see it. Could/would you repost?

    i just tried getting your code - is this your problem?

    [tex]
    \lim_{n \to \infty} \cos( \frac{2 \pi}{2n - 2} )^n = 1
    [/tex]

    okay, my version of the latex didn't take. is there a general problem with the new server setup?
     
    Last edited: Oct 20, 2008
  4. Oct 20, 2008 #3

    gel

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    No, the limit would not be 1 if you replaced the exponent by n^2 (what would it be?). You could take logarithms and apply L'hopitals rule and use cos(0)=1, cos'(0)=0.
     
  5. Oct 20, 2008 #4
    Using L'Hopitals rule I can show that \lim_{n \to \infty} \ln f(n) = 0, where f(n) is the original cosine function. If the limit of the \ln is the \ln of the limit, then I am content. Am I misunderstanding what you mean by "take logarithms"?

    Thanks for the idea!
     
  6. Oct 20, 2008 #5

    gel

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    yes. To rigorously finish off the proof you can take the exponential and use the fact that the exponential of a limit equals the limit of the exponentials - because exp is a continuous function.
     
  7. Oct 20, 2008 #6
    Excellent, I understand. Thanks.
     
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