# Find $\lim_{(x,y)\rightarrow (0,0)} \frac{e^{1-xy}}{\sqrt{2x^2+3y^2}}$.

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.

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

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
$$\frac{e^{1-xy}}{\sqrt{2x^2+3y^2}}\ge \frac 1{\sqrt{2x^2+3y^2}} \ge \frac 1{\sqrt 3\sqrt{x^2+y^2}}\to \infty$$
• 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 $1$? Commented Sep 25, 2021 at 11:18