Proof of a limit of a sequence I want to prove that $$\lim_{n\to\infty} \frac{2n^2+1}{n^2+3n} = 2.$$
Is the following proof valid? 
Proof 
$\left|\frac{2n^2+1}{n^2+3n} - 2\right|=\left|\frac{1-6n}{n^2+3n}\right| =\frac{6n-1}{n(n+3)} $ (because $n \in \mathbb N^+)$. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$
We have $n \ge 1 \implies 6n -1 > n + 3 \implies \frac{n+3}{n(n+3)} < \frac{6n-1}{n^2+3n}.$
Let $\epsilon > 0$ be given.
Note that $ \frac{6n-1}{n^2+3n}< \epsilon \iff \frac{n+3}{n(n+3)} < \epsilon \iff n > \frac{1}{\epsilon}.$  $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (**)$
By the Archimedean Property of $\mathbb R$, $\exists N \in \mathbb N^+$ such that $N > \frac{1}{\epsilon}.$
If $n \ge N$, then $n > \frac{1}{\epsilon}$, and from $(*)$ and $(**)$ it follows that $\left|\frac{2n^2+1}{n^2+3n} - 2\right| < \epsilon$.
Therefore $\lim_{n\to\infty} \frac{2n^2+1}{n^2+3n} = 2.$

Your “if and only if” are incorrect in
$$
\frac{6n-1}{n^2+3n}< \varepsilon \iff \frac{n+3}{n(n+3)} < \varepsilon \iff n > \frac{1}{\varepsilon}
$$
What you can do is
$$
n > \frac{6}{\varepsilon}\implies
\frac{6(n+3)}{n(n+3)}<\frac{6}{\varepsilon}\implies
\frac{6n-1}{n(n+3)}<\frac{6}{\varepsilon}
$$

Alternative proof
Rewrite your fraction as
$$
\frac{2n^2+1}{n^2+3n}=\frac{2n^2+6n-6n+1}{n^2+3n}=
2-\frac{6n-1}{n^2+3n}
$$
Now use partial fractions:
$$
\frac{6n-1}{n^2+3n}=\frac{A}{n}+\frac{B}{n+3}
$$
can be easily solved to give $A=-1/3$ and $B=19/3$. So your expression is
$$
2+\frac{1/3}{n}-\frac{19/3}{n+3}
$$
Now fix $\varepsilon>0$ and find an integer $k$ such that
$$
\frac{1/3}{k}<\frac{\varepsilon}{2},\qquad
\frac{19/3}{k+3}<\frac{\varepsilon}{2}
$$
that is,
$$
k>\max\Bigl(\frac{2}{3\varepsilon},\frac{38}{3\varepsilon}-3\Bigr)
$$
and, for $n\ge k$ you have
$$
\left|\Bigl(2+\frac{1/3}{n}-\frac{19/3}{n+3}\Bigr)-2\right|=
\left|\frac{1/3}{n}-\frac{19/3}{n+3}\right|\le
\left|\frac{1/3}{n}\right|+\left|\frac{19/3}{n+3}\right|<
\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon
$$
