Limit of $\frac{e^{-xy}}{1+x^2+y^2}$ in $x,y ≥ 0$ I am trying to figure out if $f(x,y)=\frac{e^{-xy}}{1+x^2+y^2}$ goes to zero, in the area x,y ≥ 0, when $x^2+y^2$ goes to infinity.
So I rewrote the function as $\frac{1}{e^{xy}}\frac{1}{1+x^2+y^2}$.
The function $\frac{1}{1+x^2+y^2} $ definitley goes to zero. But what about
$\frac{1}{e^{xy}}$? Since $x,y ≥ 0$ it looks like to me that this function should also go to zero.  
So if $x$ or $y$ is equal to zero then we have $\frac{1}{e^{0}}=1$
So in this case I would think you could argue that $f$ goes to $1\cdot0=0$ when $x^2+y^2$ goes to infinity.
When $x,y > 0$  I tried changing to polar coordinates then we have
$$0 <\frac{1}{e^{xy}}=\frac{1}{e^{r^2cos(t)sin(t)}}=\frac{1}{e^{r^2\frac{1}{2}sin(2t)}}$$
Since $- \frac{1}{2} ≤ \frac{1}{2}sin(2t) ≤ \frac{1}{2}$
I tried to use this to (maybe) see that $\frac{1}{e^{r^2\frac{1}{2}sin(2t)}}$ is a function that goes to zero. But with $\frac{1}{e^{r^2\frac{1}{2}sin(2t)}} ≤ \frac{1}{e^{-\frac{r^2}{2}}}$ we cant come to any conclusion since $\frac{1}{e^{-\frac{r^2}{2}}}$ goes to infinity when $r$ goes to infinity. 
Could I prove that the limit is zero in some other way? If not, how is it possible that $\frac{1}{e^{xy}}$ $x,y > 0$ does not approach zero when $x^2+y^2$ goes to infinity?

We have that for $x^2+y^2 \to \infty$ since
$$0< e^{-xy}\le 1$$
$$\frac{e^{-xy}}{1+x^2+y^2}\le \frac{1}{1+x^2+y^2}\to 0$$
