# Epsilon delta limit proof

• Nov 13th 2012, 04:15 PM
Delbert777
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
• Nov 13th 2012, 04:53 PM
Deveno
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.
• Nov 14th 2012, 01:48 PM
Delbert777
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
• Nov 15th 2012, 11:24 PM
Deveno
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.