# supremum, infimum - proving

Having given set $$A=\left(\frac{m^2-n}{m^2+n^2}: n,m \in \mathbb{N}, m>n\right)$$ the task is to find its supremum and infimum.
So I take $$f(m,n)=\frac{m^2-n}{m^2+n^2}\Rightarrow f(m,1) = \frac{m^2-1}{m^2+1}\geq \frac{m^2-n}{m^2+n^2}>f(1,n)=\frac{1-n}{1+n^2}$$ and from this I get $1\geq\frac{m^2-n}{m^2+n^2}>-\frac{1}{5}$ (by taking derrivative of left and right side of inequality and equaling it to $0$ and taking $E[1+\sqrt{2}]=2$ And now I am a bit concerned, because I can try to prove it with definitions, but how evaluate $1-\epsilon<\frac{m^2-n}{m^2+n^2}$ (supremum) in such a way, that it is permitted to use Archimedes rule? Thanks for your time and hints.

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Your English is a little hard to understand, but note that if you can show something for all $\varepsilon=1/N$ for all large $N\in\mathbb{N}$, then you've often showed something for arbitrarily small positive real $\varepsilon$. – Mark S. Nov 11 '12 at 18:42

## 1 Answer

Since $f(m,n)\lt1$ uniformly and $f(n^2,n)\to1$ when $n\to\infty$, the supremum is $1$.

Since $f(m,n)\gt\frac12$ uniformly and $f(n+1,n)\to\frac12$ when $n\to\infty$, the infimum is $\frac12$.

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You might have forgotten the condition that $m\gt n$ in your computations. – Did Nov 19 '12 at 23:20