Composition of Continuous Functions

The composition theorem is the quiet engine behind nearly every continuity argument in calculus and beyond. In elementary calculus, it is what justifies calling $\sin(x^2)$, $e^{\cos x}$, or $\ln(1+x^2)$ continuous without re-deriving continuity from scratch each time — each is a finite chain of compositions of functions already known to be continuous, foreshadowing the systematic treatment in Continuity of Polynomials, Rational, and Elementary Functions. In real analysis more broadly, the theorem generalizes verbatim to continuous maps between metric spaces and topological spaces (the proof is identical with $|x-a|$ replaced by a metric $d(x,a)$, or with preimages of open sets), making it a cornerstone of point-set topology: the composition of continuous maps between topological spaces is continuous, which is what makes "the category of topological spaces and continuous maps" a coherent category in the first place. In differential calculus, composing continuity lemmas at the right basepoints is the model for the chain rule, where derivative composition again requires the inner function's value to land where the outer function's derivative hypothesis is checked. In dynamical systems, iterating a continuous map $g, g\circ g, g\circ g\circ g, \dots$ relies on this theorem to guarantee that iterates of continuous maps remain continuous, underlying the study of fixed points and orbits. In numerical analysis and engineering, composed continuous transfer functions (filters, coordinate changes, normalization layers) inherit stability guarantees from this same principle — a small perturbation at the input cannot suddenly blow up through a chain of continuous stages. In probability and statistics, continuous transformations of continuous random variables (e.g. $Y = f(X)$ for continuous $f$ and a random variable $X$ with continuous distribution structure) rely on composition results to preserve regularity properties through changes of variables.

Ask the tutor: Can you show me how this composition theorem is used, step by step, to prove that $\tan(x)$ is continuous wherever $\cos(x) \ne 0$?

Think of $g$ and $f$ as two machines wired in series. The first machine, $g$, takes an input $x$ near $a$ and produces an output $g(x)$ near $g(a)$ — that is exactly what continuity of $g$ at $a$ guarantees. The second machine, $f$, takes that output and, because it is continuous at $g(a)$, produces $f(g(x))$ close to $f(g(a))$ whenever its input $g(x)$ is close enough to $g(a)$. The subtlety is the phrase "close enough": $f$'s tolerance for how close its input must be to $g(a)$ is some $\eta > 0$ determined by $f$ and by whatever error bound $\varepsilon$ we were originally handed. It is $g$'s job, using its own continuity at $a$, to guarantee that a sufficiently small perturbation $\delta$ of $x$ around $a$ forces $g(x)$ to land within that $\eta$-window around $g(a)$. The chain of implications runs backward from the desired output error to the required input error, which is why composition proofs read most naturally "outside in."

Example 1: A genuinely asymmetric chain, verified by hand. Let

so $h(x) = \sqrt{4 - \dfrac{3}{x^2+1}}$. Take the basepoint $a = 2$. Set up the pieces. We have $A = \mathbb{R}$ (domain of $g$). Since $0 < g(x) \le 1$ for all real $x$ (as $x^2+1\ge 1$), the range of $g$ lies in $(0,1] \subset B := (-\infty, 4/3]$, which is exactly the domain of $f$ (we need $4-3y \ge 0$, i.e. $y \le 4/3$). So the composition is well-defined on all of $\mathbb{R}$. Check hypothesis (1): $g$ continuous at $a=2$. By the Algebra of Continuous Functions, $x \mapsto x^2+1$ is continuous everywhere (product and sum of continuous functions) and is never $0$, so its reciprocal $g(x) = 1/(x^2+1)$ is continuous everywhere, in particular at $a=2$. Numerically, $g(2) = 1/5$. Check hypothesis (2): $f$ continuous at $g(2) = 1/5$. Write $f = \sqrt{\cdot}\circ \ell$ where $\ell(y) = 4-3y$ is continuous everywhere (polynomial), and $u \mapsto \sqrt{u}$ is continuous on $[0,\infty)$ — a standard fact provable directly from $\varepsilon$-$\delta$ (and itself an instance of continuity of elementary functions). At $y = 1/5$: $\ell(1/5) = 4 - 3/5 = 17/5 > 0$, so $\sqrt{u}$ is continuous at $u = 17/5$. By this very composition theorem applied one level down, $f$ is continuous at $y=1/5$. Apply the theorem. Since $g$ is continuous at $2$ and $f$ is continuous at $g(2) = 1/5$, the theorem gives that $h = f\circ g$ is continuous at $2$. Numerical sanity check. $h(2) = \sqrt{4 - 3/5} = \sqrt{17/5} = \sqrt{3.4} \approx 1.8439$. Perturb slightly: at $x = 2.1$, $g(2.1) = 1/(2.1^2+1) = 1/5.41 \approx 0.18484$, and $h(2.1) = \sqrt{4 - 3(0.18484)} = \sqrt{4 - 0.55452} = \sqrt{3.44548} \approx 1.8562$. The output moved by about $0.0123$ while the input moved by $0.1$ — consistent with continuity (small input change, small output change), and consistent with the chain: $g$ moved by $\approx 0.01516$, well inside any small $\eta$, and $f$'s local behavior near $17/5$ (derivative $-3/(2\sqrt{17/5}) \approx -0.813$) amplifies that by a bounded factor.

Example 2: A case where a piece is discontinuous at the wrong point, but composition still works because the theorem's exact hypothesis is met. Let

Here $f$ is discontinuous at $y=1$ (jump from the constant value $1$ nearby to $5$ at the point) but we are evaluating at $a = -1$, so $g(a) = 1$. Let's see what the theorem does and does not say. Check hypothesis (1): $g(x)=x^2$ is continuous at $a=-1$ (polynomial). Fine. Check hypothesis (2): We need $f$ continuous at $g(a) = 1$. But $f$ is discontinuous at $1$! So hypothesis (2) fails, and the theorem simply does not apply here — it makes no claim about $f\circ g$ at $a=-1$. Indeed, let's check directly: $(f\circ g)(-1) = f(1) = 5$. But for $x$ near $-1$, $x \ne -1$, $g(x) = x^2$ is near $1$ but generally $\ne 1$ (e.g. $x = -1.01 \Rightarrow g(x) = 1.0201 \ne 1$), so $f(g(x)) = 1 \ne 5$. Thus $(f\circ g)(x) \to 1 \ne 5 = (f\circ g)(-1)$ as $x \to -1$: $f \circ g$ is discontinuous at $-1$, exactly because hypothesis (2) failed. This is not a counterexample to the theorem — it illustrates that the hypothesis is necessary and that checking it at the correct point $g(a)$, not at $a$, is the crux of applying the theorem correctly. Now contrast with $a = 0$: $g(0) = 0 \ne 1$, so hypothesis (2) becomes "$f$ continuous at $0$" — and indeed $f$ is continuous at $0$ (it equals the constant $1$ on a whole neighborhood of $0$, since $f(y)=1$ for all $y\ne 1$). So the theorem applies at $a=0$ and correctly predicts $f\circ g$ continuous at $0$; direct check: $(f\circ g)(x) = f(x^2) = 1$ for all $x$ with $x^2 \ne 1$, i.e. for all $x$ near $0$ other than $x=\pm1$, and $(f\circ g)(0) = f(0) = 1$, matching.