Smooth function which is not real analytic The canonical exapmle is 
$$
 f(x) := \begin{cases}
 0 & \text{ if } x \leq 0 \\
 \exp(-1/x) & \text{ if } x > 0
\end{cases}
$$
What is the easiest way to prove that $f$ is smooth ? Can this be done by induction ?
 A: We need only worry about $x=0$.
First note that all the derivatives of $f$ for $x>0$ are sums of terms of the form $e^{-\frac{1}{x}} \frac{1}{x^n}$ (use induction if you want an explicit expression).
Second note that $e^{-\frac{1}{x}} \frac{1}{x^n} = \frac{\frac{1}{x^n}}{1+\frac{1}{x}+\frac{1}{2!x^2}+\cdots} < \frac{\frac{1}{x^n}}{\frac{1}{(n+1)!x^{n+1}}} = (n+1)!x$. It follows that all derivatives converge to $0$ as $x \downarrow 0$.
A: In this answer, it is shown inductively that since, for $x\gt0$,
$$
\frac{\mathrm{d}^n}{\mathrm{d}x^n}e^{-1/x}=R_n(x)e^{-1/x}
$$
for some rational functions $R_n$, and for any $n$
$$
\lim_{x\to0}x^{-n}e^{-1/x}=\lim_{x\to\infty}x^ne^{-x}=0
$$
we get that
$$
\lim_{x\to0}\frac{\mathrm{d}^n}{\mathrm{d}x^n}e^{-1/x}=0
$$
which, by the mean value property, says that at $x=0$
$$
\frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}}e^{-1/x}=0
$$
Thus, all derivatives from the right are $0$.
Since the function is identically $0$ from the left, all derivatives are $0$ there.
Away from $0$ there is no problem.
