Delta/Epsilon (Need opinions on my solution, I am new to these)

1. Sep 30, 2011

JPanthon

1. The problem statement, all variables and given/known data

Suppose that δ = min{1,e/7}. Show that
if |x - 3| < δ, then |x^2 - 9| < e

2. Relevant equations

Don't know.

3. The attempt at a solution

|x - 3| |x + 3| < e

Given : |x - 3| < δ => |x - 3| < e/7

|x + 3| |x-3| < e/7
|x + 3| 7|x - 3| < e

Assume δ < 1, => |x - 3| < 1 => -4 < x < -2

Sub (-4)

| (-4) + 3| 7|x-3| < e
|x-3| < e/7

Is this circular? Please give some feedback!
Thank you!

2. Sep 30, 2011

SammyS

Staff Emeritus
Have an ε (epsilon).
It's not so much circular as it is helter-skelter.

Start with 0 < |x-3| < δ , where δ = min(1, ε/7), for an arbitrary ε > 0 .

Show that |x^2 - 9| < ε .

Here's a start:
You know that 0 < |x-3| < 1
So you know that -1 < x-3 < 1
From this you can eventually conclude that |x + 3| < 7
You also know that |x-3| < ε/7.
so |x-3||x + 3| < (ε/7) |x + 3| < ...

Maybe I showed you too much ???

3. Oct 1, 2011

deluks917

I never liked the standard version of the proof. I prefer to think about it like this:

say x = 3 + s. Were s is small. Then |3-x| = |s|. So we need to show:

if |s| < δ then |(3+s)^2 - 9| < ε.

|(3+s)^2 - 9| = |s2 + 6s| <= |s2| +6|s|

Okay so s is small so s2 is going to be very small, much smaller then 6s in fact but to make this rigorous lets say |s| < 1.Then we have.
|
|(3+s)^2 - 9| = |s2 + 6s| <= |s2| +|6s| <= |s| + 6|s| =7|s|

so we just need 7|s| < ε. So clearly just make |s| < ε/7. but we still need the condition |s|<1 so we say |s| < min(1,ε/7). but |3-x| = |s|. So let δ = min(1,ε/7). Which is the same answer you got.