Q:

Infimum and supremum of the set $A\subset \mathbb Q$

Find the infimum and supremum of the set:
  $$A=\left\{\frac{m-n-1}{mn+4m+3n+12}:m,n\in \mathbb N\right\}$$

My mind was drifting and I entered a circle by factorizing the algebraic fraction:
$\frac{m-n-1}{mn+4m+3n+12}=\frac{m-n-1}{m(n+4)+3(n+4)}=\frac{m+3-n-4}{(m+3)(n+4)}=\frac{1}{n+4}-\frac{1}{m+3}$
I thought I could fix either $m$ or $n$ and examine different cases, but I haven't come up to anything formal according to the axioms of the field $\mathbb Q$, so I started believing this is alternating without a satisfying proof. The task is in the unit before the limits. What would be a wise first step?

A:

$$\lim_{n\to\infty}\frac{1}{n+4}-\frac{1}{1+3}\leq\frac{m-n-1}{mn+4m+3n+12}=\frac{1}{n+4}-\frac{1}{m+3}\leq\frac{1}{1+4}-\lim_{m\to\infty}\frac{1}{m+3}$$$$\implies \inf(A)=0-\frac{1}{4}=-\frac{1}{4},\ \sup(A)=\frac{1}{5}-0=\frac{1}{5}.$$
Remember, that the infimum of a decreasing sequence is its limit and the supremum of a increasing sequence is its limit.

