Epsilon-delta proof for the limit as x->1 of (3x(x-1))/(x-1)

Please only reply if you know the answer. I know the limit is 3. That's obvious, but this epsilon-delta stuff is for a test and it is important that I know. I got one response where the guy basically said "the limit's 3, so who cares?" Well I'm going to care when it shows up on my test so don't respond unless you know, otherwise people will think it's already been answered and I won't get the help I need.

Epsilon-delta proof for the limit as x->1 of (3x(x-1))/(x-1)?
I'm just starting to learn about the epsilon-delta definition of a limit and I need to know if I'm understanding this correctly.

limit as x approaches 1 of (3x(x-1))/(x-1), proving that the limit is 3.

I've been told to set this up with 0<|x-1|
And if that is true, is that sufficient proof that the limit of this function is 3? I hope that was understandable and if anyone can confirm that I'm on the right track with this I'd greatly appreciate it. Thank you.

I will show you the general proof of this limit, hopefully this will answer questions ( I will answer them more explicitly below).

Let f(x) = 3x(x-1)/ (x-1). We consider the limit lim x->1 f(x), which we expect to be 3 based on just taking f(1). We now want to prove that 3 is indeed the limit using the epsilon delta proof.

We want to prove that
∀ε >0 ∃δ>0 s.t |x-1|<δ => |f(x)-3|<ε (for every epsilon there is a delta such that if |x-1| is less than delta, that |f(x)-3| is less than epsilon)