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Epsilon- Delta Proof

  1. May 20, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove that
    ## lim_{x\implies 1} \frac{2}{x-3} = -1 ##

    Use delta-epsilon.

    3. The attempt at a solution
    Proof strategy:
    ## | { \frac{ 2}{x-3} +1 } | < \epsilon ##


    ## \frac{x-1}{x-3} < \epsilon ##
    , since delta have to be a function of epsilon alone and not include x. I need to restrict delta
    ## |x-1 | < 1 \leq \delta \\ -3 < x-3 < -1 ##

    I know that there's something wrong. Help?

    What if I say that
    ## -2 = x -3 \\ 1 =x \\ \frac{ 1-1}{-2} < \epsilon \\ \delta=min(1, \epsilon)##

    Does it make any sense?
     
    Last edited: May 20, 2014
  2. jcsd
  3. May 20, 2014 #2

    pasmith

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    Homework Helper

    You don't know how small [itex]\delta[/itex] may need to be, but you can decide that it's not going to be bigger than 1. That gives you [tex]|x - 1| < \delta \leq 1.[/tex] Now you need [tex]
    \left| \frac{x - 1}{x - 3 } \right| < \frac{\delta}{|x - 3|} < \epsilon.[/tex] You can ensure that by finding a constant [itex]K > 0[/itex] such that [tex]\frac{\delta}{|x - 3|} < K\delta[/tex] when [itex]|x - 1| < \delta \leq 1[/itex], and insisting that [itex]K\delta < \epsilon[/itex].
     
  4. May 21, 2014 #3

    verty

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    You're almost there, now prove that your choice for delta works.
     
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