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?

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 $\alpha$.
