In this answer the following claim is made for a $k$-th order multivariate polynomial $p:\mathbb R^n\to\mathbb R$, i.e., $p(x)=\sum_{|\alpha|\leq k} c_\alpha x^\alpha$ in multi-index notation (see remark B below).
Claim: If $$\frac{p(x)}{||x||^k}\to 0 \quad \text{ as } \quad x\to 0 \tag{1}$$ then $p=0$.
I'm trying to prove this, but I'm not gettig anywhere. The answer mentioned above justifies the claim by saying
Using (1) repeatedly with $k$ replaced by $0,1,2,…,k$ we find that $c_\alpha=0$ for all $\alpha$.
So let's see. Taking $k=0$ in (1) yields $\frac{p(x)}{||x||}\to 0 $, or,
$$ \sum_{|\alpha|\leq k} c_\alpha \frac{x^\alpha}{||x||} \to 0 $$
The argument now should probably be that for $|\alpha|>1$ we have $\frac{x^\alpha}{||x||} \to 0$ (but I truly don't see how this follows), so that we are left with
$$ \sum_{|\alpha| =1} c_\alpha \frac{x^\alpha}{||x||} \to 0 $$
which should in turn imply (but again I don't see why) that $c_\alpha=0$ for each $\alpha$ with $|\alpha|=1$. Now we can consider (1) with $k=2$, do something similar, and continue in this fashion until we have shown that $c_\alpha=0$ for all $\alpha$.
I'm really stuck so I would appreciate any help! The reason that I want to prove this claim is that I want to prove uniqueness of the polynomial explansion of a $k$ times differentiable multivariable function.
Some remarks: A) The mentioned answer did not specify what norm $||x||$ is used, but I guess this makes no difference since all norms on $\mathbb R^n$ are equivalent. So we could e.g. take $||x||^2 = ||x_1||^2+ ||x_2||^2 + \dots + ||x_n||^2$ for concreteness.
B) In the multi-index notation, $\alpha = (\alpha_1,\dots,\alpha_n)$ and $x^\alpha\equiv x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n}$ and $|\alpha| \equiv |\alpha_1|+ |\alpha_2| + \dots + |\alpha_n|$.