calculus - Infinity indeterminate form that L'Hopital's Rule: $lim_{xto0^+}frac{e^{-frac{1}{x}}}{x^{2}}$

When I tried to find the limit of
$$\lim_{x\to0^+}\frac{e^{-\frac{1}{x}}}{x^{2}}$$

by applying L'Hopital's Rule the order of denominator would increase. What else can I do for it?

Answer

Let $y = \dfrac{1}{x}$, then $x = \dfrac{1}{y} \Rightarrow L = \displaystyle \lim_{y \to +\infty} \dfrac{y^2}{e^y}= \displaystyle \lim_{y \to +\infty} \dfrac{2y}{e^y}=\displaystyle \lim_{y \to +\infty} \dfrac{2}{e^y}= 0$