TITLE: Find $\inf S$ and $\sup S$ for $S=\left\{\frac{12m-n-3mn+7}{5m-2n-2mn+5}: m,n\in \Bbb N\right\}$
QUESTION [5 upvotes]: Find the infimum and supremum (if they exist) of the following set: 
$$S=\left\{\frac{12m-n-3mn+7}{5m-2n-2mn+5}: m,n\in \Bbb N\right\}$$
My attempt: 
$$\frac{12m-n-3mn+7}{5m-2n-2mn+5}=\frac{3m(4-n)-n+7}{5(m+1)-2n(m+1)}=$$
$$\frac{3m}{m+1}\cdot\frac{4-n}{5-2n}+\frac{7-n}{5-2n}\cdot\frac{1}{m+1}$$
$S:=A+B$ where $A:=\{\frac{3m}{m+1}\cdot\frac{4-n}{5-2n}: n\in\Bbb N\}$ and $B:=\{\frac{7-n}{5-2n}\cdot\frac{1}{m+1}:m\in\Bbb N\}$
$A:=X\cdot Y$ where $X:=\{\frac{3m}{m+1}:m\in\Bbb N\}$ and $Y=\{\frac{4-n}{5-2n}:n\in\Bbb N\}$
$X$ is strictly increasing(I proved it) so its first term is the infimum $$\Rightarrow \inf X=\frac{3}{2}$$
$$\lim_{m\to\infty}\frac{3m}{m+1}=3=\sup X$$
$$\sup Y=\lim_{n\to\infty}\frac{4-n}{5-2n}=\frac{1}{2}$$
I think $\inf Y=-4$ but I'm not sure how to prove it. $Y$ is increasing for $n>2.5$ or $n<1.5$ $n_1=2, n_3=-4 \Rightarrow \inf Y=-4$
After that I use the fact that $$\inf (X\cdot Y)=\min\{\sup X\cdot \sup Y, \sup X\cdot \inf Y, \inf X\cdot\sup B, \inf X\cdot \inf B\}$$ and $$\sup (X\cdot Y)=\max\{\sup X\cdot \sup Y, \sup X\cdot \inf Y, \inf X\cdot\sup B, \inf X\cdot \inf B\}$$
And then I find $\inf B$ and $\sup B$ in a similar way, and in the end I use the fact that $$\sup(A+B)=\sup A+\sup B$$ and $$\inf(A+B)=\inf A+\inf B.$$
Is there a better or faster way to solve this problem? Is what I did above correct? Thanks!

REPLY [4 votes]: Note that
$$f(m,n):={12m-n-3mn+7\over5m-2n-2mn+5}={3\over2}-{9\over2}{1\over 2n-5}-{1\over m+1}\ .$$
It follows that
$$\eqalign{S_n&:=\sup_{m\geq1} f(m,n)={3\over2}-{9\over2}{1\over 2n-5}\ ,\cr
I_n&:=\inf_{m\geq1} f(m,n)=1-{9\over2}{1\over 2n-5}\ .\cr}$$
Now
$$(a_n)_{n\geq1}:=\left({1\over 2n-5}\right)_{n\geq1}=\left(-{1\over3},-1,1, {1\over3},{1\over5},{1\over7},\ldots\right)\ .$$
It follows that
$$\sup S=\sup_{n\geq1} S_n={3\over2}-{9\over2}\min_{n\geq1} a_n={3\over2}+{9\over2}=6\ ,$$
and
$$\inf S=\inf_{n\geq1} I_n=1-{9\over2}\max_{n\geq1}a_n=1-{9\over2}=-{7\over2}\ .$$