Prove that the limit exists.

Question

$\lim_{x\to0}$ ${e^{-1/ x^2}\over x}$

Prove that the limit exists, and show it that equals $0$ in order to show $g′(0)=0$. I'm not sure how to go about this.

Answer 1

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$1+t\le e^t\le \frac{1}{1-t} \tag 1$$

for $t<1$. We also note that $e^t>0$ for all $t$.

Then, using $(1)$ with $t=-1/x^2$ we can write

$$\begin{align} \left|\frac{e^{-1/x^2}}{x}\right|&\le\frac{1}{\left(1+1/x^2\right)|x|}\\\\ &=\frac{x^2}{|x|(x^2+1)}\\\\ &\to 0 \,\,\text{as}\,\,x\to 0\end{align}$$