Suppose $f:\mathbb{R} \to \mathbb{R}$ is $C^\infty$ and has the property that, for all $n \ge 0,$ the $n$th derivative $f^{(n)}(x)\to0$ as $x \to \pm\infty$. If we now define $g(x)=f(1/x)$ for nonzero $x$ and $g(0)=0,$ does it follow that for all $n \ge 0$ we have $g^{(n)}(0)=0$?
This came up for me when trying to verify the function $f(x)=e^{-1/x^2}$ for nonzero $x$ with $f(0)=0$ had the property that it is not equal to its power series, since the latter is the zero function. Taking the derivative of this particular $f$ using the chain and product rules became quickly involved, and then I thought that if my question above had answer "yes" it would make this (and other) example(s) easier to verify.
\begin{eqnarray} g(x) =& f(h(x))\\ g'(x) =& f'(h(x))h'(x) = -f(1/x)/x^2\\ \lim_{x\to 0} g'(x) =& -\lim_{x\to 0} f(1/x)/x^2 = -\lim_{y\to \infty} f(y)\, y^2 =0 \end{eqnarray}
In the case for n = 1 this only happens when $f(y)$ drops quicker than $y^2$. So there should be an infinite amount of other function out there which do not behave as you want.