Need help with an ε-δ limit proof

1. Sep 29, 2011

SithsNGiggles

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

Use the ε-δ definition of the limit to prove that

lim as x -> 3 of (2x - 6)/(x-3) = 2

2. Relevant equations

3. The attempt at a solution

I've started a preliminary analysis for the proof:
For any ε>0, find δ=δ(ε) such that 0 < |x - 3| < δ implies that |(2x - 6)/(x-3) - 2|< ε.
Simplifying:
|(2x - 6)/(x-3) - 2| = |(2x - 6 - 2(x - 3))/(x-3)| = |(2x - 6 - 2x + 6)/(x - 3)| = 0

Where do I go from here? Have I just proven that since the simplified form is 0, which is less than ε, and therefore proven the limit?

2. Sep 29, 2011

SammyS

Staff Emeritus
What do you get when you reduce (2x-6)/(x-3) to lowest terms?

3. Sep 29, 2011

SithsNGiggles

2. Does that mean I've proved the limit? No formal proof required?

4. Sep 29, 2011

SammyS

Staff Emeritus
No. Any δ > 0 will work, but you still need to show that for any x such that 0 < |x - 3| [STRIKE]< δ[/STRIKE] , |(2x - 6)/(x-3) - 2|< ε .

I crossed out the < δ side, because as long as (x-3) ≠ 0, (2x - 6)/(x-3) - 2 = 0 which, of course is less than ε .