## Calculus 8th Edition

$R^{2}$
Given: $f(x,y)=\frac{xy}{1+e^{x-y}}$ The given function is defined for all values of x and y except at $1+e^{x-y}=0$ This implies $e^{x-y}=-1$ But the value of $e^{x-y}$ cannot be negative; it is always greater than zero. This implies that $e^{x-y}>0$ Hence, the function $f(x,y)=\frac{xy}{1+e^{x-y}}$ is continuous on $R^{2}$.