How do I complete this $\epsilon - N$ proof that $\lim_{n\to \infty}\frac {2n-1}{3n+2}=\frac {2}{3}$?

Question

The main difficulty I'm having is the point where I have a negative part under the absolute value;

$$\left|\frac {2n-1}{3n+2}-\frac {2}{3}\right|<\epsilon$$ $$\left|\frac {3(2n-1)-2(3n+2)}{3(3n+2)}\right|<\epsilon$$ $$\left|\frac {6n-3-6n-4}{3(3n+2)}\right|=\left|\frac {-7}{3(3n+2)}\right|<\epsilon$$

How do I proceed from this point? I know that for a general absolute value problem, I'd have to consider two cases; when $n\geq 0$ and $n \lt 0$. But since this is a sequence, $n \ge 1$. Is it acceptable to take $\mathbf *$ $$\left|\frac {-7}{3(3n+2)}\right|= \frac {\left|{-7}\right|}{3(3n+2)}<\epsilon$$ since the denominator is positive?

Then, proceeding this way to get $$\frac {7}{3(3n+2)}<\frac {7}{9n}<\epsilon$$ So that we can pick $N=\frac{9}{7\epsilon}$? It seems to me that I'm done, but I'd like to confirm if my treatment of the absolute value in $\mathbf *$ is correct. That is my main question. But if there are any other errors, please point them out.

Answer 1

We may assume $n > 0$ so

$\left|\frac {-7}{3(3n+2)}\right|<\epsilon \iff $

$\frac {3(3n+2)}7 > \frac 1\epsilon \iff$

$9n + 2 > \frac {7}\epsilon \iff$

$n > \frac 7{9\epsilon} - \frac 29$.

So if $M > \frac 7{9\epsilon} - \frac 29$, then $n > M\implies \left|\frac {2n-1}{3n+2}-\frac {2}{3}\right|<\epsilon$