Which of these $\varepsilon$-$\delta$ limits definitions are right hello I've been trying to understand how $\varepsilon$-$\delta$ definition is used to demonstrate limits, but i don't get it yet.
In the book [Ron Larson, Bruce H. Edwards-Calculus_ Early Transcendental Functions-Brooks Cole (2014)]
says the following:

So I've develop mi own, and i don't know if the are right, and that's why i need your help, so here we go:
$\displaystyle{\lim_{ x \rightarrow 2}}\;(3x-2)=4$
Let $\varepsilon>0$ and $\delta>0$, then
$0<|x-2|< \delta \implies |3x-2-4|<\varepsilon$
Find a value for $\delta$:
$|3x-2-4|<\varepsilon$
$\implies -\varepsilon<3x-2-4<\varepsilon$
$\implies -\varepsilon<3x-6<\varepsilon$
$\implies -\varepsilon<3(x-2)<\varepsilon$
$\implies -\dfrac{\varepsilon}{3}<x-2<\dfrac{\varepsilon}{3}$
$\implies |x-2|<\dfrac{\varepsilon}{3}$
Choose $\delta=\dfrac{\varepsilon}{3},$  proove that $\delta$ works:
$|x-2|<\dfrac{\varepsilon}{3}$
$\implies -\dfrac{\varepsilon}{3}<x-2<\dfrac{\varepsilon}{3}$
$\implies -\varepsilon<3(x-2)<\varepsilon$
$\implies -\varepsilon<3x-6<\varepsilon$
$\implies -\varepsilon<3x-2-4<\varepsilon$
$\implies |3x-2-4|<\varepsilon$
I think in the book what they are trying to say is that:
If $|x-2|<\dfrac{\varepsilon}{3}$, then 
$0<|3x-2-4|<\delta$
$0<|3x-6|<\delta$
$0<|3(x-2)|<\delta$
$0<3|x-2|<\delta$
$0<3\dfrac{\varepsilon}{3}<\delta$
$0<\varepsilon<\delta$
 A: In any case, try to make a sketch! The idea behind the $\epsilon-\delta$ definition of limits is that as $x$ approximates $x_0$ then $f(x)$ approximates $l$. To make it more "mathematics" one says:

Regardless how close it is demanded to be to $l$, I can restrict $x$ close enough to $x_0$ so as to ensure that $f(x)$ is as close as demanded.

So, let $f$ be an arbitrary function $f$:

And also, let $\epsilon>0$. So, when we choose $\epsilon$, an $\epsilon-$tight lane around $l$ is being created, as follows:

Notice the two points $A,B$ which define a part of the curve of $f$ that remains inside this lane. Then we are requested to find a $\delta>0$ such that when $x$ is inside a $\delta-$tight lane around $x_0$, then the curve of $f$ "falls" inside the rectangle defined from the two lanes. So, we should choose a lane around $x_0$ that is "inside" ($\subseteq$) the lane that $A,B$ define. To visualise it,

we should choose any lane around $x_0$ that is inside the vertical lane of the above shape.
So, in general, you should keep in mind that the $\epsilon-\delta$ definitions are much like a game:

You are given an $\epsilon>0$ - so, this cannot change, this is fixed, you cannot alter it - and you are requested to find a $\delta>0$ - this can be whatever you want, but this is not given; you have to find it and, in general, it will not be every $\delta>0$, but something more specific - such that: $$|x-x_0|<\delta\Rightarrow|f(x)-l|<\epsilon$$

Hope this helped! :)
A: Your first sequence of $\implies$'s should be a sequence of $\iff$'s. Thereby, you would prove that$$|x-2|<\frac\varepsilon3\iff|3x-2-4|<\varepsilon.$$From this, what you deduce is that, if you take $\delta=\frac\varepsilon3$, then$$|x-2|<\delta\implies|3x-2-4|<\varepsilon,$$which is what you intend to prove.
A: Your problem is when you say : Let $\varepsilon>0$ and $\delta>0$ then 
$$|x-2|<\delta\implies |3x-2-4|<\varepsilon$$
First, you can't fix $\delta>0$ and secondly, it's wrong that for all $\varepsilon>0$ and all $\delta>0$, 
$$|x-2|<\delta\implies |3x-2-4|<\varepsilon.$$

So let $\varepsilon>0$. You remark that $$|3x-2-4|=3|x-2|.$$
Therefore, if you take $\delta\leq \frac{\varepsilon}{3}$ (as you see, $\delta$ is not unspecified and it depend on $\varepsilon$ !), then 
$$|x-2|<\delta\implies |3x-2-4|=3|x-2|\leq 3\delta\leq 3\cdot \frac{\varepsilon}{3}=\varepsilon,$$
and here we are !
A: 
Let $\varepsilon>0$ and $\delta>0$, then $0<|x-2|< \delta \implies |3x-2-4|<\varepsilon$

This statement is incorrect as $\delta$ is not given. You need to find $\delta>0$ such that 
$$
0<|x-2|< \delta \implies |3x-2-4|<\varepsilon.
$$
Instead, simply start your proof with: Let $\varepsilon>0$. 
Next, the scratch work to find $\delta$ does not have to be shown in the proof. Instead jump to

Choose $\delta=\dfrac{\varepsilon}{3}$

and proceed on with showing this $\delta$ works.
