# Using epsilon-delta to prove a multivariable limit

I'm trying to find $$\lim_{x,y \to 0,0} f(x,y)$$ where $$f(x,y)=\frac{e^\frac{-1}{\sqrt{x^2+y^2}}}{e^{x^2+y^2}-1}$$ I've tried converting to polar coordinates but it doesn't seem to be a helpful simplification. Graphically, it seems that the limit is 0 from all possible paths, and I tried to construct an epsilon-delta proof to prove this. I'm just trying to figure out if I'm on the right track.
Let $$\epsilon>0$$ be given, find a $$\delta>0$$ such that if $$|{(x,y)-(0,0)}|<\delta$$ implies $$|{f(x,y)-L}|<\epsilon$$, Assuming $$L=0$$. We know that $$\sqrt{x^2+y^2}<\delta\rightarrow x^2+y^2<\delta^2$$ which implies $$\frac{1}{e^{x^2+y^2}-1}<\frac{1}{e^{\delta^2}-1}$$ $$|{f(x,y)-0}|=|{\frac{e^\frac{-1}{\sqrt{x^2+y^2}}}{e^{x^2+y^2}-1}-0}| = |{\frac{1}{e^{x^2+y^2}-1}*e^\frac{-1}{\sqrt{x^2+y^2}}}|$$. Since $${e^\frac{-1}{\sqrt{x^2+y^2}}}<1$$ for all $$x,y\neq(0,0)$$, we get $$|{f(x,y)-0}|< |\frac{1}{e^{\delta^2}-1}*1|$$. Setting $$\delta=\sqrt{ln(\frac{1}{\epsilon}-1)}$$ We can plug in to the inequality to get $$|{f(x,y)-0}|<|\frac{1}{e^{\delta^2}-1}*1|<\epsilon$$. Thus the limit is equal to $$0$$

$$0<\sqrt {x^2 + y^2} = r < \delta$$
$$0<\frac {e^{\frac {-1}{r}}}{e^{r^2} + 1} < \frac 12 e^{\frac {-1}{r}}<\epsilon$$
We can say this because $$e^{r^2} > 1$$ and $$e^{r^2} + 1 > 2$$
$$-\frac {1}{r} < \ln 2\epsilon\\ r< -\frac {1}{\ln 2\epsilon}$$
For any $$\epsilon > 0,$$ when $$\delta < -\frac {1}{\ln 2\epsilon}$$ then $$|f(x,y)| < \epsilon$$