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The Hospital Confusion :smile:

  1. Jan 6, 2007 #1
    Hey, I have this problem where i am trying to find the limit of the equation:

    2sin(0.5*n*pi)
    ---------------
    n*pi

    when n--->0


    I used the l-hospital rule to solve it and got

    pi*cos(0.5*n*pi)
    ---------------- = 1
    pi


    The answer is supposed to be 0 and using another method it proved so!!
    So what am i doing wrong :cry:
     
  2. jcsd
  3. Jan 6, 2007 #2

    Doc Al

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    Staff: Mentor

    I'd like to see that proof, since it is well-known that (when x is in radians):
    [tex]\lim_{x\rightarrow 0}\frac{\sin (x)}{x} = 1[/tex]

    (I don't think you are wrong.)
     
  4. Jan 6, 2007 #3

    Gib Z

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    Homework Helper

    Yea me too, Looks correct.
     
  5. Jan 7, 2007 #4
    [tex] \lim_{x\rightarrow 0}\frac{2\sin (\frac{1}{2} n\pi)}{n\pi} =

    2*\frac{1}{2}\lim_{x\rightarrow 0}\frac{\sin (\frac{1}{2} n\pi)}{\frac{1}{2} n\pi} = 1 [/tex]

    hope the latex works
     
  6. Jan 7, 2007 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    In other words, WHO told you "The answer is supposed to be 0"? The limit is clearly 1.
     
  7. Jan 7, 2007 #6
    can someone tell me how to post the question form a world document as i formulated the question there!! or how can i paste formulas here?
     

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    Last edited: Jan 7, 2007
  8. Jan 12, 2007 #7
    YOu can see that also from the Taylor series: sin(x) = x-x^3/6+x^5/120-+-
     
  9. Jan 12, 2007 #8

    dextercioby

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    Science Advisor
    Homework Helper

    I don't see what prescript you used to get to C_{n} in the first place. And why you think it should work to give you C_{0} in the same manner.

    Daniel.
     
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