I am trying to calculate $\lim_{(x,y)\rightarrow(0,0)}\frac{e^{1-xy}}{\sqrt{2x^2+3y^2}}$, or show that the limit does not exist.

Thoughts: In the limit $(x,y) \to (0,0)$, it seems that the numerator is tending to a constant value $e$, while the denominator tends to zero, so the required limit might be infinity. However, I am not sure how to obtain a suitable lower bound to show this.

I also tried to show that the limit does not exist, without much luck. Approaching the origin along the $x$-axis would give: $\lim_{x\to 0}\frac{e}{x\sqrt{2}}$, and along the $y$-axis would give $\lim_{y\to 0}\frac{e}{y\sqrt{3}}$. Both these limits tend to infinity. Approaching $(0,0)$ along $y=x$ also suggests the same.

  • $\begingroup$ The denominator is at most $\sqrt{3}\sqrt{x^2+y^2}$ and you can introduce $r^2:=x^2+y^2$. $\endgroup$
    – nejimban
    Commented Sep 25, 2021 at 11:00

1 Answer 1


Yes you are right, since numerator tends to $e$ and the denominator to $0^+$ we can easily conclude that the limit is $\infty$, indeed eventually by squeeze theorem

$$\frac{e^{1-xy}}{\sqrt{2x^2+3y^2}}\ge \frac 1{\sqrt{2x^2+3y^2}} \ge \frac 1{\sqrt 3\sqrt{x^2+y^2}}\to \infty$$

  • $\begingroup$ Thank you very much for this. I am trying to understand how you obtained a numerator of $1$ in your bound: are you arguing that for $(x,y)$ sufficiently close to $(0,0)$, the numerator is sufficiently close to $e$ and therefore at least $1$? $\endgroup$
    – Michael R
    Commented Sep 25, 2021 at 11:18
  • $\begingroup$ Yes exactly, it is just a nice lower bound! $\endgroup$
    – user
    Commented Sep 25, 2021 at 11:21

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