Bounds of a set with two variables Let $$A = \left\{ \frac{n+1}{3n-5} \cdot \frac{4m-1}{2m} \quad|\quad n \in \mathbb{N}, \quad m \in \mathbb{N} \right\}$$
The problem is to find the minimum, maximum, supremum and infimum (if they exist).
The only idea I've had so far is for the supremum:
$$ \lim_{n \to \infty}{\frac{n+1}{3n-5}} = \frac{1}{3} $$
$$ \lim_{m \to \infty}{\frac{4m-1}{2m}} = 2$$
And therefore I figured that the supremum may be the product of those two, $$ \sup A = \frac{2}{3} $$
For the other stuff, I've no idea how I'd go about them.
Any ideas?
Thanks.
 A: Hint for the supremum of $A$ (which is not $2/3$) and the infimum of $A$:
$-1\leq \dfrac{n+1}{3n-5} \leq 3$ for any positive integer $n$ and $\left\{\dfrac{4m-1}{2m}\right\}_{m\geq 1}$ is strictly increasing positive sequence which tends to $2$. 
A: Graphically. Look at the hyperbolas:
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For $y_1=\frac{n+1}{3n-5}, n\in \mathbb N$:
$$\text{inf} \ \ y_1=\text{min} \ \ y_1=y_1(1)=-1; \ \text{sup} \ \ y_2=\text{max} \ \ y_2=y_2(2)=3.$$
For $y_2=\frac{4m-1}{2m}, m\in \mathbb N$:
$$\text{inf} \ \ y_2=\text{min} \ \ y_2=y_2(1)=\frac32; \ \text{sup} \ \ y_2=y_2(\infty)=2; \ \ \text{no max}.$$
For $y_1y_2=\frac{n+1}{3n-5}\cdot \frac{4m-1}{2m},n,m\in \mathbb N$:
$$\text{inf} \ \ y_1y_2=\text{inf} \ \ y_1\cdot \text{sup} \ \ y_2=y_1(1)\cdot y_2(\infty)=(-1)\cdot 2=-2;\\
\text{sup} \ \ y_1y_2=\text{sup} \ \ y_1\cdot \text{sup} \ \ y_2=y_1(2)\cdot y_2(\infty)=3\cdot 2=6;\\
\text{no min};\\
\text{no max}.$$
A: You can use multivariable calculus:
At first consider the function $\forall x \in \mathbb{R}, \forall y \in \mathbb{R},g(x,y) = \frac{x+1}{3x-5} \frac{4y-1}{2y}$ and calculate its gradient is zero. Then you can conclude about the variations of $g$ an where its extremum are. 
Then you need to switch back to $x\in \mathbb{N}, y\in \mathbb{N}$.
A: HINT
We have that
$$\frac{n+1}{3n-5}=\frac{n-\frac53+\frac83}{3n-5}=\frac13+\frac{8}{9n-15} \implies -1\le \frac{n+1}{3n-5}\le 3$$
$$\frac{4m-1}{2m}=2-\frac{1}{2m}\implies  \frac32 \le \frac{4m-1}{2m}<2$$
