# 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.