# L'hopitals rule with limits

Given the following limit,

\begin{align} \lim_{x\to 0}\frac{e^{-1/x^2}-0}{x-0}\\\\ & \end{align}

How do I calculate it? when pluggin in 0 I would get $\frac{0}{0}$

• L'hopital's rule says to take the derivative of the top and bottom, then plug in 0. Then what do you get? – RationalHusky Mar 17 '17 at 2:16

## 2 Answers

The problem here is that applying L'Hôpital's rule right of the bat yields \begin{align} \lim_{x\to 0}\frac{e^{-1/x^2}-0}{x-0}&=\lim_{x\to 0}\frac{2e^{-1/x^2}}{x^3}, \end{align} which isn't particularly helpful. However, we notice that \begin{align} \lim_{x\to 0^+}\frac{e^{-1/x^2}-0}{x-0}&=\lim_{x\to 0^+}\frac1{xe^{1/x^2}}\\ &=\lim_{x\to\infty}\frac x{e^{x^2}}\\ &=\lim_{x\to\infty}\frac1{2xe^{x^2}}\\ &=0; \end{align} similar manipulations can be used to confirm that the left-handed limit is zero as well. Thus \begin{align} \lim_{x\to 0}\frac{e^{-1/x^2}-0}{x-0}&=0. \end{align}

We don't need to use L'Hospital's Rule here.

Simply note that $e^x\ge 1+x$, which I proved in This Answer. Hence, we have

$$\left|\frac{e^{-1/x^2}}{x}\right|=\left|\frac{1}{xe^{1/x^2}}\right|\le \frac{|x|}{1+x^2}$$

whereupon taking the limit yields

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