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 proof is incorrect: you want to show that
$$\frac{6n-1}{n^2-3n}<\varepsilon$$
so what you need is
$$\frac{6n-1}{n^2-3n}<{\rm something}$$ or $6n-1<$something, that is an estimate of $6n-1$ from above, not from below.
