Consider $\big\{\frac{3n}{n+4}$: $n \in \mathbb{N}\big\}$. We have to use $\varepsilon$. For $\inf(A)= \frac{3}{5}$
$$\frac{3n}{n+4} > \frac{3n}{n+4n}=\frac{3n}{5n}=\frac{3}{5}.$$
For $\sup(A)$ we have to show $n-\varepsilon$. We know $$\frac{3n}{n+4}= 3-\frac{12}{n+4}.$$ We assume $\varepsilon < \frac{12}{n+4}$.
I dont know how to finish this proof. Thank you
On
It looks like your convention is $\mathbb{N} = \{1,2,3,\dots\}$. Well, with your first steps you are proving that $\frac{3}{5}$ is a lower bound, not exactly the infimum. In order to prove the latter, you can observe that $$\frac{3}{5} = \frac{3\cdot1}{1+4}$$ so it corresponds to $n=1$. This shows $\frac{3}{5}$ is the minimum of your set, not only the infimum.
If you want to prove that $3$ is an upper bound, then your observation looks ok. To prove that it is the supremum, you should show now that for any fixed $\varepsilon > 0$ you can find $n \in \mathbb{N}$ such that $$3-\varepsilon < \frac{3n}{n+4}. $$ This gives $n > \frac{12-4\varepsilon}{\varepsilon}$ and you can conclude that $3$ is the supremum.
I don't really understand what you are doing so here is my solution:
By definition the $\sup$ is the LEAST upper bound and similarly, the $\inf$ is the LARGEST lower bound for the set.
In your case, the set is a monotonic sequence - it is easily seen that $\frac{3n}{n+4}\leq \frac{3(n+1)}{n+5}$ for $n\in \mathbb{N}$
Hence, the infimum is $\frac{3}{5}$ since it is not-bigger than any of the numbers and it is in the set itself so the $\inf$ can't be bigger than it
As for the supremum - since the sequence tends to $3$ (try to show it using $\epsilon-\delta$ argument or as in your work, say that $\frac{12}{n+4}$ converges to $0$ and use algebra of limits) and is monotonic increasing, a reasonable guess should be that $3$ indeed is the supremum. This is actually the case and you can show it using the convergence of the sequence and the fact, the supremum is the least upper bound.