Find the Limit: Finf simple limit Find \lim_{x \rightarrow 1}

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Discussion Overview

The discussion centers around finding the limit \(\lim_{x \rightarrow 1} \frac{\sin(\pi x)}{x-1}\). Participants explore various methods and substitutions to evaluate this limit, with a focus on mathematical reasoning and potential techniques applicable to the problem.

Discussion Character

  • Mathematical reasoning, Homework-related, Exploratory

Main Points Raised

  • One participant expresses difficulty in finding the limit and seeks guidance on suitable substitutions for \(x\) or expressions for \(t\).
  • Another participant suggests using L'Hopital's rule as a potential method for evaluating the limit.
  • A third participant notes that L'Hopital's rule is not covered in their current textbook section, indicating a preference for methods that do not rely on it.
  • A later reply provides a detailed manipulation of the limit expression, proposing a substitution \(t = x - 1\) and relating the limit to the known limit of \(\frac{\sin(u)}{u}\) as \(u\) approaches 0.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method to use for finding the limit, with some suggesting L'Hopital's rule while others prefer to avoid it due to their current study materials. The discussion remains unresolved regarding the best approach to take.

Contextual Notes

Participants express uncertainty about the applicability of certain mathematical techniques based on their current knowledge and textbook coverage. There are also unresolved steps in the manipulation of the limit expression.

John O' Meara
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Find [tex]\lim_{x \rightarrow 1} \frac{\sin(\pi x)}{x-1}[/tex]. I have tried to find this limit by letting [tex]t= \pi x[/tex] t= x-1, etc. All I get is 1/0 or [tex]\pi/0[/tex] or etc., but not the answer the graph of the function suggests. I cannot find the substitution for x that will work. Is there some rule that I can use to find a suitable substitution for x or expression for t. I am doing this for my own interest. Can anyone point me along the correct line of reasoning that will allow me find the expression for t= ? . Thank you.
 
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Have you tried l'hopital's rule?
 
No, the book does not mention L'Hopital's rule until page 470, I am on page 135, so I don't think it means us to use that rule.
 
[tex]\frac{sin(\pi x)}{x-1} = \frac{sin(\pi(x-1) + \pi)}{x-1} = \frac{sin(\pi(x-1))cos\pi + sin(\pi)cos(\pi(x-1))}{x-1} = \frac{-sin(\pi(x-1))}{x-1}[/tex].

Let [tex]t = x - 1[/tex]. Then [tex]\frac{-sin(\pi(x-1))}{x-1} = \frac{-sin(\pi t)}{t} = \frac{-\pi sin(\pi t)}{\pi t}[/tex].

As x goes to 1, t = x - 1 goes to 0. You know the limit of sin(u)/u as u goes to 0 right? Now use that.
 
Thanks very much.
 

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