Let $\displaystyle A= \left\{\frac{m^2-n}{m^2+n^2} : n,m \in N, m>n \right\}$ find $\sup(A)$ and $\inf(A)$ of set $A$
My idea is to transform $\displaystyle \frac{m^2-n}{m^2+n^2}= \frac{1}{1+(\frac{n}{m})^2}-\frac{1}{\frac{m^2}{n}+n}$ and since I can show that $1+(\frac{n}{m})^2<\frac{m^2}{n}+n$ because then we have $m^2>n$ it implies $\displaystyle \frac{m^2-n}{m^2+n^2}>0$ so $\inf(A)=0$? and I don't have idea how to find $\sup(A)$
As $0<\frac{m^2-n}{m^2+n^2}<1$ [as m>n] So dividing the numerator & denominator by $m^2$ we get $\frac{1-n/m^2}{1+n^2/m^2}$ take $m\to \infty$ we have sup is $1$.
Next $\frac{m^2-m+1}{m^2+(m-1)^2}$ taking $m\to \infty$ it will be $\frac12$. So inf is $\frac12$.