$\sum_{n=0}^{\infty} \frac{f^{(n)}(x)}{n!}(x-c)^n \qquad \qquad(1)$
Theorem: Let $I\subseteq \mathbb{R}$ be open interval, $c\in I$ and $f: I \rightarrow \mathbb{R}$, $f\in C^{\infty}(I)$. If there exist $n_0\in \mathbb{N}, \delta \gt 0, M\gt 0$ and $C\gt 0$, such that $$|f^{(n)}(x)|\le CM^nn!,\quad \forall x\in I'=\langle c- \delta,c + \delta\rangle \cap I,\quad \forall n\ge n_0$$
then $(1)$ converges to $f(x)$, $\forall x \in I' \cap \langle c- \frac1{M},c + \frac1{M} \rangle$.
Proof:
For every $x\in I$, by Taylor theorem, there exists $c_x$ between $x$ and $c$ such that $$f(x)=T_n(x)+\frac{f^{(n+1)}(c_x)}{(n+1)!} (x-c)^{n+1}$$
From here we have
$$|f(x)-T_n(x)| = \frac{|f^{(n+1)}(c_x)|}{(n+1)!}|x-c|^{n+1}\le C\frac{M^{n+1}(n+1)!}{(n+1)!}|x-c|^{n+1}=C(M|x-c|)^{n+1}$$
For $x\in I' \cap \langle c-\frac1{M},c+ \frac1{M} \rangle$ is $M|x-c|\lt 1$ so $\lim_{n\to \infty}|f(x)-T_n(x)|=0$.
Hence, $$ \lim_{n\to \infty} T_n(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n=f(x) $$
My question is this: $c_x$ doesn't have to be from $I'$, so why is this ok:
$$\frac{|f^{(n+1)}(c_x)|}{(n+1)!}|x-c|^{n+1}\le C\frac{M^{n+1}(n+1)!}{(n+1)!}|x-c|^{n+1}$$
Is there a mistake in theorem?