The Hospital Confusion :smile:

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Discussion Overview

The discussion revolves around finding the limit of the equation \( \frac{2\sin(0.5n\pi)}{n\pi} \) as \( n \) approaches 0. Participants explore different methods to evaluate the limit, including L'Hôpital's rule and Taylor series expansion, while addressing conflicting interpretations of the expected result.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant claims to have used L'Hôpital's rule and obtained a limit of 1, expressing confusion over the expected answer of 0.
  • Another participant references the well-known limit \( \lim_{x\rightarrow 0}\frac{\sin (x)}{x} = 1 \) to support the claim that the limit should be 1.
  • A third participant agrees with the previous claims, suggesting that the initial calculations appear correct.
  • One participant provides a step-by-step evaluation of the limit, arriving at 1 through manipulation of the sine function.
  • Another participant questions the source of the assertion that the limit should be 0, emphasizing that the limit is clearly 1.
  • There is a mention of using the Taylor series expansion for sine as a method to analyze the limit, although details are not fully elaborated.
  • One participant raises a question about the prescript used in the initial formulation, indicating a potential misunderstanding or lack of clarity in the problem setup.

Areas of Agreement / Disagreement

Participants generally disagree on the expected limit value, with some asserting it is 1 and others suggesting it should be 0. The discussion remains unresolved regarding the correct limit.

Contextual Notes

There are unresolved assumptions regarding the methods used to evaluate the limit and the definitions applied in the context of the problem. The discussion reflects differing interpretations of the limit's behavior as \( n \) approaches 0.

eng_pro
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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:
 
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eng_pro said:
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.)
 
Yea me too, Looks correct.
 
[tex]\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[/tex]

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