The Hospital Confusion :smile:

1. Jan 6, 2007

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

2. Jan 6, 2007

Staff: Mentor

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.)

3. Jan 6, 2007

Gib Z

Yea me too, Looks correct.

4. Jan 7, 2007

theperthvan

$$\lim_{x\rightarrow 0}\frac{2\sin (\frac{1}{2} n\pi)}{n\pi} = 2*\frac{1}{2}\lim_{x\rightarrow 0}\frac{\sin (\frac{1}{2} n\pi)}{\frac{1}{2} n\pi} = 1$$

hope the latex works

5. Jan 7, 2007

HallsofIvy

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

6. Jan 7, 2007

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?

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Last edited: Jan 7, 2007
7. Jan 12, 2007

robert Ihnot

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

8. Jan 12, 2007

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.

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