Say we have a set $F=\left\{4+\frac{2}{n^3}:n\in\mathbb{N}\right\}$
We assume $F$ is bounded below by $4$, since
$4\lt4+\frac{2}{n^3}$ for $n\in\mathbb{N}$
We now show $m=4$ is the greatest lower bound of $F$ by proving that if $m'\gt4$ there is an element $4+\frac{2}{n^3}$ in $F$ such that $4+\frac{2}{n^3}\lt m'$. If $m'\gt4$
$$ \begin{align} 4+\frac{2}{n^3}\lt m'&\Leftrightarrow \frac{2}{n^3}\lt m'-4\\ &\Leftrightarrow\frac{n^3}{2} \gt \frac{1}{m'-4}\\ &\Leftrightarrow n^3 \gt \frac{2}{m'-4}\\ &\Leftrightarrow n \gt \sqrt[3]{\frac{2}{m'-4}} \end{align} $$
If we take a positive integer $n$ that $n\gt \sqrt[3]{\frac{2}{m'-4}}$, this gives a number $4+\frac{2}{n^3}$ in $F$ such that $4+\frac{2}{n^3} \lt m'$.
Therefore $4$ is the greatest lowest bound of $F$
Am I right in my conclusion?
If $m'>4$, since $\frac{2}{n^3}\rightarrow 0$, there exists $N$ such that $\forall n\geq N$, $$ 0< \frac{2}{n^3} \leq \frac{m'-4}{2}$$
In particular $m'>4+\frac{2}{N^3}\in F$, so we can't have $m'>4$ and $4$ is the greatest lower bound of $F$.
No calculus is needed...
But if you really want to : $$ \begin{align} 4+\frac{2}{n^3}\lt m'&\Leftrightarrow \frac{2}{n^3}\lt m'-4\\ &\Leftrightarrow \frac{2}{m'-4}\lt n^3 \\ &\Leftrightarrow n \gt \sqrt[3]{\frac{2}{m'-4}} \end{align} $$