Confusing limit problem within proof of the chain rule Let $f:G\to G'$, $G\subset \Bbb{R^n}$ and $G'\subset \Bbb{R^m}$ be differentiable at $x_0 \in G$ and let $g:G' \to \Bbb{R^p}$ be differentiable at point $y_0 = f(x_0)$. Then the mapping $g\circ f$ is differentiable at $x_0$ and $D_{g\circ f}(x_0)=D_g(y_o)\cdot D_f(x_0)$
Proof.
$g(f(x_0 + u)=g(f(x_0)+D_f(x_0)u+\vert u\vert\epsilon_1(u))$ , let $h:=D_f(x_0)u+\vert u\vert\epsilon_1(u)$
$=g(y_0)+D_g(y_0)(D_f(x_0)u+\vert u \vert\epsilon_1(u))+\vert h \vert\epsilon_2(h)$
$=g(y_0)+D_g(y_0)\cdot D_f(x_0)u+D_g(y_0)\vert u\vert\epsilon_1(u)+\vert h\vert\epsilon_2(h)$
At this point, I substract $g(y_0)+D_g(y_0)\cdot D_f(x_0)u$ and divide by $\vert h \vert$.
The proof is done, if I could solve the limit $$ \lim_{h\to 0} \frac{D_g(y_0)\vert u\vert\epsilon_1(u)+\vert h \vert\epsilon_2(h)}{\vert h \vert}$$
How can I do this?
 A: You have things a little mixed up. You want the error term to be $o(|u|)$, not $o(|h|)$, since $u$ was the pertubation term you used for the composition $g\circ f$. With that in mind we have
\begin{equation}\tag{1}
\frac{D_g(y_0)\vert u\vert\epsilon_1(u)+\vert h \vert\epsilon_2(h)}{\vert u \vert} = D_g(y_0)\epsilon_1(u)+\frac{|h|}{|u|}\epsilon_2(h).\end{equation}
On the other hand,
\begin{equation}\tag{2}
|h| = |D_f(x_0)u+|u|\epsilon_1(u)|\leq \big(\|D_f(x_0)\|+|\epsilon_1(u)|\big)|u|.
\end{equation}
Thus, in norm, (1) becomes
$$\frac{|D_g(y_0)\vert u\vert\epsilon_1(u)+\vert h \vert\epsilon_2(h)|}{\vert u \vert}\leq \|D_g(y_0)\||\epsilon_1(u)|+\big(\|D_f(x_0)\|+|\epsilon_1(u)|\big)|\epsilon_2(h)|.$$
Since $u\to 0$ implies $h\to 0$ as well by (2), the right-hand side of the above tends to $0$ as $u\to 0$. This completes the proof.
One last note: I've kept the same notation as you used in your statement of the problem, but $D_f(x)$ is really nonstandard. $D_{x}f$, $(Df)(x)$, and $df_x$ are all much more common.
