Q.State whether True or False: The composition of two continuous functions is a continuous function.

Reasoning: If $g$ is continuous at $x = a$ and $f$ is continuous at $g(a)$, the limit-based definition of continuity alone settles this — no epsilon-delta chase is needed.

Step 1: Continuity of $g$ at $a$ means $\lim_{x \to a} g(x) = g(a)$.

Step 2: Continuity of $f$ at the point $g(a)$ means the limit can be taken through $f$:

$\lim_{x \to a} f\big(g(x)\big) = f\Big(\lim_{x \to a} g(x)\Big) = f\big(g(a)\big).$ …

Setting Up the Statement

The claim is: if $f$ and $g$ are both continuous functions, and the composite $f \circ g$ (i.e., $(f \circ g)(x) = f(g(x))$) is defined, then $f \circ g$ is also continuous wherever it is defined. To decide True/False, it is enough to check this at an arbitrary point $x = a$ in the domain of $f \circ g$.

Recall the definition used throughout this chapter: $f$ is continuous at a point $c$ if

Step-by-Step Reasoning

1. State what continuity of $g$ at $a$ gives us.

Since $g$ is continuous at $x = a$:

2. State what continuity of $f$ at $g(a)$ gives us.

Since $f$ is continuous at the point $y = g(a)$:

3. Let the limit pass through $f$.

Because $f$ is continuous at $g(a)$, as $x \to a$ the inner quantity $g(x) \to g(a)$ (from Step 1), and $f$ applied to a quantity approaching $g(a)$ approaches $f(g(a))$. In limit notation:

4. Recognise this as continuity of the composite.

The left side, $\lim_{x \to a} f(g(x))$, is exactly $\lim_{x \to a} (f \circ g)(x)$, and the right side, $f(g(a))$, is exactly $(f \circ g)(a)$. So:

which is precisely the statement that $f \circ g$ is continuous at $x = a$.

5. Since $a$ was arbitrary, this holds at every point of the domain, so $f \circ g$ is continuous throughout its domain. …

Method: Proving a Composition of Continuous Functions Is Continuous (Limit-Substitution Argument)

This method solves "is $g\circ f$ continuous" proof/justification questions, using the limit-substitution reasoning taught at this level (not a first-principles epsilon-delta argument, which is outside the syllabus).

Steps

Step 1: State what needs to be shown

To prove $g\circ f$ is continuous at $x=a$, you must show

Step 2: Use the continuity of the inner function $f$

Since $f$ is continuous at $a$, $\displaystyle\lim_{x\to a} f(x)=f(a)$ — as $x\to a$, the inner value $f(x)$ approaches $f(a)$.

Step 3: Use the continuity of the outer function $g$ at that landing value

Because $g$ is continuous at $f(a)$, the limit can be passed through $g$: …

Common Mistakes

Mistake 1: Overgeneralizing the domain-mismatch warning

The concept notes flag that a composite like $\sqrt{x-4}$ needs the inner function's output to land in the outer function's domain. Students sometimes stretch this into thinking the general statement about composing continuous functions must be False because such mismatches CAN occur. Why it's wrong: the statement is about functions where the composition IS well-defined (the standard classroom convention); when it is, continuity of each piece really does guarantee continuity of the whole. Correct approach: don't import an edge-case caveat about domain-matching into rejecting the core theorem itself.

Mistake 2: Assuming composition behaves like division …