It is simple, and I'll bet your thinking is right, but I have a few corrections.
There's a good paper on these epsilon-delta proofs here at MHF - it's one of the four "sticky" threads in this forum (the Calculus forum).
To prove :
For the scratch work, you start with and work backwards to some function of . That tells you what you need to use for . In your case, this involves no steps (!), as you start with and it is already in the right format. So you set , which is what you said. I'm not sure why you had earlier.
Now you need to write the proof. It looks like you might have skipped this step - which would be very important on an exam.....
I'm also a little confused as to what you mean by you've only worked with numbers.
- Hollywood
what hollywood is saying is: to prove a limit exists, GIVEN epsilon, you have to somehow produce delta. you can't "start with delta".
a proper proof (which hides the "scratch work", but that's ok, if sometimes mysterious) would go something like:
let ε > 0. then for δ = ε,
we have: 0 < |x - a| < δ implies |x - a| < ε.
that is, for f(x) = x, we have:
. since this also equals f(a), this shows that the identity function f(x) = x, is continuous (which really shouldn't be that surprising...if x is near a, then ...x is near a!).