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Epsilon-Delta proofs, once again

  1. May 9, 2008 #1
    I'm trying to understand [tex]\epsilon-\delta[/tex] proofs, but I'm having some trouble. For example, if we want to prove that [tex]\lim_{x\rightarrow2}x^3=8[/tex], starting from [tex]|x^3-8|[/tex] we get to something like

    [tex]|x-2||x^2+2x+4|[/tex]

    And this is what confuses me: we conjecture that [tex]|x-2|<1[/tex], then [tex]|x|<3[/tex], so we get

    [tex]|x-2||x^2+2x+4| \leq |x-2|||x|^2+2|x|+4|<|x-2||3^2+2*3+4|=19|x-2|[/tex] then we can easily find out what [tex]\delta[/tex] is. But, why do we make that assumption when we're working with every [tex]\epsilon>0[/tex]? That might not be true for some [tex]\epsilon[/tex] ([tex]|x-1|<\delta<1[/tex]. I might be missing something big here, thank you in advance.
     
  2. jcsd
  3. May 9, 2008 #2
    I think I sorted this out. We're taking [tex]\delta<1[/tex], so it doesn't matter since we only have to find one [tex]\delta[/tex] to prove that the limit holds.
     
  4. May 9, 2008 #3

    HallsofIvy

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    ?? It is not relevant what [itex]\epsilon[/itex] is, you are requiring that [itex]\delta[/itex] be less that 1. That, you are free to choose. As long as there is some [itex]\delta[/itex] works, then any smaller [itex]\delta[/itex] also works. That's why we can require that [itex]\delta[/itex] be less than 1.

    In this particular case, what we would have to do is choose [itex]\delta[/itex] less than the smaller of [itex]\epsilon/19[/itex] and 1.
     
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