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

ok, you're leaving some things out.

we want to prove that:



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.

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

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.