The limit of $e^{\frac{1}{x^2 + y^2}}$ as $(x, y) \to (0, 0)$

Question

$$\lim_{(x, y) \to (0, 0)} e^{\frac{1}{x^2 + y^2}}$$

This really should be a simple limit question, I've done similar things many times before, but I'm very out of practice with limits and cannot for the life of me figure this one out. Can anyone lend a hand?

Answer 1

The denominator of the exponent tends to zero from above, no matter how $(x,y)$ tends to $(0,0)$. So the exponent grows without bound to $+\infty$. Hence the expression itself does as well. (Some may say the limit doesn't exist since it is infinite, but that's a matter of convention.)

If the numerator were $-1$ instead of $1$, the exponent would tend to $-\infty$, so the expression would tend to $0$.

Answer 2

As $(x,y)\rightarrow 0$, $x^2+y^2 \rightarrow 0$ while always remaining positive. Since $\lim_{t\to 0^+} \frac{1}{t} = +\infty$, we find that $\lim_{(x,y)\to 0} \frac{1}{x^2 + y^2} = +\infty$. Finally, since $\lim_{t\to +\infty} e^t = +\infty$, we string it all together to get:

$$ \lim_{(x,y)\to (0,0)} e^{\frac{1}{x^2+y^2}} = e^{\lim_{(x,y)\to (0,0)} \frac{1}{x^2+y^2}} = +\infty $$

Answer 3

Note that $x^2+y^2=r^2$. This is the equation of a circle of radius $r$. Saying that $(x,y)\to (0,0)$ here means that the radius of this circle is approaching zero. So we can transform this to a one variable limit:

$$ \lim_{r\to 0} y=\lim_{r\to 0} e^{\frac{1}{r^2}}$$

Now note that $$\log y=\frac{1}{r^2}\rightarrow \infty,$$ as $ r\to 0$. So $y \to \infty$.