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

**Example**

Claim: The sequence $a_n=(4n+3)/n$ has limit $4$.

Note that this intuitively obvious since $a_n=4+3/n$ and $3/n\to 0$ is reasonable.

For a rigorous proof we need to use the definition (applying a definition to find something is often called **using first principles ).**

Given an $\varepsilon $ we will need to find an $N$ which depends on that $\varepsilon $.

First we simplify $|a_n-a|$ as it plays a central role in the definition. This simplification follows from a straightforward chain of equalities:

\[

|a_n-a|=|a_n-4|

= \left| \frac{4n+3}{n} -4 \right|

=\left| \frac{4n+3-4n}{n} \right|

= \left| \frac{3}{n} \right|

= \frac{3}{n} .

\]

Note that the last equality occurs because $n>0$ and hence $3/n>0$.

Now we can investigate $|a_n-a|\lt \varepsilon $:

\begin{align*}

|a_n-4| \lt \varepsilon &\iff \frac{3}{n} \lt \varepsilon \\

&\iff \frac{3}{\varepsilon } \lt n .

\end{align*}

Reading from *bottom* to *top*, we have the condition

\[

n>\frac{3}{\varepsilon } \Longrightarrow |a_n-4| \lt \varepsilon .

\]

This is almost what we want. We want a $N$ such that when $n>N$ we get $|a_n-4| \lt \varepsilon $.

Hence take $N$ to be a natural number greater than $\frac{3}{\varepsilon } $. Then if $n>N$ we obviously have $n>\frac{3}{\varepsilon } $.

Let’s recap what we have proved:

- Let $N$ be a natural number greater than $\frac{3}{\varepsilon } $.
- If $n>N$, then $n > \frac{3}{\varepsilon } $.
- If $n>\frac{3}{\varepsilon } $, then $ |a_n-4| \lt \varepsilon $.

This shows that $a_n$ has limit $4$ but we do not write the solution in this way. We rewrite it without showing where we got $N$ from. We just show that our $N$ gives us what we want. (No doubt you will have had teachers who say that you should show your working. Well, here we are not showing all our working, just the bit that proves the result!)

Here is the polished version that proves that $(4n+3)/n\to 4$. Note that we use the working from above but rearrange it into a different format.

**Polished version: ** The sequence $a_n=(4n+3)/n$ has limit $4$.

Given $\varepsilon >0 $, let $N$ be a natural number greater than $\frac{3}{\varepsilon } $. If $n>N$, then $\frac{3}{n} \lt \varepsilon $ and we have

\begin{align*}

|a_n-4| &= \left| \frac{4n+3}{n} -4 \right| \\

&= \left| \frac{4n+3-4n}{n} \right| \\

&= \left| \frac{3}{n} \right| \\

&= \frac{3}{n} \\

&\lt \varepsilon .

\end{align*}

Notice that in our working we had a string of equivalences marked by $\iff $ and that these involved $\varepsilon $. In the write up, we did not use $\varepsilon $ and the implication $\Longrightarrow $ was used but not explicitly. It is used implicitly in the logic of our argument.