Assume that $f:\mathbb R \rightarrow \mathbb R$ is continuous and $h\in \mathbb R$. Let $\Delta_h^n f(x)$ be a finite difference of $f$ of order $n$, i.e
$ \Delta_h^1 f(x)=f(x+h)-f(x), $ $ \Delta_h^2f(x)=\Delta_h^1f(x+h)-\Delta_h^1 f(x)=f(x+2h)-2f(x+h)+f(x), $ $ \Delta_h^3 f(x)=\Delta_h^2f(x+h)-\Delta_h^2f(x)=f(x+3h)-3f(x+2h)+3f(x+h)-f(x), $ etc. There is an explicite formula for $n$-th difference: $ \Delta_h^n f(x)=\sum_{k=0}^n (-1)^{n-k}\frac{n!}{k!(n-k)!} f(x+kh). $
Assume now that $n\in \mathbb N$ and $f:\mathbb R \rightarrow \mathbb R$ are such that for each $x \in \mathbb R$: $ \frac{\Delta_h^n f(x)}{h^n} \rightarrow 0 \textrm{ as } h \rightarrow 0. $ Is it then $f$ a polynomial of degree $\leq n-1$?
It is clear if $n=1$, because then $f'(x)=0$ for $x\in \mathbb R$.
Edit. Without continuity assumption about $f$ it is not true, because for $n-1$-additive function $F$ which is not $n-1$-linear we have $\Delta_h^nf(x)=0$, where $f(x)=F(x,...,x)$.