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

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?

• put $x= 1/y$ so that $y \to \infty$ and then note that the given function $f(x) = e^{-1/x}/x^{2} = y^{2}/e^{y}$. Then use the fact that $$e^{y} = 1 + y + \frac{y^{2}}{2!} + \frac{y^{3}}{3!} + \cdots > \frac{y^{3}}{6}$$ for all $y > 0$. Then we get $0 < f(x) < 6/y$ and hence by Squeeze theorem $f(x) \to 0$ as $y \to \infty$. – Paramanand Singh Jun 23 '15 at 11:02

Let $y = \dfrac{1}{x}$, then $x = \dfrac{1}{y} \Rightarrow L = \displaystyle \lim_{y \to +\infty} \dfrac{y^2}{e^y}= \displaystyle \lim_{y \to +\infty} \dfrac{2y}{e^y}=\displaystyle \lim_{y \to +\infty} \dfrac{2}{e^y}= 0$
Hint : $\frac{e^{-\frac{1}{x}}}{x^{2}}=\frac{1}{x^{2}e^{\frac{1}{x}}}$
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 $\alpha$.
• Good one. +1. I normally add these two limits $$\lim_{x \to \infty}\frac{\log x}{x^{a}} = \lim_{x \to \infty}\frac{x^{a}}{e^{x}} = 0$$ for any $a > 0$ in my list of standard limits like $\lim_{x \to 0}(\sin x)/x =1$. – Paramanand Singh Jun 23 '15 at 11:14
• That's right, but I sticked to the particular question. I would add $$\lim_{x\to 0+}x^{\alpha}\lvert\ln x\rvert^{\beta}=0\quad(\alpha,\beta>0)$$ and $$\lim_{x\to+\infty}\Bigl(1+\frac cx\Bigr)^x=\mathrm e^{c}.$$ – Bernard Jun 23 '15 at 11:23