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Math Help - Epsilon delta limit proof

  1. #1
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    Epsilon delta limit proof

    Need Help

    on the internet i found the following epsilon delta generic limit proof
    f(x) = Ax^2 + Bx +C as limit of x approaches p

    Delta = Min 1 or E/ /a/ +/2ap+b/


    A=3 B=4 C=5 P=2-----------------How did they come up with the part E divided by /a/ + /2ap+B/
    They continued and proved the limit but i cannot see how they got the E divided by the constants first.
    Could someone show me the algebra or the reasoning on how they got the bound this way.

    Thank You
    Del
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  2. #2
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    Re: Epsilon delta limit proof

    ok, you're leaving some things out.

    we want to prove that:

    \lim_{x \to p} (ax^2 + bx + c) = ap^2 + bp + c

    this means that for ANY ε > 0, we can find SOME δ > 0 so that:

    |x - p| < δ will guarantee that |ax2+bx+c - (ap2+bp+c)| < ε.

    first, let's see what we can do in terms of re-arranging |ax2+bx+c - (ap2+bp+c)|:

    |ax2+bx+c - (ap2+bp+c)| = |a(x2 - p2) + b(x - p)|

    = |(x - p)||a(x + p) + b|

    now, suppose we could show that: |a(x + p) + b| ≤ M, for some constant M.

    then picking δ = ε/M would give:

    |ax2+bx+c - (ap2+bp+c)| = |(x - p)||a(x + p) + b| < (ε/M)(M) = ε.

    now if one δ works, replacing it by an even SMALLER δ will work, too. so we can set an upper limit on how large we'll let δ be.

    so suppose we insist that δ ≤ 1.

    what does this mean for our x's?

    |x - p| < 1 means:

    -1 < x - p < 1, so p-1 < x < p+1.

    suppose p ≥ 0. then -(|p| + 1) = -p-1 < p-1 < x < p+1 = |p| + 1, so in this case, |x| < |p| + 1.

    on the other hand, suppose p < 0.

    then -(|p| + 1) = p-1 < x < p+1 < -p+1 = |p| + 1, so in this case as well, |x| < |p| + 1.

    and THIS means that when |x - p| < 1,

    |a(x + p) + b| = |ax + ap + b| ≤ |ax| + |ap| + |b| = |a||x| + |a||p| + |b| < |a|(|p| + 1) + |a||p| + |b| = 2|a||p| + |a| + |b|

    so that is what we choose for M (so that it serves as an upper bound for the |a(x + p) + b| factor of |f(x) - f(p)|).

    so, choosing δ = min(1,ε/(2|a||p|+|a|+|b|)) will definitely work.
    Last edited by Deveno; November 13th 2012 at 04:56 PM.
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  3. #3
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    Re: Epsilon delta limit proof

    Thanks you very much. I worked thru the algebra & found that was a correct Delta. I thought he might have a way of looking at the equation
    & coming up with the delta using those constants without the scratch work. I thought he might have a shortcut.

    DEL
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  4. #4
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    Re: Epsilon delta limit proof

    mathematicians have a habit of making themselves look smarter then they are, by omitting their "scratch work", so it looks like they pulled a rabbit out of their...erm, hat.
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