$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$

Example

Show that $a_n = \frac{3n^2}{n^2-5} $ has limit equal to $3$.

We do the usual and consider $|a_n-a|$:

\begin{align*}
|a_n-3|&= \left| \frac{3n^2}{n^2-5} – 3 \right| \\
&= \left| \frac{3n^2-3n^2+15}{n^2-5} \right| \\
&= \left| \frac{15}{n^2-5} \right| \\
&= \frac{15}{\left| n^2-5 \right|} .
\end{align*}
Now for $n^2-5>0$, i.e., $n>\sqrt{5}$ we have $\left| n^2-5 \right| >n^2-5$. Thus,
\[
|a_n-a| = \frac{15}{n^2-5} {\text{ for }} n>\sqrt{5} .
\]
Now, let’s introduce $\varepsilon$ (and keep assuming $ n>\sqrt{5}$):
\begin{align*}
\frac{15}{n^2-5}\lt \varepsilon
&\iff \frac{15}{\varepsilon} \lt n^2-5 \\
&\iff \sqrt{\frac{15}{\varepsilon} +5} \lt n .
\end{align*}
We have two conditions that we need $n$ to satisfy: We want $n>\sqrt{5}$ and $n>\sqrt{\frac{15}{\varepsilon} +5}$. Thus we take
\[
N>\max \left\{ \sqrt{5} , \sqrt{\frac{15}{\varepsilon} +5} \right\} = \sqrt{\frac{15}{\varepsilon} +5}.
\]

As usual let’s rewrite everything in a polished version.
Polished version:

Show that $a_n = \frac{3n^2}{n^2-5} \to 3$.

Given $\varepsilon >0$, let $N$ be an integer such that $N> \sqrt{\frac{15}{\varepsilon} +5}$. Then for $n>N$

\begin{align*}
|a_n-3|&= \left| \frac{3n^2}{n^2-5} – 3 \right|
= \left| \frac{3n^2-3n^2+15}{n^2-5} \right| \\
&= \left| \frac{15}{n^2-5} \right|
= \frac{15}{\left| n^2-5 \right|} \\
&=\frac{15}{n^2-5} , {\text{ as }} n>\sqrt{5} ,\\
&\lt \varepsilon ,{\text{ as }} n> \sqrt{\frac{15}{\varepsilon} +5} .
\end{align*}
Hence $a_n\to 3$.

Using the Algebra of Limits there is far superior way to prove that the limit is $3$.

Remark

It should be noted that if we have found a value for $N$, then any value bigger than that will also make the important condition hold.
So, for example, if $|a_n-a|\lt \varepsilon $ is true for all $n>5$, then clearly it is true for $n>6$, $n>7$, and $n>1,000,000$. In the above example, this means that we can remove the square root symbol. This is because
\[
N>\frac{15}{\varepsilon} +5 \Longrightarrow N>\sqrt{\frac{15}{\varepsilon} +5}
\]
We can go bigger still:
\[
N>\frac{15}{\varepsilon} +15 \Longrightarrow N>\frac{15}{\varepsilon} +5
\]
and so we could use any natural number $N$ greater than
$15 \left( 1+ \frac{1}{\varepsilon} \right) $.