The Hospital Confusion :smile:

  • Thread starter eng_pro
  • Start date
7
0
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:
 

Doc Al

Mentor
44,815
1,078
The answer is supposed to be 0 and using another method it proved so!!
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.)
 

Gib Z

Homework Helper
3,344
4
Yea me too, Looks correct.
 
184
0
[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
 

HallsofIvy

Science Advisor
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In other words, WHO told you "The answer is supposed to be 0"? The limit is clearly 1.
 
7
0
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:
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YOu can see that also from the Taylor series: sin(x) = x-x^3/6+x^5/120-+-
 

dextercioby

Science Advisor
Homework Helper
Insights Author
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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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