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An Epsilon-delta Proof

  1. Sep 29, 2007 #1
    Hey there everyone. I was looking at an epsilon-delta proof I did and realized that I wasn't exactly sure why one of my statements was true:

    http://img255.imageshack.us/img255/7356/proofmy5.jpg [Broken]

    On the third last line, why is it fine to assume that .5 | x - 4 | < e is true? Isn't there a chance that .5 | x - 4 | may be greater than e?

    Thanks in advance!

    Edit: Hmm, wonder why the IMG tags don't work ...
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 29, 2007 #2
    it's not okay to assume that.

    The inequality [tex]\left|\sqrt{x}-2\right|\leq \frac{1}{2} \left|\ x - 4 \right|\ [/tex] shows that if you restrict the x-values so that [tex] \left|\ x - 4\right|\ <2\epsilon [/tex] for the given [tex]\epsilon[/tex], then you get

    [tex]\left|\sqrt{x}-2\right|\leq \frac{1}{2} \left|\ x - 4 \right|\ < \frac{1}{2} (2\epsilon) =\epsilon[/tex]

    Also, the last line, 'And since . . .' is a bit weird. Remember that you're showing that such a delta exists in the first place.
    Last edited: Sep 29, 2007
  4. Sep 29, 2007 #3
    How does [tex]\left|\sqrt{x}-2\right|\leq \frac{1}{2} \left|\ x - 4 \right|\ [/tex] show that [tex] \left|\ x - 4\right|\ <2\epsilon [/tex] without assuming [tex]\frac{1}{2} \left|\ x - 4 \right|\ < \epsilon [/tex] first? Sorry. I don't see how the first inequality connects with it being less than epsilon. Thanks for the help!

    Oh and yeah, I typed this up haphazardly without thinking what I meant by the last line. Thanks again!
  5. Sep 29, 2007 #4
    remember that you fixed [tex]\epsilon[/tex] as a positive number.

    the inequality shows that IF you make [tex]\left|\ x - 4\right|\ < \epsilon[/tex] , THEN you get [tex]\left|\sqrt{x}-2\right|\<\epsilon[/tex].

    Read over the definition of a limit carefully and see how it applies to this particular problem:

    We say that [tex]\lim_{x\to\\a}f(x)=v [/tex] whenever,

    for all [tex] \epsilon > 0 [/tex], there is some [tex]\delta >0[/tex] such that,

    whenever we have [tex]0<\left|\ x-a \right|\ < \delta[/tex], it follows that [tex]\left|\ f(x) - v\right|\ < \epsilon [/tex]
  6. Sep 29, 2007 #5


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    The reason that last line seems a bit "peculiar" is that it is really a peculiarity of the way we do proofs of limits.

    The definition of limit requires that we show that, for a specific [itex]\delta[/itex], if [itex]|x-a|< \delta[/itex], then [itex]|f(x)- L|< \epsilon[/itex]. But what we do is work the other way- assuming a value of [itex]\epsilon[/itex], we calculate the necessary [itex]\delta[/itex]. The point is that every step is "reversible"- you could start from, in this case, |x-4|< [itex]\delta[/itex] and, by just going through the derivation "in reverse" arrive at [itex]|\sqrt{x}- 2|< \epsilon[/itex].

    That's sometimes referred to as "synthetic" proof. Again, we go from the conclusion we want to the hypothesis- but it is crucial that every step be "reversible"!
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