<s>[INST] 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 $\lim_{n\to\infty} \frac{2n^2+1}{n^2+3n} = 2.$
 [/INST] From $\lim_{n\to\infty} \dfrac{2n^2+1}{n^2+3n}$, divide top and bottom by $n^2$ to get:
$\lim_{n\to\infty} \dfrac{2+1/n^2}{1+3/n}$, which, as $n\to\infty$ becomes
$\dfrac{2+0}{1+0}=\dfrac{2}{1}=2$.
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