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

Question

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

Let $t=\sqrt{x^2+y^2+z^2}$

$$\lim_{t\to 0}\frac{1}{t^2}e^{-\frac{1}{t}}$$

Let $r=\frac{1}{t}$

$$\lim_{r\to \infty}r^2 e^{-r}=\lim_{r\to \infty}\frac{r^2}{e^r}$$

using l'hopital
$$\lim_{r\to \infty}\frac{2r}{e^r}$$

using l'hopital again
$$\lim_{r\to \infty}\frac{2}{e^r}=0$$

Is the following valid? is there is a simpler way?

Answer 1

HINT

By spherical coordinates $\sqrt{x^2+y^2+z^2}=r\to 0$

$$\frac{1}{ x^2+y^2+z^2 }e^{-\frac{1}{\sqrt{x^2+y^2+z^2}}}=\frac1{r^2}e^{-\frac1r}$$

and for $y=\frac1r\to \infty$ since eventually $e^y\ge y^3$

$$\frac1{r^2}e^{-\frac1r}=\frac{y^2}{e^y}\le\frac{y^2}{y^3}=\frac1y\to 0$$

Answer 2

It is valid, yes, and I suppose that that's how I would have done it. Also, I would have written $\lim_{t\to0^+}$ and $\lim_{r\to+\infty}$ instead of $\lim_{t\to0}$ and $\lim_{r\to\infty}$ respectively.

Answer 3

All is well.

A comment, instead of using L'Hopital twice:

Let $r \gt 0:$

$e^r \ge 1+ r +r^2/(2!) + r^3/(3!)+....\ge r^3/(3!).$

Then $0\lt \dfrac{r^2}{e^r} \lt \dfrac{r^2}{(r^3/(3!))}= \dfrac{3!}{r}.$

Now consider $r \rightarrow \infty$ , squeeze .