Rudin phrases Taylor's Theorem in the following way:
Suppose $f$ is a real function on $[a, b]$, $n$ is an integer, $f^{(n-1)}$ continuous on $[a, b]$, $f^{(n)}(t)$ exists for every $t \in (a, b)$. Let $\alpha$ and $\beta$ be distinct points of $[a, b]$, and define \begin{equation} P(t) = \sum_{k = 0}^{n-1}\frac{f^{(k)}(\alpha)}{k!}(t - \alpha)^k \end{equation} Then there exists a point $x$ between $\alpha$ and $\beta$ such that \begin{equation}\tag{*} f(\beta) = P(\beta) + \frac{f^{(n)}(x)}{n!}(\beta - \alpha)^n \end{equation}
It seems to me then that $f^{(n)}(\alpha) =\lim_{x \to \alpha}f^{(n)}(x) = 0$, simply by shuffling terms across and using repeated l'Hopitals, on (*).
Specifically, since $\beta \to \alpha$ forces $x \to \alpha$, from (*) \begin{align} \lim_{x \to \alpha}\frac{f^{(n)}(x)}{n!} &= \lim_{\beta \to \alpha}\frac{f(\beta) - P(\beta)}{(\beta - \alpha)^n}\\ &= \lim_{\beta \to \alpha}\frac{f'(\beta) - P'(\beta)}{n(\beta - \alpha)^{n-1}}\\ &= \ldots\\ &= \frac{1}{n!}\lim_{\beta \to \alpha}(f^{(n)}(\beta) - P^{(n)}(\beta))\\ &= 0 \end{align} by l'Hopital. Hence $f^{(n)}(\alpha) = \lim_{x \to \alpha}f^{(n)}(x) = 0$. What is wrong?