Request for Clarification : Differentiability vs Analytic Consider the function $f(x)=e^{-1/x}$ if $x>0$ and $f(x)=0$ otherwise, now previously I managed to show that $f$ is infinitely differentiable in $\mathbb{R}$ so that includes differentiable at $x>0$, $x<0$, and the most important one $x=0$. The problem lies with showing that $f$ is not analytic at $x=0$ this confuses me since I have shown that $f$ is differentiable at $x=0$. How could this be interpreted?
 A: Look back at the definition of analyticity. It doesn't just say that all of the derivatives of $f$ exist, it says that "Taylor series work well" for $f$:

$f$ is analytic iff for each $a$, the Taylor series $T^f_a$ agrees with $f$ around $a$ - that is, there is some $\epsilon>0$ such that $T_a^f$ is defined and equals $f$ on the interval $(a-\epsilon, a+\epsilon).$

Intuitively, analyticity doesn't just say that all the derivatives of $f$ exist at each point, but that the various derivatives contain "all the information about $f$:" there's no point where knowing the derivatives of $f$ won't let you figure out what $f$ is, at least "locally."
So to show that your $f$ is not analytic, what you want to do is find some point where the Taylor series for $f$ "goes wrong" somehow. Now your $f$ consists of three regions: negative $x$, positive $x$, and $x=0$ where the first two regions meet. On each of the first two regions $f$ is very "tame:" $f$ is equal to some function we already know to be analytic. So the only possible "weird point" for $f$ is $a=0$. You know that $f$ is infinitely differentiable here too, but again analyticity is more than just infinite differentiability:

What is $T_0^f$? Does this "look like $f$" even in a small neighborhood of $0$?

What we'll see is that, while the infinite differentiability of $f$ at $0$ does allow us to build the series $T_0^f$, somehow information is lost when we do so.
