$f$ differentiable at $a$, $\Longleftrightarrow$ $f(a+h)-f(a)=f'(a)h+\varepsilon(h)h$ In a proof of the chain rule, the following equality is used for a function $f$ differentiable at $a$:
\begin{align}
f(a+h)-f(a)=f'(a)h+\varepsilon(h)h
\end{align}
with $\varepsilon(h)\to 0$ as $h\to 0$. I would like to see the equivalence of the two statements. I can do the converse:
If $f(a+h)-f(a)=f'(a)h+\varepsilon(h)h$ we get $f'(a)+\varepsilon(h)=\frac{f(a+h)-f(a)}{h}$ and taking limits on both sides we get (because $\varepsilon(h)\to 0$, and $f'(a)$ does not involve $h$) $f'(a)=  \lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$, which is true by the definition of differentiability at $a$.
However, the other implication is giving me some trouble, perhaps due to my unfamiliarity with $\varepsilon$-$\delta$ proofs.
If $f$ is differentiable at $a$, we have $\forall \varepsilon >0 \exists \delta$ such that, if $0<\lvert h \rvert<\delta \Longrightarrow \lvert \frac{f(a+h)-f(a)}{h}-f'(a)\rvert < \varepsilon$. Multiplying both sides by $\lvert h\rvert$ gives us $\lvert f(a+h)-f(a)-f'(a)\lvert h\rvert\rvert<\varepsilon \lvert h\rvert$. Using the reverse traingle ienequality, we get $\lvert f(a+h)-f(a)\rvert-\lvert h\rvert\lvert f'(a)\rvert<\varepsilon \lvert h\rvert \Longleftrightarrow \lvert f(a+h)-f(a)\rvert <\lvert h\rvert\lvert f'(a)\rvert+\varepsilon \lvert h\rvert$. This is quite close to what we wanted to prove, but with a strict inequality and not an $\varepsilon(h)$ function. Translating this last bit into something formal is my actual question.
 A: You went a little too far. You have 
$$\left|\frac{f(a+h)-f(a)}h - f'(a)\right|< \epsilon \quad\text{when } 0<|h|<\delta.$$
Set $$\varepsilon(h) = \dfrac{f(a+h)-f(a)}h - f'(a).$$ 
By definition, $\lim\limits_{h\to 0} \varepsilon(h) = 0$. But we have, by definition, $f(a+h)-f(a)-f'(a)h=\varepsilon(h)h$, so you're done.
A: A more precise statement is:
Claim. The function $f\colon \mathbb R\to\mathbb R$ is differentiable at $a$ iff there exists a number $c$ and a function $h\mapsto \epsilon(h)$ such that $\lim_{h\to0}\epsilon(h)=0$ and $$\tag1 f(a+h)-f(a)=ch+\epsilon(h)h$$
for all $h\ne 0$. 
Proof. Assume $f$ is differentiable at $a$. Let $c=f'(a)$ and $\epsilon(h)=\frac{f(a+h)-f(a)}{h}-f'(a)$. Then $(1)$ holds for all $h$ and by taking limits we see that $\epsilon(h)\to 0$.
On the other hand assume that $c$ and $\epsilon$ exist. Then from $(1)$ $$\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{h\to 0}(c+\epsilon(h))=c $$
which means that $f$ is differentiable at $a$ (and collaterally $f'(a)=c$). $_\square$
