Epsilon delta proof of a limit problems I am attempting to decipher the word of Stewart but I can't really understand any of the epsilon delta stuff. I watched several videos online and I have a better understanding but I still don't quite get how to do the math as no one really describes that part. As far as I understand I need to find the distance between x and delta that is less than delta and greater than zero that will coorespond to an epsilon (y).
So I have the problem
$$
\lim_{x \to 1}\frac{2+4x}{3}=2
$$
so I do some algebra magic and I get $x=1$ but from there I am not sure what to do.
 A: In order to calculate the limit using $\epsilon-\delta$ we have to prove that:
$ \lim_{x \to 1} \frac{2+4x}{3}=2 \Longleftrightarrow ( \forall \epsilon >0) (\exists \delta >0) ( \forall x) |x-1|< \delta \Longrightarrow |\frac{2+4x}{3}-2|<\epsilon$
So, choosing $ \delta= \frac{3}{4}\epsilon$ we get:
$ |\frac{2+4x}{3}-2|=\frac{4|x-1|}{3} <\frac{4 \delta}{3} =\epsilon$.
A: The construction of an $\epsilon$-$\delta$ proof is usually exactly opposite its presentation.  This should be a good example.  Given $\epsilon$, you need to be able to produce some $\delta$ such that for all $x$ with $|x-1|<\delta$, $\frac{2 + 4x}{3} - 2|<\epsilon$.
How do you do this?  You (usually) need some formula to produce $\delta$ in terms of $\epsilon$.  So start with the expression you need, 
$$\bigg|\frac{2 + 4x}{3} - 2\bigg|<\epsilon,$$
and start solving for $x$ in terms of $\epsilon$.
I'll let you work out the details here; it's a simple algebraic exercise.  At the end, you'll get something like 
$$\bigg|x - 1\bigg| < \frac{3}{4}\epsilon.$$
Ah-hah! Now you have produced a constraint on $|x-1|$ in terms of $\epsilon$.  If I give you $\epsilon$, you can pick $\delta < \frac{3}{4}\epsilon$ and, as you can quickly verify, this $\delta$ will satisfy the $\epsilon$ bound. In fact, it has to --- that's how you made it.
The take-home lesson here, again, is that the way you build the proof is backwards. Start with what you want and backsolve for what you need.  Then when you write the proof out, you know how to choose $\delta$, so you can quickly verify that $\lim_{x\to 1}\frac{2 + 4x}{3} = 2.$
