How to find $N(\varepsilon)$ such that $f(n)N(\varepsilon)$ Where $f(n)=\dfrac{2}{8n^2-5}$ and $n\in \mathbb{N}$, $\varepsilon\in\ \mathbb{R}^{+}$, $N:\mathbb{R}^{+}\rightarrow \mathbb{N}$.
I realise that this is related to epsilon-delta limit proofs but I can't figure out how to get the answer. 
I have attempted this by just solving for $f(n)=\varepsilon$, however I bieleve that I need to solve it using epsilon delta proof which I have no idea how to do. 
I've researched the epsilon-delta limit proofs as an attempt to understand the problem but all the example problems give you the limit and make you prove, whereas this seems very different. 
Sorry for my bad English.
Let $n \ge 3$.

$f(n)= \dfrac{2}{8n^2-5} \lt$ 

$\dfrac{2}{8n^2-n^2}\lt \dfrac{2}{2n^2} \lt 1/n.$

Let $\epsilon \gt 0$ be given.

Archimedean principle:

There is a $n_0$, positive integer, such that 

$n_0&gt; 1/\epsilon$.

For $n \ge n_0$ :

$f(n) \lt 1/n \le 1/n_0 \lt \epsilon$.
We have $0 &lt;f(n) \le \frac{1}{n^2}\le \frac{1}{n}$ for all $n$.

Can you proceed ?