# Evaluating $\lim \limits_{x \to 0} \frac {e^{-1/x^2}}{x}$ without using L' Hôpital's rule [duplicate]

Is there a way to evaluate

$$\lim \limits_{x \to 0} \frac {e^{-1/x^2}}{x}$$

without applying L' Hôpital's rule? I've tried subbing $y = \frac {1}{x}$ and it did not work, and I also tried evaluating the expression by the exponent's Taylor's series expansion and still got no idea finding an upper limit to it.

Thank you!

HINT:

Enforcing the substitution $x=1/y$ reveals

$$\lim_{x\to 0}\frac{e^{-1/x^2}}{x}=\lim_{y\to \infty}\frac{y}{e^{y^2}}$$

Then, recall the elementary inequality $e^x\ge 1+x$ with $x=y^2$.

• Hey there! ${}{}{}{}{}$ – copper.hat Sep 28 '16 at 4:55
• @copper.hat Hey there back! Good to see you've returned Joe. -Mark – Mark Viola Sep 28 '16 at 5:09
• Just intermittently - my current work requires a lot of commuting which cuts into fun time :-). Take care! – copper.hat Sep 28 '16 at 5:11

$\lim \limits_{x \to 0} \frac {e^{(-x^{-2})}}{x}$

If $f(x) = \frac {e^{(-x^{-2})}}{x}$, $\ln f(x) =-x^{-2}-\ln x$.

If $0 < x < 1$, $\ln x =\int_1^x \frac{dt}{t} =-\int_x^1 \frac{dt}{t} \gt -\frac{1-x}{x}$ so

$\begin{array}\\ \ln f(x) &=-x^{-2}-\ln x\\ &<-x^{-2}+\frac{1-x}{x}\\ &=\frac{-1+x-x^2}{x^2}\\ &\to -\infty \text{ as } x \to 0 \end{array}$

so $f(x) \to 0$ as $x \to 0$.