Higher derivatives and the classes $C^k$ and $C^\infty$

Let $I\subseteq\mathbb R$ be an interval and $f:I\to\mathbb R$. Put $f^{(0)}:=f$. Recursively, wherever $f^{(j)}$ is differentiable, put $f^{(j+1)}:=(f^{(j)})'$, with derivatives at endpoints understood in the one-sided sense fixed by The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$, and differentiability on a set and The left and right limits of $f$ at $c$, as limits of the restrictions of $f$ to $A \cap (-\infty, c)$ and $A \cap (c, \infty)$.

For $k\in\mathbb N$, the function is $k$-times differentiable on $I$ if $f^{(j)}$ exists on $I$ for every $j\le k$. It is of class $C^k$ on $I$ if these derivatives exist and every $f^{(j)}$, $0\le j\le k$, is continuous on $I$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point). It is smooth, or $C^\infty$, if it is $C^k$ for every $k\in\mathbb N$.

Since $0\in\mathbb N$ (The natural numbers $\mathbb{N}$ (von Neumann)), $C^0$ means continuity. The definitions also give $C^{k+1}\subseteq C^k$. Existence of $f^{(k)}$ alone does not assert that $f^{(k)}$ is continuous.