TITLE: How to prove that a smooth function is NOT analytic?
QUESTION [9 upvotes]: For class we are exploring an example of a function that is smooth at $x=0$ but not analytic in any open interval centered at $0$. My question is, how can one prove that a function is not analytic? I am unaware of what tools are available to do so.

REPLY [14 votes]: An explicit example might help:
$$f(x) =\begin{cases}e^{-1/x} \text{ for } x >0 \\
0 \text{    for } x \leq 0\end{cases}$$
This is smooth but not analytic at $x=0$. Note that $f^n(0)=0$ for all $n$, so the Taylor series at $x=0$ is just $0$, which is clearly not $f(x)$ for any neighborhood.
However if you don't have smoothness, or even continuity, you don't have analyticity. 
So ways you can tell is by if it's continuous/differentiable/smooth. If it IS smooth, you can check to see if it actually equals its Taylor series.