Let $U$ be an open subset of $R^n$ and $f:U\rightarrow \mathbb{R}$ a function and $x\in U$ such that in a small neighbourhood of $x$ and for $\epsilon \in \mathbb{R^b}$ sufficiently small we have the following:
$$\begin{equation} f(x+ \epsilon) = \sum\limits_{|\alpha| \leq k} a_{\alpha}\epsilon^{\alpha} + g(\epsilon) \\ f(x+ \epsilon) = \sum\limits_{|\alpha| \leq k} b_{\alpha}\epsilon^{\alpha} + h(\epsilon) \end{equation}$$
where $\alpha \in \mathbb{N_0}^n$ a multiindex, $k$ a natural number, $g$ and $h$ functions such that $\lim \limits_{\epsilon \rightarrow 0} \frac{g(\epsilon)}{||\epsilon||^k} = \lim \limits_{\epsilon \rightarrow 0} \frac{h(\epsilon)}{||\epsilon||^k} = 0$.
My goal is to prove that for all $\alpha$ with $|\alpha|\leq k$ it holds that:
$a_{\alpha}$ = $b_{\alpha}$ but I am very confused my the multiindex notation because I am not sure how the limits would behave since I don't know that $f$ is differenctiable.
I would appreciate some help. Thank you