If $f$ vanishes sufficiently fast at $x_0$ does this imply that all derivatives vanish as well? Let $f: U \rightarrow \mathbb{R} $, with $U \subset \mathbb{R}^n$ open, be infinitely differentiable.
Suppose there is a $x_0 \in U$ such that 
$$
\lim_{x \rightarrow x_0} \frac{f(x)}{|x-x_0|^k}=0
$$
for all $k \in \mathbb{N}$.
How can we prove that every derivative $D^{\alpha} u(x_0)$, $\alpha \in \mathbb{N}^n$, is zero?
For $|\alpha|=1$ we can just use the definition of the partial derivative. But in the case $n=2$ this already gets messy.
 A: This is a consequence of Taylor's theorem. Suppose we're on the line and $x_0=0.$ Suppose we have $f(0) = f'(0) = 0$ but $f''(0) \ne 0.$ By Taylor we have
$$f(x) = f(0) +f'(0)x + f''(0)x^2/2 +o(x^2)= f''(0)x^2/2 +o(x^2).$$
That implies $|f(x)| \ge (|f''(0))|/3 )x^2$ for small nonzero $x,$ contradicting the hypothesis. Hence $f''(0)=0.$ This reasoning can be continued to show all derivatives of $f$ at $0$ are $0.$
For $n>1$ the same ideas apply, although Taylor polynomials get more complicated. Suppose we know $D^\alpha f(0,0) =0$ for $|\alpha| \le 1.$ Then
$$f(x) = \sum_{|\alpha| = 2} [D^\alpha f(0,0)]/\alpha !]x^\alpha + o(|x|^2).$$
This follows from the one variable result. Let $T_2(x)$ denote the above sum. Note that $T_2$ is homogeneous of degree $2,$ so for $r> 0$ and $|\zeta| =1,$ $T_2(r\zeta) = r^2T_2(\zeta).$ Suppose $T(\zeta_0) \ne 0.$ Then $|T_2(r\zeta_0)|$ is on the order of $r^2$ as $r\to 0.$ This implies the same is true of $f.$ That contradicts the given hypothesis. Thus we have $T_2(\zeta)=0$ for all $\zeta, |\zeta| =1.$ That implies $T_2\equiv 0,$ which in turn implies $D^{\alpha}f(0) = 0$ for $|\alpha|=2.$
For more on this topic, search on Taylor polynomials in several variables, either on MSE or google.
A: I take it from your comment about $|\alpha| = 1$, that you already know the first derivatives vanish quickly at $x_0$.  You do realize that all the $|\alpha| = 2$ derivatives are first derivatives of things you have already shown vanish quickly...
