Limit of $\frac{e^{1/x}}{x^2}$ as x approaches 0 negatively

Question

I know the following:

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

I cannot, however, see why. Is there a method that makes this result intuitively clear?

Answer 1

Note that if $x\to 0^-$ and $t=-1/x$ then $x=-1/t$ and $t\to+\infty$, hence $$\lim_{x\to 0^-}\frac{e^{1/x}}{x^2} = \lim_{t\to\infty}\frac{e^{-t}}{(-1/t)^2} = \lim_{t\to\infty}\frac{t^2}{e^t}.$$ And the last limit should be a proud member of the collection of limits you know.