Derivative in 1D as a linear transformation with reminder There are many topics with the derivative definition, but I couldn't find a precise answer to my doubts. In one of the formulation the derivative of a function in a given point $x_0$ is a number $a\in\mathbb{R}$ such as:
$$f(x_0+h)=f(x_0) + a\cdot h +r(x_0,h)$$
In this, the $f(x_0) + ah$ term is the "best" linear approximation of $f(x_0+h)$, and $r(x_0,h)$ is some reminder (or correction). Now, if we make $h \to 0$ we want the $r(x_0,h) \to 0$. However, such an approach will not provide the proper derivative definition, and we must make the following:
$$\lim_{h \to 0} \frac{r(x_0,h)}{h}=0$$
which means the $r(x_0,h)$ vanishes "faster" than $h$ when $h \to 0$. Is there are clear explanation why this entire fraction must vanish, rather than the reminder itself? With many thanks.
 A: As you noted it is necessary that:
$$\lim_{h \to 0} \frac{r(x_0,h)}{h}=0$$
to guarantee that the $r(x_0,h)$ vanishes "faster" than $h$ when $h \to 0$.
In this case we say that f(x) is differentiable at $x_0$ and $a=f'(x_0)$.
Indeed:
$$\lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h}=\lim_{h \to 0}\frac{a\cdot h+r(x_0,h)}{h}=\lim_{h \to 0} \left(a+\frac{r(x_0,h)}{h}\right)=a+\lim_{h \to 0} \frac{r(x_0,h)}{h}$$

A: You write "Now, if we make $h \to 0$ we want the [that] $r(x_0,h) \to 0$."
Not quite. What you want is that $a$ is the derivative $f'(x_0)$, which is given by
$$
\lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}h=\lim_{h \to 0}\frac{f(x_0) + a\cdot h +r(x_0,h)-f(x_0)}{h}=a+\lim_{h \to 0} \frac{r(x_0,h)}{h}.
$$
So that this is really $a$ we need that
$$
\lim_{h \to 0} \frac{r(x_0,h)}{h}=0.
$$
A: I have been thinking a lot yesterday, and following the discussion with user Gimusi (thanks! see above) I was able to clarify my doubts and now pose the question with more precision. My aim is to define the derivative differently than the "standard" approach with the difference quotient. For that reason I consider a bunch of straight lines of the following form $y(x)=ax+b$. Let us now consider the line must pass through the point of the $(x_0,f(x_0))$ coordinates we get the following expression for every line as $y(x)=f(x_0)+a\cdot(x-x_0)$. This form already encodes the line will pass through the selected point and it can be controlled with the slope parameter $a$. Now, let us consider the change of the function $f$ can be expressed as the sum of the linear part and some remainder (as in the original question):
$$f(x_0+h)=f(x_0)+a\cdot h+r(x_0,h)$$
Now, the discussion concluded we need to have $r(x_0,h)$ vanishing faster than $h$ when $h \to 0$, that is:
$$\lim_{h \to 0} \frac{r(x_0,h)}{h}=0$$
and not just:
$$\lim_{h \to 0} r(x_0,h) = 0$$
My impression is that in the latter case will not provide the definision of the unique $a$ value. My feeling is that if we let just $r(x_0,h) \to 0$ this will be true for any $a$. Only by letting the $r/h \to 0$ we will be able to define a single $a$ out of many for different lined passing through the point. Can we somewhat prove such a statement? With many thanks in advance.
