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Trouble computing this limit

  1. Sep 29, 2004 #1
    Hi i'm having trouble computing this limit
    lim n-> infinity tan(pi/n)/(n*sin^2(2/n))

    Any hints would be great
     
  2. jcsd
  3. Sep 29, 2004 #2

    Tide

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    How about using the small angle approximations to the sine and tangent (Taylor series - if you're familiar with that)? Otherwise, l'Hopital might come in handy.
     
  4. Sep 29, 2004 #3
    "How about using the small angle approximations to the sine and tangent"

    not sure about that.

    And wouldn't l'hopitals be a little nasty? I don't think it'll work out
    (tan(pi/n)/n)/sin^2(2/n)
    this has form 0/0 so I apply l'hopitals and get something real nasty that won't simplify. I'll try simplifying it sometime tonight though thanks for your help
     
  5. Sep 29, 2004 #4

    Hurkyl

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    These problems can often be done by using known limits to "replace" messy things with simple ones.

    For example, you know that the limit of tan(pi/n) / (pi/n) = 1 in this case (I hope!) For your more complicated limit, you could simplify it by pulling the tan(pi/n) off to the left and dividing it by (pi/n), then multiplying the other stuff by (pi/n).
     
  6. Sep 29, 2004 #5
    so I'd get (tan(pi/n)/(pi/n))*[(pi/n)/(n(sin^2(2/n))]

    I don't really see how that helps.

    [(pi/n)/(n(sin^2(2/n))]
    Not sure how to computer the limit of that thing.

    Any ideas?
     
  7. Sep 29, 2004 #6

    Hurkyl

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    Would you agree that, at least, [(pi/n)/(n(sin^2(2/n))] looks simpler than [tan(pi/n)/(n*sin^2(2/n))]?

    While this new expression can be simplified somewhat (for instance, pulling the 1/n in the numerator into an n on the denominator), the big thing to realize is you can keep applying the trick I mentioned...
     
  8. Sep 30, 2004 #7

    Tide

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    You might be able to see your way through l'Hopital if you replace 1/n with x and look at the limit as x -> 0.

    [tex]\lim_{x \rightarrow 0} \frac {x \tan \pi x}{\sin^2 \pi x}[/tex]
     
  9. Sep 30, 2004 #8
    thanks hurkyl and tide I was able to compute the limit both ways. You were of great help!
     
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