Infinity indeterminate form that L'Hopital's Rule: $\lim_{x\to0^+}\frac{e^{-\frac{1}{x}}}{x^{2}}$

Question

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 1

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

Answer 2

You can do without the application of the L'Hospitals rule.

Hint : $ \frac{e^{-\frac{1}{x}}}{x^{2}}=\frac{1}{x^{2}e^{\frac{1}{x}}} $

Answer 3

Don't use L'Hospital's rule. It won't work here, and when it works, it is equivalent with Taylor's polynomial at order $1$, which is much less error-prone.

It is a problem of Asymptotic analysis: set $u=\dfrac1x$. Then $$\lim_{x\to 0^+}\frac{\mathrm e^{-\tfrac1x}}{x^2}=\lim_{u\to+\infty}\frac{u^2}{e^u}=0$$ since a basic result in Asymptotic analysis is $\,u^{\alpha}=_{+\infty}o\bigl(\mathrm e^u\bigr)$ for any $\alpha$.