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

Let’s try a more complicated example.

**Example**

Show that $a_n=\frac{2n+3}{n^2}$ has limit $0$.

We proceed as before by simplifying $|a_n-a|$ via equalities:

\[

|a_n-0|=\left| \frac{2n+3}{n^2} – 0 \right|

= \left| \frac{2n+3}{n^2} \right|

= \frac{2n+3}{n^2} .

\]

Now what we want is to find $N$ such that

\[

\frac{2n+3}{n^2} <\varepsilon {\text{ for all }} n>N.

\]

It is possible but messy to make $n$ the subject of this inequality so that we have $n$ greater than something. However, we shall adopt a simpler method by allowing ourselves to use $\leq $ and $<$ as follows:
\begin{align*}
\frac{2n+3}{n^2} &\leq \frac{2n+3n}{n^2}, {\text{ as obviously }} 3\leq 3n {\text{ for }} n\geq 1,\\
&=\frac{5n}{n^2}\\
&=\frac{5}{n} .
\end{align*}
This is much simpler. It doesn't have a quadratic term in $n$ and it is easy to find $N$ such that $n>N$ implies that $5/n\lt \varepsilon $, we just need $N>5/\varepsilon $.

Let’s put everything together in a coherent argument.

**Polished version: **

Show that $a_n=\frac{2n+3}{n^2}$ has limit $0$.

Answer: Given $\varepsilon >0$, let $N$ be an integer greater than $5/\varepsilon$. Then for all $n>N$ we have $n>5/\varepsilon $, i.e., $\varepsilon >5/n$ and hence

\[

|a_n-0|=\left| \frac{2n+3}{n^2} – 0 \right|

= \frac{2n+3}{n^2}

\leq \frac{2n+3n}{n^2}

=\frac{5n}{n^2}

=\frac{5}{n}

\lt \varepsilon .

\]

Notice that the written answer is short and the important choice, that of $N$, appears to come from nowhere. The reader does not need to know the working to find it. For the working we ensure that $|a_n-a|$ is less than, or less than or equal to, some simple expression.