Numerical Analysis and Big O

Question

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 $x\to0$?

Answer 1

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 $$\lim_{x\to0}\frac{e^x-1}{x^2}=\infty$$

Answer 2

The power series is probably the best general approach. However, if you know that the derivative of $e^x$ at $0$ is $1$, you can use the definition of the derivative to conclude that $$\lim_{x\to 0}\frac{e^x-1}{x}=1,$$ which does what you need.

Answer 3

A related problem.

Hint:

$f=O(g)$ iff $\limsup_{x\to x_0}\frac{f(x)}{g(x)}=c$,

where $c$ is finite.

Added: In your case the limit goes to infinity which implies that $e^x-1$ is not a big O of $x^2$.