Suppose that $\displaystyle f: A \subset \mathbb {R}^n \rightarrow \mathbb {R}^m $ is differentiable at $\displaystyle x_0 \in Int(A) $. Prove that for every $\displaystyle v \in \mathbb {R}^n $, $\displaystyle \lim _{t \rightarrow 0} \frac {f(x_0+tv)-f(x_0)}{t} $ exists and is equal to $\displaystyle Df_{x_0}(v) $

My work:

Let $\displaystyle x_0 \in int(A) $, then there exists an open set $\displaystyle U$ such that $\displaystyle x_0 \in U \subset \mathbb {R}^n $. Let $\displaystyle v \in \mathbb {R}^n $

Find $\displaystyle \epsilon > 0 $ such that $\displaystyle ||x_0- \epsilon || \subset U $

Pick $\displaystyle t \in \mathbb {R} $ such that $\displaystyle |t| ||v|| < \epsilon $

Then there exists $\displaystyle z_1,z_2,...,z_n \in \mathbb {R}^n$ such that $\displaystyle ||x-z_k||<||tv||=|t|||v|| \ \ \ \ \ \forall k \in \{ 1,2,...,n \} $ and $\displaystyle f(x+tv)-f(x)= \sum ^N _{k=1}tv_k \frac { \partial f }{ \partial x_k } (z_k) $

Am I doing this right so far? Thanks.