$\frac{1}{n}$ does not converge to $0$ in $\mathbb{R}(t)$?

Question

Let $F = \mathbb{R}(t)$ be a field of rational functions in an indeterminate $t$. Set $ε = \frac{1}{t}$.

I have to show that $0 < ε < \frac{1}{n}$ for all $n ∈ \mathbb{N}$ and that $\frac{1}{n}$ does not tend to the limit $0$ in $F$.

Could the reason for $ε < \frac{1}{n}$ for any $n$ be, that $t$ is an upper bound for $\mathbb{N} \subset \mathbb{R}$ in $\mathbb{R}(t)$ and thus no matter what $n$ we choose, $ε < \frac{1}{n}$ for all sufficiently large $t$?

And that $\frac{1}{n}$ does not converge to 0, is it because taking the formal definition of a limit, we can always find sufficiently large $t$ such that $|\frac{1}{n} – L| > ε$?

All in all I would like to ask for a hint at how to approach the question.

Answer 1

You have\begin{align}\varepsilon<\frac1n&\iff\frac1t<\frac1n\\&\iff n<t\\&\iff t-n>0,\end{align}which is true, by the definition of the order relation in $\Bbb R(t)$.

And you don't have $\lim_{n\to\infty}\frac1n=0$ precisely because there is a $\varepsilon>0$ such that $(\forall n\in\Bbb N):\frac1n>\varepsilon$.