I believe I have more knowledge about this concept, I will try to solve some problems and hopefully I actually got the message.

Prove using an $\displaystyle \epsilon - \delta$ argument that $\displaystyle \lim_{x \to 3} 2x+1 = 7$

My solution:

a = 3, L = 7, f(x) = 2x+1

$\displaystyle |f(x) - L | < \epsilon$=

$\displaystyle |(2x+1)-7|<\epsilon$=

$\displaystyle |2x-6|<\epsilon$ =

$\displaystyle 2|x-3|<\epsilon$ =

$\displaystyle |x-3|<\epsilon/2$=

$\displaystyle \delta = \epsilon/2$

After reading this document I feel like I still haven't completed the proof.

http://www.ocf.berkeley.edu/~yosenl/...ilon-delta.pdf