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 $x=y^2$.
$\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 $x \to 0$.
