Composition of limits holds under either hypothesis: $f$ is defined at $L$ with value $M$, or $g$ avoids $L$ on a punctured neighbourhood of $c$

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 $c$ be a limit point of $A$ and $L$ a limit point of $B$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$), and suppose the limits

$\lim_{x \to c} g(x) = L \qquad \text{and} \qquad \lim_{y \to L} f(y) = M$

both exist, with the stated values (The $\varepsilon$-$\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$). Suppose in addition that at least one of the following holds:

• (i) $L \in B$ and $f(L) = M$;
• (ii) there is a real $\eta > 0$ with $g(x) \ne L$ for every $x \in A$ satisfying $0 < |x - c| < \eta$.

Then the limit of $f \circ g$ at $c$ exists, and

$\lim_{x \to c} f\bigl(g(x)\bigr) \;=\; \lim_{y \to L} f(y) \;=\; M .$

At least one extra hypothesis is necessary. With both omitted the statement is false, and FALSE: $\lim_{x \to c} f(g(x)) = M$ whenever $\lim_{x \to c} g = L$ and $\lim_{y \to L} f = M$ refutes it with a two-line witness in which (i) fails because $f(L) \ne M$ and (ii) fails because $g$ is constantly equal to $L$.

Why an extra hypothesis is needed at all. The inner limit controls $g(x)$ only up to $|g(x) - L| < \rho$; it does not prevent $g(x)$ from equalling $L$. But The $\varepsilon$-$\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ says nothing about $f$ at the point $L$, so the outer estimate is unavailable exactly at the values $g(x) = L$. Hypothesis (i) supplies the missing value directly; hypothesis (ii) excludes those values.

• The hypothesis that $L$ is a limit point of $B$ is what makes $\lim_{y \to L} f(y)$ meaningful at all (The $\varepsilon$-$\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$); it is not an extra assumption of convenience. Note that it does not follow from $\lim_{x \to c} g(x) = L$: a constant $g$ has that limit while $B$ may be a set for which $L$ is isolated.

• Hypothesis (i) is the continuity hypothesis in disguise. Saying $L \in B$ and $f(L) = M = \lim_{y \to L} f(y)$ is exactly saying that $f$ is continuous at $L$ in the sense the next page of this track will define; that is the form in which this theorem is usually quoted, and it is why textbook statements of "the limit of a composition" almost always assume continuity of the outer function.

• Hypothesis (ii) is the one that survives without continuity, and it is the hypothesis under which substitutions such as $y = 1/x$ are legitimate: there the inner function omits the critical value on a punctured neighbourhood for a structural reason, not by assumption on $f$.

• The two hypotheses are genuinely different, neither implying the other. The companion page exhibits a pair satisfying neither, and the same pair with the inner function replaced by the identity, which satisfies (ii) but not (i).