How $\lim_ {x\to 0} \frac{e^{-x^{-2}}}{x^{r}}=0$ for $r>0$?

Question

How to show that $$\lim_ {x\to 0} \frac{e^{-x^{-2}}}{x^{r}}=0$$ for $r\in \mathbb N.$

My attempt: If I apply l'hospital rule then $\lim_ {x\to 0} \frac{e^{-x^{-2}}}{x^{r}}=\lim_ {x\to 0} \frac{e^{-x^{-2}} 2x^{-3}}{rx^{r-1}}$. But it is again $(0/0)$ form...

Motivation: This has a connection to mollifier

$e^{-y}y^{a} \to 0$ as $ y \to \infty$ for any real number $a$. — Kavi Rama Murthy, Oct 03 '22 at 04:57
Related: Examples of applying L'Hôpitals rule ( correctly ) leading back to the same state? — Dave L. Renfro, Oct 03 '22 at 18:56

score 1 · Answer 1 · answered Oct 03 '22 at 15:27

If you change variables and let $y = 1/x$ then your limit becomes

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

and L'hospital will work with that more easily. Since $r$ is a positive integer, you can repeat until $r=0$ or $1$ and then check those cases.

How $\lim_ {x\to 0} \frac{e^{-x^{-2}}}{x^{r}}=0$ for $r>0$?

1 Answers1