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.
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 .
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
Also, since and , have that , so .
Thus implies implies .
This proves that, given any , if you choose by , then
That has proven that .