Bounds of a Bivariate Function

Question

I am given that $h(x, y) = \frac{x}{(x+y)}$ , $x > 0$ , and $y > 0$.

I am supposed to deduce that the bounds for $h(x, y)$ are $0 < h(x, y) < 1$, but I do not understand how to arrive at this conclusion.

I have tried using limits to solve for the min and max

$lower bound = \lim_{(x,y) \to (0,0)} h(x, y)$ and $upper bound = \lim_{(x,y) \to (\infty,\infty)} h(x, y)$

My reasoning for this is that 0 and 0 are the lowest values x and y can take, and $\infty$ and $\infty$ are the highest values that x and y can take. However, both of these limits do not exist.

As a little background on this problem, I am dealing with multivariable transformations using jacobians. $Y_1$ and $Y_2$ are gamma distributions (so $Y_1 > 0$ and $Y_2 > 0$), and $u_1 = h_1(y_1, y_2) = \frac{y_1}{y_1+y_2}$.

Answer 1

You don't need any calculus: it's a matter of simple algebra.

If $x>0$ and $y>0$, then $0<x<x+y$, and since $x+y>0$, we can divide by $x+y$ to get $0<\frac{x}{x+y}<1$.

(And you don't mean the minimum and maximum of a bivariate "equation." You mean bounds on the values of a bivariate function.)