Does the set $$ \left\{\frac{n}{m+n}: n,m\in\mathbb{N}\setminus\left\{0\right\}\right\} $$ have an infimum, supremum, maximum or minimum? If yes, what are the values?
My results:
1) The set does not have maximum and minimum since we always find an element within the set that is larger resp. smaller.
2) The supremum is 1.
3) The infimum is 0.
Your reasoning is correct but you need to formalize your argument.
Here is a more rigorous but not so epsilon-delta-ish way to do it:
The set is clearly bounded above by $1$ and:
$$\lim_{n \to \infty} \frac{n}{n+m} = 1$$
So the supremum is $1$. If it's attained then $n+m = n$ for some $n,m \neq 0$, which is impossible. So the maximum doesn't exist.
The set is bounded below by $0$ and:
$$\lim_{m\to \infty}\frac{n}{n+m} = 0$$
So the infimum is $0$ and it's clearly not attained. So the minimum also doesn't exist.
To do it epsilon-delta style, let $\epsilon >0$. We have to prove that we can find $n,m>0$ such that:
$$\frac{n}{n+m} > 1 - \epsilon$$
i.e.
$$\frac{m}{n+m} < \epsilon$$
Take $m=1$. By the Archimedean property there exists $n$ such that $\frac1{n+1}<\epsilon$. So we are done.
Once again for $\epsilon >0$, we have to find $n,m$ such that:
$$\frac{n}{n+m} < \epsilon$$
Repeat the above argument, swapping the roles of $n$ and $m$.