Numerical Analysis and Big O How can I show that $e^x -1$ is not $O(x^2)$ as $x\to0$ 
I'm not sure where to start. We can use Taylor's Theorem with remainder:
\begin{equation}
e^x = \sum\limits_{k=0}^n\dfrac{x^n}{n!} +\dfrac{f^{n+1}(\xi)}{(n+1)!}x^{n+1}
\end{equation}
Where $\infty < x < \infty$ and $\xi$ is between $0$ and $x$. However, when and why would you cut off at $n=1$? Also, what is the significance that $x\to0$?

Saying $e^x-1=_0O(x^2)$ is similar to say that the function $x\mapsto\frac{e^x-1}{x^2}$ is bounded as $x$ tend to $0$ which isn't the case since 
$$\lim_{x\to0}\frac{e^x-1}{x^2}=\infty$$
