While arguing about (essentially a prelude) of the fact that embeddings are open in $C^r$, Hirsch makes the following claim about $C_1$ functions $g_n$ such that $|g_n(x)-f(x)| \to 0 $ and $\Vert Dg_n(x) - Df(x) \Vert\to 0$, where $f$ is also $C^1$:
By uniformity of Taylor expansion (with remainder in integral form), $$\frac{|g_n(a_n)-g_n(b_n)-Dg_n(b_n)(a_n-b_n)|}{|a_n-b_n|} \to 0.$$
(here, $a_n,b_n \to p \in K$, a compact set on which each $g_n$ is defined). However, it is not clear to me what he means by "integral form", because the integral form I know would need the second derivative of $g_n$. And even if we suppose $g_n$ is $C^2$, we could have the second derivative of $g_n$ blowing up, and the estimate would not hold.
My question is: What exactly is happening, and how to make this work?
OBS: For the sake of completeness, I put the entire lemma and hypotheses:
Let $U \subset \mathbb{R}^n$ be an open set and $W \subset U$ an open set with compact closure $\overline{W} \subset U$. Let $f:U \to \mathbb{R}^n$ be a $C^1$ embedding. There exists $\epsilon>0$ such that if $g:U \to \mathbb{R}^n$ is $C^1$ and $$\Vert Dg(x)-Df(x) \Vert < \epsilon ~\text{ and }~|g(x)-f(x)| < \epsilon$$ for all $x \in W$ then $g|_W$ is an embedding.
The $g_n$ arise by supposing the result isn't true and using the fact that immersions are open.
What about using that $$ g_n(a_n) - g_n(b_n) = \int_0^1Dg_n(ta_n + (1 - t)b_n)(b_n - a_n)dt? $$ Then \begin{eqnarray*} \frac{|g_n(a_n) - g_n(b_n) - Dg_n(b)(a_n - b_n)|}{|a_n - b_n|} &\leq& \bigg(\int_0^1 ||Dg_n(ta_n + (1 - t)b_n) - Dg_n(b_n)||dt\bigg) \bigg|\frac{a_n - b_n}{|a_n - b_n|}\bigg| \to 0. \end{eqnarray*} Recall that on Hirsch's argument $(a_n - b_n)/|a_n - b_n|$ converges.