Equivalent definitions of differentiable I am trying to show:
The two statements are equivalent: 
(i) $f$ is diﬀerentiable at $a$, 
(ii) $f(a + h) = f(a) + ch + o(h)$,
where c is some constant (depending on $a$) and $o(h)$
denotes some function of $h$ (also depending on $a$), with the property that
$$\lim_{h\to 0} \frac{o(h)}{|h|} = 0$$
(That is $o(h) = h\alpha(h)$; where $\lim_{h\to0} \alpha(h) = 0$)
What is the relation between $c$ and $f′(a)$?
For (i) I have given the standard definition of differentiable in terms of limit. I see this is not too different from the statement in (ii) but I cannot make them equivalent.
Any help would be much appreciated.

By definition, function $f$ is differentiable at $a$ if there exists a scalar $f'(a)$ such that for any $\varepsilon>0$, such that there exists $\delta>0$, for any $x$ satisfying $|x-a|<\delta$, $|f(x)-f(a)-f'(a)(x-a)|<\varepsilon|x-a|$. Let $x=a+h$ and $c=f'(a)$, we have $|f(a+h)-f(a)-ch|<\varepsilon|h|$ for any $|h|<\delta$. Equivalently, $\lim_{h\to0}\frac{f(a+h)-f(a)-ch}{h}=0$ and hence $f(a+h)=f(a)+ch+o(h)$. Every step above is revertible.
