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How to calculate this limit?

  1. Aug 2, 2010 #1
    Hello guys. I'm trying to create a formula here, and I got stuck at this step, where I have to calculate the following limit (see attachment). I have no idea how to do this. Any help would be appreciated. Thanks a lot.

    Attached Files:

  2. jcsd
  3. Aug 3, 2010 #2


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    Attachments require approval before being accessible, so we can't see it just yet. For future reference, it would be quicker if you uploaded a picture of it which doesn't require approval by a mod.
  4. Aug 3, 2010 #3


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    Staff: Mentor

    Simplest and fastest approach is to post the expression in Latex.
  5. Aug 3, 2010 #4
    I think this may lead to a correct answer:

    Notice that for [tex]n > 2[/tex]

    [tex] \int_0^\pi \csc^2(\frac{x}{n})-1dx = n\tan (\frac{\pi}{n})-\pi [/tex]

    and that,

    [tex] \int_{0}^{2\pi}\cos(\frac{x}{n})-1dx=n\sin(\frac{2\pi}{n})-2\pi [/tex].

    Finally, I am pretty sure that the integrands in the above converge uniformly as [tex]n\rightarrow\infty[/tex], although you should check it.


    [tex]\lim_{n\rightarrow \infty} \int_{0}^{2\pi}\cos(\frac{x}{n})-1dx = \int_{0}^{2\pi} \lim_{n\rightarrow \infty} \cos(\frac{x}{n})-1dx = \int_{0}^{2\pi} 0 dx = 0 [/tex]

    Hope that helps.
  6. Aug 3, 2010 #5
    The easiest way would probably be to use a substitution u = 1/n, then use l'Hôpital's rule.
  7. Aug 4, 2010 #6
    That substitution is particularly good because it can by done in your head and then the Taylor expansion of sin and tan can be done by inspection and the answer is clear. It's "good practice" (using both definitions of the term) to verify it with l'Hôpital's rule, however.
  8. Aug 4, 2010 #7

    Gib Z

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    We can interpret the limit geometrically as the difference in the area of a regular n-gon that circumscribes the unit circle and the area of a regular n-gon inscribed inside that circle.
  9. Aug 4, 2010 #8
    Use Taylor up to second order.
    Your limit is 0.
  10. Aug 4, 2010 #9


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  11. Aug 4, 2010 #10
    Thank you very much for all the help, you guys are great!
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