Compute $\lim_{x\to 0} \frac{e^{-1/x^2}}{x}$ without using L'Hopital?

Question

Purely out of curiosity, is it possible to compute the following limit without using L'Hopital?

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

Using L'Hopital and realizing that we cannot get what we want if we simply take derivatives of $e^{-1/x^2}$ and $x$ directly, we have that

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

I am curious about whether there is a different way of computing this.

Answer 1

By the change of variables: $t=\frac{1}{x}$, $$\lim_{x\to 0^+}\frac{\exp(-\frac{1}{x^2})}{x}=\lim_{t\to\infty}\frac{t}{e^{t^2}}$$ now observe that $$\lim_{t\to\infty}\frac{e^{t^2}}{t}=+\infty$$ since $e^{t^2}\ge 1+t^2$ ($e^x\ge 1+x$ for all $x$ is a well known thing)

Hence $$\lim_{t\to\infty}\frac{t}{e^{t^2}}=0$$

Answer 2

Here is a different method:

$$\lim_{x\to 0} \frac{e^{-\frac{1}{x^2}}}{x}=\lim_{x\to 0} \frac{e^{-\frac{1}{x^2}}}{e^{\ln(x)}}=\lim_{x\to 0} e^{-(\frac{1}{x^2}+\ln(x))}=\lim_{x\to 0} e^{-\frac{1}{x}(\frac{1}{x}+x\ln(x))}=0$$