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Finf simple limit

  1. Oct 9, 2009 #1
    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.
  2. jcsd
  3. Oct 9, 2009 #2
    Have you tried l'hopital's rule?
  4. Oct 9, 2009 #3
    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.
  5. Oct 9, 2009 #4
    [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.
  6. Oct 9, 2009 #5
    Thanks very much.
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