# How to do a delta-epsilon proof by contradiction?

1. Dec 11, 2011

### shawli

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

Prove that lim x→1 of x2 does not equal 1+10-10. You could use a proof by contradiction.

(It is question 2.b here)

2. Relevant equations

δ-ε proofs!

3. The attempt at a solution

Given ε > 0, there is some number δ > 0 such that if:

|x - a | < δ
|x - 0 | < δ
|x| < δ

Then:

| f(x) - L | < ε
| x2 - 1+10-10 | < ε

...and here's where I get stuck. Delta-epsilon proofs always seemed a bit circular to me, and what confuses me about proving "by contradiction" here is the fact that I should be able to choose some δ and the limit WOULD approach 1+10-10 :s... I'm a bit lost on where to go from here!

Last edited: Dec 11, 2011
2. Dec 11, 2011

### SammyS

Staff Emeritus
So the (sketch of a) proof would go something like this ...
Assume that the limit, L = 1+10-10.

Then pick ε small enough so that no matter how small δ is, it's impossible to find an x for which both of the following are true:
0 < |x - 1| < δ​
and
| x2 - (1+10-10) | < ε​
How small do you suppose ε must be?

3. Dec 11, 2011

### shawli

Is it a matter of picking something that is smaller than 10-10?

4. Dec 11, 2011

### SammyS

Staff Emeritus

5. Dec 11, 2011

### shawli

One half of that... so 0.5-10 (or 5-11)?
Did you choose this for some specific reason or just arbitrarily?

Another question -- If I'm choosing a specific value of ε here that isn't based on δ (so far as I can tell!), then do I still have to go through those initial steps of simplifying |f(x) - L | < ε to make the left-hand side look like the expression for delta (i.e., | x - 1| < δ )? Or am I missing the point of something here completely...

Hmm... Because to show that something isn't true, I just need one example that works to disprove it. So I really can just choose a small ε (like, an actual number that isn't in terms of δ) and show that it doesn't work once and that's it?

6. Dec 11, 2011

### SammyS

Staff Emeritus
That's the idea.