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Limit of cosine question

  1. Feb 20, 2014 #1
    1. The problem statement, all variables and given/known data

    What would be the limit of {cos(m!*pi*x}^{2n} as 'n' and 'm' go to infinity along with proof? Also x is real. I have no clue as to how to start the question. Please help! Thanks.



    3. The attempt at a solution

    I tried to write the original function as e^[(ln(cos(m!*pi*x)/(1/2n)], but since not both the numerator and denominator go to zero, I couldn't use L'Hopital's Rule.

    Other method I tried was using the power series expansion of cosine, but I am not sure how to deal with m!

    Any hints or help will be greatly appreciated. Thanks!
     
  2. jcsd
  3. Feb 20, 2014 #2

    vela

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    Consider the fact that cosine is bounded between -1 and 1.
     
  4. Feb 20, 2014 #3
    Actually I should clarify the question and mu doubt:

    the actual question is Evaluate lim m→infinity [lim n→ infinity ((cos (m!*pi*x))^2n) ].

    So if we went with -1<= cos (m!*pi*x)<= 1, take powers of 2n on all sides and take limit as n goes to infinity, we get the limit of the original function as 1. Is that correct? I am not sure how to deal with the part where m goes to infinity.
     
  5. Feb 20, 2014 #4

    vela

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    Are you sure that you're supposed to take the limit as m tends to infinity last? The order will make a difference in how the problem works out.

    No, that's not correct. Suppose m=1 and x=1/2. What would be the limit as n→∞? How about if x=1/3 instead?
     
  6. Feb 20, 2014 #5
    I see your point. The limit varies as we vary m and x, but I am still stuck. So do we just say that there is no limit because the cosine function keeps oscillating between -1 and 1? I still have no clue how to deal with the m!
     
  7. Feb 20, 2014 #6

    vela

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    Like I said, the order of the limits matters. When you take the limit as n tends to infinity, you're doing so for a fixed m. And then you take the result of that and see what happens when m goes to infinity.
     
  8. Feb 20, 2014 #7
    Could you please show me how you would do the problem? I have tried a variety of things like converting cosine squared into sine squared, or using e^ln form, but nothing seems to work. I am sorry, I am completely blank.
     
  9. Feb 20, 2014 #8

    vela

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    I don't think you did actually, because I gave you the wrong examples. :redface:

    Say m=1. What's the limit equal to when x=1 as n→∞? What's the limit equal to when x=1/3? You should be able to convince yourself there are only two possible outcomes for the inner limit. Which one you get depends on m and x.
     
  10. Feb 20, 2014 #9
    Okay let's do it the way you suggest. When m=1, x=1, then cos(pi)^2n=1 as n→∞. When m=1, x=1/3, then cos(pi/3)^2n= (sqrt3/ 2)^infinity = 0. So the two possible outcomes are zero and 1. So since m is an integer, it all comes down to x. If x=integer, then the limit is 1 and if x=/ an integer, then the limit is 0. Am I right yet?
     
  11. Feb 20, 2014 #10
    Of course, if m=x, then we will have the limit as 1.
     
  12. Feb 20, 2014 #11

    vela

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    What if x=1/12? Is the limit always 0 regardless of the value of m?
     
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