I have $A=\left\{{|k-n| \over k^2+n}: k,n \in \mathbb{N}, k \neq n\right\}$. I want to conclude about $\text{sup}A$ and $\text{inf}A$.
Edit
First, we notice that the terms of the original expression are bound from below by $0$. Then, we take the subsequence for $n=1$, ${k-1 \over k^2+1} \to 0$. So we have a subsequence that converges to a lower bound. This means $\text{inf}A=0$.
For the supremum, we use the triangle inequality: $${|k+(-n)| \over k^2+n} \le {|k| + |-n| \over k^2+n}={k + n \over k^2+n} < 1$$ So $1$ is an upper bound. We choose the subsequence for $k=1, {|1 - n| \over 1 + n}=1$ and notice that for large $n$s the numerator is $n - 1$. Then ${n - 1 \over 1 + n} \to 1$, $\text{sup}A=1.$
Hint:
Take $k=1$ and $n=2$ to see that $\frac15$ is not the supremum.
Then take $k=1,n=100$, and then $k=1, n=1000$, to get an idea about what the supremum might be.