I am preparing for my exam and need help with the following task:
Let B={$\frac{m^2-n}{m^2+n^2}: n,m \in \mathbb{N}, m>n$}. Prove $supA=1$ and $infA=\frac{1}{2}$.
Well at first, I thought that we could show that $1$ is an upper bound and $\frac{1}{2}$ a lower bound. $$\frac{m^2-n}{m^2+n^2}<1 \leftrightarrow -n>n^2 \leftrightarrow-1<n$$ This is obviously true. Also $supA=1$ if for all $\epsilon>0$ there is a $m_*$ and $n_*$ such that $$\frac{m_*^2-n_*}{m_*^2+n_*^2}>1-\epsilon$$. But I don't know how to continue.
Unfortunately I also have problems proving that $infA=\frac{1}{2}$. Because if we say $$\frac{m^2-n}{m^2+n^2}>\frac{1}{2}$$ we get $$ (n+1)<\sqrt{m^2+1}$$ But what information does this give us. Then we have to show that for all $\epsilon$ there is an $m_*$ and $n_*$ such that $$\frac{m_*^2-n_*}{m_*^2+n_*^2}<\frac{1}{2}+\epsilon$$ Unfortunately I don't know how.
Is there anyone who could give me an advice? I would be very grateful.
Hints: For the supremum take $n_{*}=1$. You get $$\frac{m_*^2-n_*}{m_*^2+n_*^2}>1-\epsilon$$ if $m_{*}>\sqrt {\frac {2-\epsilon} {\epsilon}}$. [There is no loss of generality in taking $\epsilon <2$. Why?]
For the infimum first note that $m >n$ implies $m \geq n+1$. From this you can see that $\frac 1 2 $ is a lower bound. Take $m_{*}=n_{*}+1$. Now you should be able to see that the given ratio is less than $\frac 1 2 +\epsilon$ if $m_{*}$ is sufficinetly large.