Help trying to prove the existence of sequence limit. So I'm trying to prove that
$$\lim_{n \to +\infty} \frac{2n - 4}{3n - 7} = \frac{2}{3}$$
using the formal definition of the limit of a sequence. Applying it I've got 
$$\left|\frac{2n - 4}{3n - 7} - \frac{2}{3}\right| \lt \varepsilon,$$
$$\frac{2}{\left|9n-21\right|} \lt \varepsilon.$$
From here I don't know how to continue, I guess I have to get ripped of the absolute value but I don't know how should I do it. Thank you so much.
 A: Now, we need $$|3n-7|>\frac{2}{3\epsilon},$$ which gives that it's enough to take
$$n>\frac{7}{3}+\frac{2}{9\epsilon}.$$
A: You should be more clear in what you want to show.
By definition you have to show that for every $\varepsilon > 0$ it exists $N\in\mathbb{N}$ such that for every $n\geq N$ it is $|x_n-a|<\varepsilon$.
So let $\varepsilon >0$ be arbitrary.
Then we have to find for that given $\varepsilon$ our $N$ such that the inequality holds.
$\left|\frac{2n - 4}{3n - 7} - \frac{2}{3}\right| <\varepsilon$
We get to the expression
$\left|\frac{2}{9n-21}\right|<\varepsilon$
For $n\geq 3$ it is $9n-21>0$
So $\frac{2}{9n-21}<\varepsilon$
Now we have to find $n$ by simple calculations:
$$\frac{2}{9n-21}<\varepsilon $$ $$\Leftrightarrow 9n-21>\frac2\varepsilon$$
$$\Leftrightarrow n > \frac{\frac{2}{\varepsilon}+21}{9}=\frac2{9\varepsilon}+\frac73$$
Now choose $n=\lceil \frac2{9\varepsilon}+\frac73\rceil$ (because $n$ has to be a natural number) and conclude the proof.
For example. If $\varepsilon=\frac{1}{100}$ then we could choose $N=25$ and for every $n\geq N$ would the inequality hold.
It does not matter if the inequality also holds for smaller $n$ as long as we can find some $N$ where the inequality holds for every larger $n$.
A: We need to find $$M>n \text{ such that }\bigg\vert\frac{2n-3}{3n-7}-\frac{2}{3}\bigg\vert=\frac{2}{|9n-21|}<\epsilon$$
$$\implies3|3n-7|>\frac{2}{\epsilon} \implies |3n-7|>\frac{2}{3\epsilon} \implies -\frac{2}{9\epsilon}+\frac73>n \text{ or } n>\frac{2}{9\epsilon}+\frac73$$
A: Almost done:
Let $n >21$;
$0 < \dfrac{2}{9n-21} < \dfrac{2}{ 9n -n}= \dfrac{1}{4n}.$
Let $\epsilon >0$ be given.
Archimedean principle:
There is a $n_0 \in \mathbb{Z^+}$ s.t.
$n_0 > 1/(4\epsilon)$.
For $n \ge n_0$
$0 <\dfrac{2}{9n -21} <\dfrac{1}{4n} \le \dfrac{1}{4n_0} <\epsilon$.
