Problem: Let $I$ be an interval, $f \in C^n(I,\mathbb R), x_0 \in I,$ and $T$, a polynomial of degree $n$ with
$$\lim_{x\to a}\frac{f(x)-T(x)}{(x-x_0)^n}=0.$$
Prove that T is the Taylor polynomial of $f$ of degree $n$ in $x_0 $.
I have come across several questions proving that if T is a Taylor polynomial then we have the above equation. However, I can't seem to figure out how to make my way towards this proof as it is basically the asking for the other direction of the Taylor polynomial theorem. I see a pattern between how the above limit is similar to what is the derivative of $f(x_0)$, but I need some help seeing the overall picture.
I know we are done once I prove that:
$$T(x) = \sum_{k=0}^{n}\frac{f^{k}(x)}{k!}(x-x_0)^k$$
Any help is appreciated.
Best Answer
Hint: The Taylor polynomial $p(x)=\sum^n_{k=0}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$ satisfies $$\lim_{x\rightarrow x_0}\frac{f(x)-p(x)}{(x-x_0)^n}=0$$ See for example this posting
This, combined with your assumptions on the polynomial $T$ yields $$\lim_{x\rightarrow x_0}\frac{T(x)-p(x)}{(x-x_0)^n}=0$$
Since $T$ and $p$ are polynomials of degree $n$, it follows that $T(x)=p(x)$.
For the last statement, show that if $\phi(x)=a_0+a_1x+\ldots +a_nx^n$ and $\lim_{x\rightarrow0}\frac{\phi(x)}{x^n}=0$, then $\phi(x)\equiv0$ and so $a_j=0$ for $0\leq j\leq n$.