Limit $\lim_{(x,y)\to\infty} e^{-e^{xy}}$ with polar coordinates

Question

can i use polar to solve this limit? $$\lim_{(x,y)\to\infty} e^{-e^{xy}}=$$ $$\frac{1}{e^{e^{r^2\cos\theta\sin \theta}}}=$$ but i'm quite stuck here i think the denominator goes to infinity but should i show that? or can i just write $0$ as a solution after the step above like $$\frac{1}{e^{e^{r^2\cos\theta\sin \theta}}}=0$$ is it correct? any help or suggestion would be very helpful Thank's in advance

Answer 1

We can use polar coordinates but the final evaluation is not correct indeed by $x=y=t \to \infty$ we obtain

$$e^{-e^{xy}}=e^{-e^{t^2}} \to 0$$

but for $x=-y=t \to \infty$ we obtain

$$e^{-e^{xy}}=e^{-e^{-t^2}} \to e^0=1$$

therefore the limit doesn't exist.