A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs

Let $A, B \subseteq \mathbb{R}$, let $g : A \to \mathbb{R}$ with $g(A) \subseteq B$, and let $f : B \to \mathbb{R}$, so that the composite $f \circ g : A \to \mathbb{R}$ is defined. Let $c \in A$. If $g$ is continuous at $c$ and $f$ is continuous at $g(c) \in B$, then $f \circ g$ is continuous at $c$ (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).

Consequently, if $g$ is continuous on $A$ and $f$ is continuous on $B$, then $f \circ g$ is continuous on $A$.

No side hypothesis is needed, and that is the whole point. The composition theorem for limits, Composition of limits holds under either hypothesis: $f$ is defined at $L$ with value $M$, or $g$ avoids $L$ on a punctured neighbourhood of $c$, must assume one of two extra conditions: either $L \in B$ with $f(L) = M$, or $g \ne L$ on a punctured neighbourhood of $c$; with both dropped the statement is false, which is FALSE: $\lim_{x \to c} f(g(x)) = M$ whenever $\lim_{x \to c} g = L$ and $\lim_{y \to L} f = M$. The first of those conditions is exactly continuity of $f$ at $L$ written out, so under the hypotheses above it holds automatically and nothing has to be assumed. The mechanism is visible in the proof: 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 quantifies over $|y - b| < \rho$ rather than over $0 < |y - b| < \rho$, so the value $y = b$ that the limit version cannot control is precisely the one the continuity hypothesis does control.

• The contrast with limits, in one sentence. Composition of limits holds under either hypothesis: $f$ is defined at $L$ with value $M$, or $g$ avoids $L$ on a punctured neighbourhood of $c$ cannot control $f$ at the single value $L$, because The $\varepsilon$-$\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ never evaluates $f$ there; continuity of $f$ at that value is exactly the missing information, and it is hypothesis (i) of that theorem. So the present theorem is not a strengthening obtained by cleverness: it is the composition theorem under the hypothesis that makes the obstruction vanish.

• What is not claimed. Continuity of $f \circ g$ at $c$ does not follow from continuity of $g$ at $c$ alone together with $f$ merely having a limit at $b$; nor does it follow from continuity of $f$ at $b$ together with $g$ merely having a limit at $c$, since then $g(c)$ need not be $b$ at all. Both hypotheses above are hypotheses about the same two points, $c$ and $g(c)$.