Let $ I\subseteq\mathbb R$ be an open interval, $ f\in\mathcal C^n(I)$, and $ x_0\in I$. Let $ p $ be a polynomial of degree at most $ n $. Prove that if $$ \lim_{h\to0}\frac{f(x_0+h)-p(h)}{h^n}=0,$$ then $p$ is the $n$-th Taylor polynomial of $f$ at $x_0$.
I was thinking one might be able to do this with induction on $n$? Let us write $$p(x)=a_nx^n+\ldots+a_1+a_0$$ such that we have to show $$ a_0=f(x_0),\,a_1=f'(x_0),\,a_2=\frac{f''(x_0)}2,\,a_3=\frac{f'''(x_0)}6,\ldots$$
Then for $n=0$, the condition is equivalent to $$\lim_{h\to0}f(x_0+h)=p(h)=a_0$$ and since $f$ is continuous, the limit is equal to $f(x_0)$. Hence $p(x)=f(x_0)$ which is indeed the zero-th Taylor polynomial of $f$ at $x_0$.
Any ideas for the induction step?
