Bounds for $\frac{x-y}{x+y}$ How can I find upper and lower bounds for $\displaystyle\frac{x-y}{x+y}$? So I do see that
$$\frac{x-y}{x+y} = \frac{1}{x+y}\cdot(x-y) = \frac{x}{x+y} - \frac{y}{x+y} > \frac{1}{x+y} - \frac{1}{x+y} = \frac{0}{x+y} = 0$$
(is it correct?) but I don't get how to find the upper bound.
 A: Look at pairs $(x,y)$ such that $x = 1-y$. For these pairs, $\frac{x-y}{x+y} = 1-2y$ and we have $\mbox{lim}_{y\rightarrow \pm\infty}(1-2y) = \mp \infty$.
A: Suppose $\dfrac{x-y}{x+y} = c$, where $c \neq -1$.
Then
$$\begin{eqnarray}
 x - y &=& c(x + y) \\
 x - cx &=& y + cy \\
 (1 - c)x &=& (1 + c)y
\end{eqnarray}$$
Therefore $y = \dfrac{1-c}{1+c} x$.
But if $\dfrac{x-y}{x+y} = -1$, then $x - y = -1(x + y)$,
and from this we conclude that $x = 0$ and $y$ can be anything you want.
So you can set $\dfrac{x-y}{x+y}$ to any value you like by choosing
suitable values of $x$ and $y$. There are no limits on $\dfrac{x-y}{x+y}$.
A: so we have $f(x,y)=\frac{x-y}{x+y}$, we want to show that $f(x,y)$ is unbounded, therefore we fix $y=y_0>0$ and then have
$$
f(x,y_0)=f(x)=\frac{x-y_0}{x+y_0}
$$
now choose $x=x^*-y_0$ which gives us
$$
f(x)=\frac{x^*-2y_0}{x^*}
$$
and now observe that for $\lim_{x^*\to0+}\frac{x^*-2y_0}{x^*}=-\infty$ and for $\lim_{x^*\to0-}\frac{x^*-2y_0}{x^*}=\infty$.
So $f(x,y)$ is indeed unbounded as we thought.
