Show that limit of $e^{e^{-xy}}$ as $ x^2+y^2 \to \infty$ does not exist

Question

The problem consists of showing that $\lim\limits_{x^2+y^2 \to \infty}e^{e^{-xy}}$ does not exist. My initial approach was to set $x = t, y = 0$ and show that the function converges to different limits when I let $t \to \infty$ compared to when $t \to -\infty$. From that, I concluded that the function does not have a limit. However, I am not sure whether it is a valid approach to set $y=0$ in this case. Is this the right way to go about it, or should I try some other strategy? I tried using polar coordinates but that does not get me far.

Thanks,

Answer 1

Note that

• for $x=y$

$$e^{e^{-xy}}=e^{e^{-x^2}}\to1$$

• for $x=-y$

$$e^{e^{-xy}}=e^{e^{x^2}}\to+\infty$$

thus the limit does not exist.

Answer 2

Let $(x,y)=r(\cos \theta,\sin \theta)$, then

\begin{align} \exp e^{-xy} &= \exp e^{-r^2\cos \theta \sin \theta} \\ &= \sum_{n=0}^{\infty} \frac{e^{-nr^2\cos \theta \sin \theta}}{n!} \end{align}

If $\cos \theta \sin \theta<0$,

$$\lim_{r\to \infty} \exp e^{-xy} = \infty$$

If $\cos \theta \sin \theta=0$,

$$\lim_{r\to \infty} \exp e^{-xy} = e$$

If $\cos \theta \sin \theta>0$,

$$\lim_{r\to \infty} \exp e^{-xy} = 1$$

The required limit is path dependence and hence limit doesn't exist.