Evaluate $\lim_{n \to \infty} \frac{2n^2 + 3}{n^2 - 7}$ directly from the $\epsilon$-definition.
So we must show that given $\epsilon > 0$, $\frac{2n^2 + 3}{n^2 - 7}$ exists with an $\epsilon$ neighbourhood of 2 for $n \geq N$.
(Sequence in $(x_n)\ in\ \mathbb{R}$ converges to $x \in \mathbb{R}$ if $\forall \epsilon > 0, \exists N \in \mathbb{N}\ \forall n \in \mathbb{N}, n \geq N \implies |x_n - x| < \epsilon$)
$|\frac{2n^2 + 3}{n^2 - 7} - 2| = |\frac{2n^2 -2n^2 + 14}{n^2 - 7}| = |\frac{14}{n^2 - 7}| $
Take $n \geq 3$, $\frac{14}{n^2 - 7} < \epsilon \implies \frac{1}{n^2 - 7} < \epsilon * 14 \implies {n^2} \geq \epsilon * 14 + 7 \implies n \geq \sqrt{\epsilon * 14 + 7} \implies \frac{1}{n} < \frac{1}{\sqrt{\epsilon * 14 + 7}}$, so by Archimedes $\exists$ such an $n$ and so the limit converges to $2$.
New to this rigorous definition of sequences and not sure my proof quite holds. Would appreciate clarification.