I was doing the proof to show that given $\epsilon > 0$, there exist an $n_ 0 \in \mathbb{N}$ such that etc.. And I chose my $n_0 > \max\lbrace2,\sqrt{(13+ \epsilon)/\epsilon} \rbrace$. But, is this correct?
I have that given an $\epsilon > 0$, choose $n_0 > \max\lbrace2,\sqrt{(13+ \epsilon)/\epsilon} \rbrace$. Then for all $n \ge n_0$, we have
$$\bigg|\frac{3n^2+5}{2n^2-1} - \frac{3}{2}\bigg| = \bigg| \frac{13}{2(2n^2-1)}\bigg| < \frac{13}{n^2-1}$$
but I am little confused for what to do next... What is the method?
And if not, how should I go about choosing my $n_0$ for situations similar to this?
Trusting your computations, you have $$ \bigg|\frac{3n^2+5}{2n^2-1} - \frac{3}{2}\bigg| = \bigg| \frac{13}{2(2n^2-1)}\bigg| \le \frac{13}{n^2} < \epsilon $$ with the last equality being true as soon as $$ n > n_\epsilon := \sqrt\frac{13}\epsilon $$ This is not the most accurate bound, but this is quick to find.