What is the geometric meaning of $f(x)=f(c)+f'(c)(x-c)+r(x)(x-c)$? I see a few, but not many, books give a characteristic theorem about derivatives of real-valued real functions. That is the following:

If $f$ is differentiable at $c$, then there is a function $r$
  continuous at $c$ such that $f$ can be written as
  $f(x)=f(c)+f'(c)(x-c)+r(x)(x-c)$ in an appropriate neighborhood of $c$, where
$r(x)=\begin{cases}\frac{f(x)-f(c)-f'(c)(x-c)}{x-c},\qquad x\neq c \\0,\qquad x=c\end{cases}$

However, I can't find any explanation that says what is the geometric meaning of $f$ being written in this way and how did we come up with this somewhat technical function $r$. And what can we do with this function $r$? Does it have any theoretical benefit?
 A: Let's pick $c=0$ for simplicity. Then $f(x)\approx f(0)$ around $x\approx 0$. To get a best linear fit, we need to use the derivative: $f(x)\approx f(0)+f'(0)x$. Then the difference $f(x)-f(0)-xf'(0)$ is already tending to $0$ faster than any linear function (since otherwise that linear term could be absorbed into the $xf'(0)$ term for an even better linear fit), so we can "factor out" an $x$ from the difference to get
$$f(x)-f(0)-xf'(0)=xr(x) \tag{$\circ$}$$
for some "remainder" function $r(x)$.
This is the preliminary thinking behind the Taylor series expansion of a function. Let's say we apply the principle $(\circ)$ to $r(x)$ itself instead of $f(x)$, and then incorporate that into an even better approximation for $f(x)$. This would give $f(x)\approx f(0)+xf'(0)+\frac{1}{2}x^2f''(0)+x^2 s(x)$ for yet another error term $s(x)$. If we continue on in this way, we get the complete expansion
$$ f(x)=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{6}f'''(0)x^3+\cdots=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n. $$
This is no longer an approximation, but an equality in some interval containing $0$. (At least assuming $f$ is sufficiently "regular," i.e. nice.) If we apply this to $f(x)=\exp(x)$ for instance, we get the expansion $e^x=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\cdots$ which you might have seen before.
This process can be done around any point, not just $0$, yielding
$$ f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(h)}{n!}(x-h)^n $$
valid in an interval containing $x=h$ assuming $f$ is "nice."
