Before the current problem I work on, I proved the following:



> 
> Let $q$ be a polynomial with $\deg(q) \le n$. If $q(x)=o(x^n)$ for $x \to 0$, then $q$ is the zero polynomial. 
> 
> 
> 


I **have** to use the above for the problem I'm working on currently now, which is



> 
> Let $f:I \to \mathbb{R}$ be a $C^n$ function, defined on a open interval $I$ and assume $0\in I$. Let $p(x)=\sum\_{k=0}^{n} b\_{k} x^{k}$ be a polynomial with $\deg(p) \le n$. Show if $f(x)=p(x)+o(x^n)$ for $x \to 0$, then $b\_{k}=\frac{f^{(k)}(0)}{k !} \quad \text{for} \quad k=0,1, \ldots, n$.
> 
> 
> 


**My attempt**


I know that $o(x^n)$ is 0 for $x\to 0$. So in this limit $f(x)=p(x)$. And (I assume) since $p$ is the Taylor polynomial it's equal to $f$ for $x\to 0$. So I only have to calculate the Taylor polynomial of order $n$ with center point $0$. That will end the proof, right? Also, which theorem am I applying? I can find anything in my textbook which looks like what I have done.


Also, there is another problem with my method. I assume that $f$ is $n$ times differential in the center point $x=0$ when I calculate the Taylor polynomial $p$ of order $n$. But it is only given that $p$ is $n$ times continuous. I'm not sure about what to do.



Proof of: If $f(x)=p(x)+o(x^n)$ for $x \to 0$, then $b_{k}=\frac{f^{(k)}(0)}{k !} $ for $ k=0,1, \ldots, n$
Let $T$ be the [Taylor polynomial](https://en.wikipedia.org/wiki/Taylor%27s_theorem#Statement_of_the_theorem) for $f$ of order $n$ at $0$, so that $f(x)=T(x)+o(x^n)$ (as $x\to0$). Comparing this with the condition $f(x)=p(x)+o(x^n)$, we see that $(p-T)(x)=o(x^n)$. Using now what you have proved, we see that $p=T$, and hence the polynomial $p$ has the same coefficients as $T$. 


