Uniform bound on the error from the derivative Suppose $f: \mathbb{R}^m \to \mathbb{R}^n$ is differentiable. For each $x$, the function 
$$h \mapsto f(x+h) - f(x) - Df_x(h)$$
is $o(|h|)$ at $h=0$. Assuming $f$ is $C^1$ (or more, if we need), I'm interested in finding a neighborhood $N_\epsilon$ of a given point $x_0$ such that if $z_1, z_2 \in N_\epsilon$, then 
$$|f(z_1) - f(z_2) - Df_{z_2}(z_1-z_2)| \leq \epsilon |z_1 - z_2|. \tag{1} $$
Would my best bet be to take the function $z \overset{\phi}{\mapsto} \delta_{\epsilon, z}$, where $\delta_{\epsilon, z}$ is the tolerance on $h$ needed at $z$ to make $|f(z+h)-f(z)-Df_z(h)| \leq \epsilon |h|$, argue that $\phi$ is continuous, and take its maximum on a compact set? Or is there an easier way?
This came from trying to show that given $x_0$ and $\epsilon$, there is a neighborhood of $x_0$ such that for $z_1, z_2$ in that neighborhood, we have
$$|f(z_1)-f(z_2)-Df_{x_0}(z_1-z_2)| \leq \epsilon |z_1 - z_2|.\tag{2}$$ This came up in doing Newton iteration in order to prove the Inverse Function theorem when $m=n$. Usually IFT is shown assuming the derivative is the identity at the relevant point; this simplifies things but sort of hides exactly what we're doing in the general case. 
So if you come up with a way to show (2) without using (1), even better.

Edit: Updated proof to omit the $C^1$ on a neighborhood assumption.
Assume $f$ is differentiable near $x_0$ and $f'$ is continuous at $x_0$.
Let $\epsilon > 0$. Then, there exists some $\delta > 0$ such that for every $x$ with $\| x-x_0 \| < \delta$ follows $\max_{1\le i \le n}\| \nabla f_i(x) - \nabla f_i(x_0) \| < \epsilon$.
Let $N_\epsilon$ be the ball around $x_0$ with radius $\delta$. Let $x,x'\in N_\epsilon$. Then, for $0\le t\le 1$ we have 
$$ \| x + t(x' - x) - x_0 \| = \| (1-t)(x-x_0) + t(x' - x_0) \| \le (1-t)\delta + t\delta = \delta. $$
Thus, by mean value theorem, for every $1\le i \le n$ there exists a $t_i\in (0, 1)$ such
\begin{align}  
  | f_i(x') - f_i(x) - \nabla f_i(x_0)^T (x' - x) |
  &= | (\nabla f'(x + t_i(x' - x)) - \nabla f'(x_0))^T (x' - x) | \\
  &\le \| \nabla f_i(x + t(x' - x)) - \nabla f_i(x_0) \| \|x' - x\| \\
  &\le \epsilon \|x' - x\|.
\end{align}
