Determine whether $\lim_{(x,y) \to (0,0)} e^{\frac{-1}{(x^{2} + y^{2})}}$ exists Determine whether $$\lim_{(x,y) \to (0,0)} e^{\frac{-1}{(x^{2} + y^{2})}}$$
exists.
Claim: The limit does exist.
Justification:  This is where I'm having problems. I know in one variable calculus that $$\lim_{t \to 0}e^{\frac{-1}{t^{2}}} = 0$$
The first thing I tried was building a composition of previously defined limits. In particular I tried using $f(x,y) = x + y$ and $g(t) = e^{\frac{-1}{t^{2}}}$. With this idea I was hoping to be able to compare $(x+y)^{2} = x^{2} + 2xy + y^{2}$ to $x^{2} + y^{2}$ and get something I could then use the squeeze principle on. But after manipulation I end up with
$$ e^{\frac{-1}{x^{2} + y^{2}}} \leq e^{\frac{-1}{x^{2} + 2xy + y^{2}}}$$
Which doesn't give me much.
The only other idea that comes to mind is to "try" and find a satisfactory $\delta$, but that looks like it will be hell. I must be missing a comparison to apply to this. Could I get some assistance?
EDIT: Ooooh...I just noticed polar coordinates will make it alot easier to solve. But I want to leave the question here. Assuming I didn't "know" about the polar coordinates approach what may work?

If you want to do it without polar coordinates, one way is to leverage the fact $$e^p \geq 1 + p$$ for $p \geq 0$ to conclude that $$0 \leq e^{-p} \leq \frac{1}{1 + p}$$ for $p > 0$ also. Now substitute $p = \frac{1}{x^2 + y^2}$ and (after simplifying) notice that the denominator is always strictly positive, so the simplified expression is continuous on all of $\Bbb{R}^2$. You can then use a two-variable version of Squeeze Theorem.
