Show that $p(x)=T_{n,x_0}(x)$ Let $f\in C^n(I)$ and $x_0\in I$. Suppose that, for some polynomial $p(x)$ of degree $n$, the inequality $|f(x)-p(x)|\leq c|x-x_0|^{n+1}$ holds for all $x\in\ I$ and some constant $c$. Show that $p(x)=T_{n,x_0}(x)$, i.e. the Taylor polynomial of $f$ of degree $n$ centered at $x_0$.

I'm having a really hard time trying to solve the previous  exercise. The hypothesis reminds me of Lipschitz continuity, however; I don't know how to use the hypothesis. How should I proceed?
 A: The trick here is to use characterizations of the remainder in Taylor's theorem. The most basic version of Taylor's theorem applied to $f$ on the interval $I$ with expansion centered at $x_0$ gives $f(x) = T_{n,x_0}(x)+h_n(x)(x-x_0)^n$ where $\lim_{x\to x_0} h_n(x)=0$. Thus
$$|f(x) - p(x)| = |(T_{n,x_0}(x)-p(x))+h_n(x)(x-x_0)^n|\leq c|x-x_0|^{n+1}.$$
Of course, this implies
\begin{align*}|T_{n,x_0}(x)-p(x)|&\leq |(T_{n,x_0}(x)-p(x))+h_n(x)(x-x_0)^n|+|h_n(x)(x-x_0)^n|\\
&\leq c|x-x_0|^{n+1}+|h_n(x)||x-x_0|^n,\end{align*}
so $g(x):=T_{n,x_0}(x)-p(x)$ is a polynomial of degree $n$ that is $o(|x-x_0|^n)$ as $x\to x_0$. Such a polynomial must be identically $0$. Indeed, let $k\leq n$ be the smallest nonnegative integer such that $a_k\neq 0$ in the expansion $g(x) = \sum_{j=0}^n a_j(x-x_0)^j$. Then $g(x)/(x-x_0)^k$ is a polynomial with constant term $a_k$, and we should have
$$|a_k| = \lim_{x\to x_0}|g(x)|/|x-x_0|^k.$$
On the other hand, the limit on the right must be zero as $g(x)/(x-x_0)^k$ is $o(|x-x_0|^{n-k})$ as $x\to x_0$, a contradiction. This completes the proof.
