The Hospital Confusion :smile:

[eng_pro]

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

 

[Doc Al]

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):
\lim_{x\rightarrow 0}\frac{\sin (x)}{x} = 1

(I don't think you are wrong.)

 

[Gib Z]

Yea me too, Looks correct.

 

[theperthvan]

\lim_{x\rightarrow 0}\frac{2\sin (\frac{1}{2} n\pi)}{n\pi} =<br /> <br /> 2*\frac{1}{2}\lim_{x\rightarrow 0}\frac{\sin (\frac{1}{2} n\pi)}{\frac{1}{2} n\pi} = 1

hope the latex works

 

[HallsofIvy]

In other words, WHO told you "The answer is supposed to be 0"? The limit is clearly 1.

 

[eng_pro]

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?

 

[robert Ihnot]

YOu can see that also from the Taylor series: sin(x) = x-x^3/6+x^5/120-+-

 

[dextercioby]

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.