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Homework Help: One more try at epsilon-delta limits

  1. Jul 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that lim[x->5] 3x^2 - 1 = 74

    2. Relevant equations
    Epsilon-delta stuff.

    3. The attempt at a solution
    Ok, so first we have to find delta in terms of epsilon, and this is particularly sucky because it's a nonlinear function. Here's my attempt:

    |3x^2 - 1 - 74| = |3||x - 5||x + 5| < ε → |x - 5||x + 5| < ε/3
    Define C > |x + 5| → |x + 5||x - 5| < C|x - 5|
    Define |x - 5| < ε/(3C) = δ

    Now we have to find an actual value for C, so let's put a boundary of 1 on δ:

    |x - 5| < 1 → |x + 5| < 11
    C = 11

    Now here's my attempt at the actual proof:

    Let ε > 0, δ=min{1, ε/(3C)}=min{1, ε/33}
    |3x^2 - 75| = |3||x + 5||x - 5| < 3 * 11 * ε/(3C) = 33 * ε/33 = ε
    Therefore |x - 5| < δ implies |(3x^2 - 1) - 74| < ε
    Ergo lim[x->5] 3x^2 - 1 = 74

    Does that sound about right? I have the feeling that my proof might be missing a few bits. I also don't completely understand why I have to say δ=min{1, ε/33}. I suppose there's a specific reason δ=ε/33 won't work, but I don't see it.
     
  2. jcsd
  3. Jul 9, 2012 #2

    HallsofIvy

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    Why do you "need to find" a value for C? Showing that C exists is sufficient.
     
  4. Jul 9, 2012 #3
    *scratches head* Now that you mention it, I don't know. I take it I don't need to know a value for C because it ends up as 3*|x + 5||x - 5| < 3*C*ε/C, where the C's cancel each other out?

    But, uhm, I don't really know. As I've said in a few posts before this one, I'm very unfamiliar with mathematical proofs, as most of what I've done before was of the 'just plug in numbers'-kind. I tried to do this as close to the book (Stewart's Calculus) as I could. I find it all very vague. On the one hand you need to *find* numbers, but on the other people keep talking about how you simply *define* all constants to mean I-don't-know-what.
     
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