# How does this proof on differentiation work?

This is a proof in my book on introduction to real analysis I don't quite understand. I had to translate it so there could be some translation errors.

Theorem:

$f$ is differentiable in $a \iff$ $(\exists \alpha \in \mathbb{R})(\exists r:\mathbb{R} \rightarrow \mathbb{R}) : (f(a+h) = f(a) + \alpha h + hr(h) \;\text{with}\; r(h) \rightarrow 0 \;\text{if}\; h \rightarrow 0)$

Proof:

From the given follows: for $h \neq 0$ and sufficiently small, $\frac{f(a+h) - f(a)}{h} = \alpha + r(h)$. Conversely, suppose that $\lim_{h\rightarrow0}{\frac{f(a+h)-f(a)}{h}}$ exists and let $\alpha = \lim_{h\rightarrow0}{\frac{f(a+h)-f(a)}{h}}$. Then we can define, for $h \neq 0$ and sufficiently small: $r(h) = \frac{f(a+h)-f(a)}{h} - \alpha$. This means $f(a+h) = f(a) + \alpha h + hr(h) \;\text{with}\; r(h) \rightarrow 0 \;\text{if}\; h \rightarrow 0$

What I do not understand is the first sentence of the proof. How can that sentence prove that $\lim_{h\rightarrow0}{\frac{f(a+h)-f(a)}{h}}$ exists?

-

$$\lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h} = \lim_{h \rightarrow 0} \alpha + r(h) = \alpha.$$