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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!

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    At n=2, Y=2, so its sup is at least that. – 2017-01-14
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    @martycohen Oh, you're right, thanks. – 2017-01-14

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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 $$\inf S=\inf_{n\geq1} I_n=1-{9\over2}\max_{n\geq1}a_n=1-{9\over2}=-{7\over2}\ .$$