1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:


    when n--->0

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

    ---------------- = 1

    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

    User Avatar

    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

    User Avatar
    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


    User Avatar
    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?

    Attached Files:

    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


    User Avatar
    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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook