First of all, I can't use limits. I need to use $0 < \epsilon < \epsilon_{0}$ method.
So when $n=1$, I get $\frac{k}{3+3k} = \frac{1}{3} - \frac{1}{3(1+k)}$
and we all know that for very large k the limit is $\frac{1}{3}$ so because $0 < \epsilon < \epsilon_{0}$, let $k = \frac{1}{\epsilon}$
Then we have $\frac{1}{3}-\frac{1}{3(\frac{1}{\epsilon}+1)} > \frac{1}{3} - \epsilon$
$-\frac{1}{3(\frac{1}{\epsilon}+1)} > - \epsilon$
$-\frac{1}{3} > -\epsilon\ (\frac{1}{\epsilon}+1)$
$\frac{1}{3} < 1 + \epsilon$
$ \epsilon > -\frac{2}{3} $ Which is alway true since $0 < \epsilon < \epsilon_{0}$
Thus $\frac{1}{3} = sup A$
So few questions:
since $k \in \mathbb{N}$ can I say $k = \frac{1}{\epsilon}$ ?
"and we all know that for very large k the limit is $\frac{1}{3}$" how can i put this is mathematical language? Or is it okey to leave it how it is?
I think the sup is $\infty$. Try dividing by just n and letting n to infinity. You are left with k / 2, which can be as large as you like.
The inf is 1/6 ? (n=k=1)