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.
 A: First we see how i) implies ii).
$f$ is differentiable at $a$ so that the limit $$\lim_{h \to 0}\dfrac{f(a + h) - f(a)}{h} = f'(a)$$ exists. This means that $$\lim_{h \to 0}\dfrac{f(a + h) - f(a) - hf'(a)}{h}= 0$$ or in other words if we let $g(h) = f(a + h) - f(a) - hf'(a)$ then $g(h)/h \to 0$ as $h\to 0$. Thus $g(h) = o(h)$. Now we can see that $$f(a + h) = f(a) + hf'(a) + g(h) = f(a) + hf'(a) + o(h)$$ which is of the form $f(a) + ch + o(h)$ where $c = f'(a)$ is a constant dependent on $a$.
Next we see how ii) implies i).
Let $f(a + h) = f(a) + ch + o(h)$ so that $o(h) = f(a + h) - f(a) - ch$ and by definition of $o(h)$ we see that $o(h)/h \to 0$ as $h \to 0$. Thus we see that $$\lim_{h \to 0}\frac{f(a + h) - f(a) - ch}{h} = 0$$ or $$\lim_{h \to 0}\frac{f(a + h) - f(a)}{h} - c = 0$$ or $$\lim_{h \to 0}\frac{f(a + h) - f(a)}{h} = c$$ and we get the usual definition of derivative as a limit and $c$ is now denoted by $f'(a)$.
A: 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.
A: Hint: First try to prove that i) follows from ii). Knowing that $f(a+h)=f(a)+ch+o(h)$, calculate $f'(a)$ by definition. 
