I am currently working on Chapter 14 - local approximations of function and Taylor polynomials - in Königsberger Analysis 1
Background: Königsberger introduced the Taylor Polynomial of order (magnitude) $n$ around the point $a \in \mathbb{R}$ as $$T_nf(x,a)=T_nf(x)= f(a) + \frac{f'(a)}{1!}(x-a)+ \frac{f''(a)}{2!}(x-a)^2+ \dots + \frac{f^{(n)}(a)}{n!}(x-a)^n $$ discussed its properties and defined the approximation error as $$R_{n+1}(x):=f(x)-T_nf(x,a) $$ for which he introduced and proved the theorem of Lagrange:
Theorem (Lagrange Form of $R_{n+1}$) Let $f$ be a real $C^{n+1}$ function on an Interval $I$. Then there exists a $\xi$ between $a$ and $x$ such that $$R_{n+1}(x)= \frac{f^{(n+1)}( \xi)}{(n+1)!}(x-a)^{n+1} $$
The above is clear to me, now Königsberger wants to introduce the Landau Symbol $\mathcal{o}$ by first showing this theorem:
Theorem (Qualitative Taylor Formula) Let $f: I \to \mathbb{R}$ be $n$-times continuous differentiable. Then there exists on $I$ a continuous function $r$ with $r(a)=0$ and $$f(x)=T_nf(x) + (x-a)^n r(x)\tag{*} $$
Proof (as in Königsberger page 285): He manipulates (*) (solves for $r$) and obtains $$r(x)=\frac{1}{(x-a)^n}(f(x)-T_nf(x)) \overset{?}{=} \frac{1}{n!}(f^{(n)}(\xi) - f^{(n)}(a)) $$ which would complete the proof because $\xi \in (x,a)$ and $f$ being $n$-times continuous differentiable just gives the desired property. However I do not see how he obtains the right hand side of this equation.
My intuitive approach would have been to substitute in the $f(x)-T_nf(x)=:R_{n+1}$ in the middle expression and then use the Lagrange Theorem, however I cannot use it in this form because Lagrange requires a $n+1$-times continuous differentiable function.
So what Königersberger surely did was make use of $R_n$, but I still don't see how he obtains the right hand side.
He wrote
$$f(x) = T_{n-1}f(x) + R_n(x),$$
which yields
$$\begin{align} f(x) - T_nf(x) &= \sum_{k=0}^{n-1} \frac{f^{(k)}(a)}{k!}(x-a)^k + \frac{f^{(n)}(\xi)}{n!}(x-a)^n - \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k\\ &= \frac{(x-a)^n}{n!}\left(f^{(n)}(\xi) - f^{(n)}(a)\right). \end{align}$$