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Homework Help: Rules for evaluating limits?

  1. Feb 12, 2007 #1

    danago

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    Gold Member

    Hi. We havnt started limits in class yet, but ive come across one, which i need to evaluate.

    [tex]
    \mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n}
    [/tex]

    For a limit like that, can i just substitute infinity into the equation? Or is there some rule against that? If so, would i be right in saying:

    [tex]
    \mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n} = 2\pi r
    [/tex]

    Thanks for the help.
    Dan.
     
  2. jcsd
  3. Feb 12, 2007 #2
    Well sir dango

    What your saying to me seams perfectly correct, however, limits are not my specialties. For the limit which you defined, that is the correct answer

    unique_pavadrin

    EDIT: dear sir dango what i had previously sated is incorrect, as n approaches n infinity, the function appraocd 2(pi)(r+R)
     
    Last edited: Feb 12, 2007
  4. Feb 12, 2007 #3

    Gib Z

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    Homework Helper

    [tex]
    \mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n} = 2\pi r
    [/tex]

    2pi r is a constant, take out of the limit. We have

    [tex]
    \mathop {\lim }\limits_{n \to \infty } 2nR\sin \frac{\pi }{n} = 2\pi r
    [/tex]

    Remember for small x, sin x=x? You can see that from the fact the tangent at x=0 is y=x (on the sine curve). So as n gets large, pi/n gets small.

    The n's cancel so we get that limit is equal to 2Rpi, so the solution unique_pavadrin posted is correct.
     
  5. Feb 12, 2007 #4

    danago

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    Gold Member

    Ahh thanks very much for that. That answer is exactly what im after, considering the context im using the equation in :biggrin:
     
  6. Feb 12, 2007 #5

    HallsofIvy

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    Science Advisor

    No, you have cancel the 2 [itex]2\pir[/itex] terms:
    [tex]
    \mathop {\lim }\limits_{n \to \infty } 2nR\sin \frac{\pi }{n} = 0
    [/tex][/quote]
    Now what you say further makes sense.

     
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