Let $f: \mathbb{R}^N \rightarrow \mathbb{R}$ be a function such that for all $x_0 \in \mathbb{R}^N$ and $\epsilon > 0$, there exists $\delta > 0$ such that if $a, b, c, d \in B(x_0, \delta)$ with $a \neq b$ and $c \neq d$, then it follows that: $ \left|\frac{f(a)-f(b)}{\|a-b\|}-\frac{f(c)-f(d)}{\|c-d\|}\right| < \epsilon. $
(a) Prove that the partial derivatives of $f$ exist.
(b) Now prove that they are continuous.
My attempt:
The first idea I have is to choose $a, b, c,$ and $d$ in a way that allows us to break down the inequality into a definition I already know. Let's review the inequality again:
$$ \left|\frac{{f(a) - f(b)}}{{\|a - b\|}} - \frac{{f(c) - f(d)}}{{\|c - d\|}}\right| < \epsilon. $$
Now, let $\mathbf{e}_j$ be a unit vector belonging to $\mathbb{R}^N$. Let $h$ be small enough so that $(x_0 + h\mathbf{e}_j)$ belongs to $B(x_0, \delta)$. We can then do the following:
$ a = x_0 + h\mathbf{e}_j, \quad b = x_0, \quad c = x_0 - h\mathbf{e}_j, \quad d = x_0. $
Let's substitute these values into the inequality:
$ \left|\frac{{f(x_0 + h\mathbf{e}_j) - f(x_0)}}{{\|h\mathbf{e}_j\|}} - \frac{{f(x_0 - h\mathbf{e}_j) - f(x_0)}}{{\|h\mathbf{e}_j\|}}\right| < \epsilon. $
Now, I have the definition of partial derivatives on the left side approaching from the "right side" (thanks to $+h$) and the definition of partial derivatives on the right side approaching from the "left side" ($-h$). My idea is that as $h$ approaches 0, the partial derivative actually exists and is zero since this holds for every $\epsilon$. I am also considering using this as a way to state that the partial derivative is continuous.
my proof is ok? i did something wrong? can i do something better? Any help will be much appreciated. Thanks!