prove this is not true. $e^x-1 = O(x^2)$ as $x \rightarrow 0$

Question

Hello I need to show that this assertion is not true and I ran into an issue. I was taught that I could prove this by doing $\lim_{x\to0} \frac{(e^x-1)}{x^2}=c\neq0$. I used l'hoptial rule once but then was stuck at $\frac{1}{0}$. When I plugged the limit into wolfram alpha online it used a laurent series to expand it but not quite sure how to do that. Is there an easier way to calculate this limit. Thanks

Answer 1

You can easily compute the limit that you mentioned: $$ \lim_{x\to 0}\dfrac{e^x-1}{x^2} = \underbrace{\lim_{x\to 0}\dfrac{e^x-1}{x}}_{=1}\cdot \lim_{x\to 0} \dfrac{1}{x}= \infty $$

This means that $x^2$ tends to zero much faster than $e^x-1$ (as $x\to 0$). In fact, using Taylor's formula (as pointed out in comments), you see that $e^x-1 = O(x)$ as $x \to 0$.