I didn't follow parts of your work, but it seemed that you were making |x| small (<12e). The problem requires looking at x-3 small (positive). If you make x small (i.e. close to 0), then of course you'll have a hard time showing that x is also close to 3.

means

The first step is, given an , to find a that works. The second step is to show that your discovered does in fact work.

1st) Find :

As always, we'll look at the epsilon expression of the limit, and try to manipulate it into looking like , where that "something" will become the delta.

The first thing to note is that if , then implies , so that .

So, given , we'll make sure that, whatever else we decide about , we'll also have . This "gets rid of" that awkward function [x].

Want , so, since , want .

iff

iff iff .

Now is the desired form. Thus want to set . However, must also include the stipulation that . So set .

2nd) Prove that works:

Given any , let . Note that .

Then implies implies

implies implies

implies .

Also, since and , have that , so .

Thus implies implies .

This proves that, given any , if you choose by , then

implies .

That has proven that .