The fact is you do not choose the epsilon. That is given at the start of the process.
Your task is to a delta that makes it all work out.
So you control delta, in this case the size of $\displaystyle |x-3|.$
Look at$\displaystyle \left| {\frac{{2x + 4}}{5} - 2} \right| = \left| {\frac{{2x - 6}}{5}} \right| = \frac{2}{5}\left| {x - 3} \right|$.
Ah, there is that factor you control. To get $\displaystyle \frac{2}{5}\left| {x - 3} \right|<\varepsilon$ we pick $\displaystyle \delta=\frac{5\varepsilon}{2}$
Right - you're given $\displaystyle \epsilon$ and you need to find a $\displaystyle \delta$ (which will be a function of $\displaystyle \epsilon$) that works. So you're doing the algebra backwards. Then when you write your proof, doing the algebra forward, it works. There's an excellent paper on how to do these proofs right here on the MHF Calculus forum - it's one of the "sticky" threads at the top.
- Hollywood