Continuity of Combined Functions

9 min read

Reference: Stewart §2.5

Before You Start: Prerequisite Check

📋 Can you do these? (Click to reveal self-test)

Test yourself on these prerequisite skills:

Continuity definition: Is $f(x) = x^2 + 1$ continuous at $x = 3$?

Check

Yes, polynomials are continuous everywhere. (But can you explain why using the definition?)
Function composition: If $f(u) = u^2$ and $g(x) = \sin x$, what is $(f \circ g)(x)$?

Check

Answer: $f(g(x)) = (\sin x)^2 = \sin^2 x$
Domain identification: What is the domain of $h(x) = \sqrt{x - 1}$?

Check

Answer: $x \geq 1$, or $[1, \infty)$

If you struggled:

Review Continuity at a Point for the definition
Review Function Composition for inner/outer functions

Building Complex from Simple

Here’s a powerful principle: if you know certain “basic” functions are continuous, you can build infinitely many continuous functions by adding, multiplying, dividing, and composing them.

Why is this valuable? Because it means you rarely need to verify continuity from scratch. Instead of checking the three conditions for $f(x) = \sin(x^3 + e^x)$, you can simply observe: polynomials are continuous, $e^x$ is continuous, $\sin$ is continuous, sums and compositions of continuous functions are continuous, so $f$ is continuous everywhere.

This “Lego-block” approach to continuity is your most efficient tool for limit evaluation. If a function is continuous at $a$, then $\lim_{x \to a} f(x) = f(a)$: just plug in.

Prerequisite Map

Quick Reference

Property	Value
Chapter	1.8
Course	MATH161
Difficulty	Intermediate
Time	~20 minutes

Continuity Rules Summary

Operation	If $f$, $g$ continuous at $a$...	Result is continuous at $a$?
$f + g$	None	✅ Yes
$f - g$	None	✅ Yes
$c \cdot f$	None	✅ Yes
$f \cdot g$	None	✅ Yes
$\frac{f}{g}$	AND $g(a) \neq 0$	✅ Yes
$f \circ g$	AND $f$ continuous at $g(a)$	✅ Yes

The Power Move: If you can build a function from continuous “building blocks” (polynomials, trig, exp, log) using these operations, it’s continuous on its natural domain.

Key Concepts

Arithmetic Rules for Continuity

If $f$ and $g$ are continuous at $a$, then:

Operation	Result	Continuity
Sum	$f + g$	Continuous at $a$
Difference	$f - g$	Continuous at $a$
Constant multiple	$cf$ (where $c \in \mathbb{R}$)	Continuous at $a$
Product	$f \cdot g$	Continuous at $a$
Quotient	$\dfrac{f}{g}$	Continuous at $a$ if $g(a) \neq 0$

Key insight: These rules follow directly from the corresponding limit laws. Since continuity means $\lim = f(a)$, and limits obey these arithmetic operations, so does continuity.

The Composition Rule

If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then:

$$\boxed{f \circ g \text{ is continuous at } a}$$

In other words: A continuous function of a continuous function is continuous.

Visualizing composition:

         g                    f
    a ────────→ g(a) ────────→ f(g(a))

    g continuous    f continuous    f∘g continuous
    at a           at g(a)         at a

Example: $h(x) = \sin(x^2)$

Inner function: $g(x) = x^2$ (polynomial → continuous everywhere)
Outer function: $f(u) = \sin(u)$ (sine → continuous everywhere)
At any $a$: $g$ is continuous at $a$, and $f$ is continuous at $g(a) = a^2$
Therefore $h = f \circ g$ is continuous everywhere

The “Basic Building Blocks”

These fundamental functions are continuous on their entire domains:

Function Type	Examples	Domain
Polynomials	$x^3 - 2x + 1$	All $\mathbb{R}$
Rational functions	$\dfrac{x^2+1}{x-3}$	Where denominator $\neq 0$
Root functions	$\sqrt{x}$, $\sqrt[3]{x}$	Where radicand makes sense
Trigonometric	$\sin x$, $\cos x$, $\tan x$	Their natural domains
Exponentials	$e^x$, $2^x$	All $\mathbb{R}$
Logarithms	$\ln x$, $\log x$	$(0, \infty)$

The power of this list: By combining these with arithmetic operations and composition, you can determine continuity of almost any function you’ll encounter.

Domain Matters

$$f(x) = \sqrt{x^2 - 4}$$

This function is continuous on its domain: $(-\infty, -2] \cup [2, \infty)$.

Why?

$g(x) = x^2 - 4$ is a polynomial (continuous everywhere)
$f(u) = \sqrt{u}$ is continuous for $u \geq 0$
The composition is continuous where $x^2 - 4 \geq 0$

Important: A function can be continuous on its domain even if that domain has gaps. “Continuous” doesn’t mean “defined everywhere.”

Quick Continuity Check Procedure

Goal: Determine where $f(x)$ is continuous.

Step 1: Identify the “building blocks” in the function (polynomials, trig, exp, log, roots, etc.)

Step 2: Note their domains of continuity.

Step 3: Apply arithmetic and composition rules.

Step 4: State: “$f$ is continuous on [domain], which is...”

Example: $f(x) = \ln(\cos x)$

$\cos x$ is continuous on $\mathbb{R}$
$\ln u$ is continuous for $u > 0$
$\cos x > 0$ when $x \in \left(-\frac{\pi}{2} + 2\pi k, \frac{\pi}{2} + 2\pi k\right)$ for integer $k$
Therefore $f$ is continuous on those intervals

Common Pitfalls

Mistake	Why It’s Wrong	Correct Approach
“Sum of continuous = continuous everywhere”	True, but domain matters!	Continuous on the intersection of domains
Forgetting quotient restriction	$\frac{f}{g}$ undefined where $g = 0$	Always exclude $g(x) = 0$ points
Wrong composition order	$(f \circ g)(x) = f(g(x))$, not $g(f(x))$	$g$ is inner (applied first), $f$ is outer
Assuming $\sqrt{f}$ continuous if $f$ continuous	Only where $f(x) \geq 0$	Find domain where $f(x) \geq 0$ first
“Continuous on domain” = “continuous everywhere”	These are different statements	Be explicit about the domain

💡 The "Inside-Out" Strategy for Compositions

When analyzing $h(x) = f(g(k(x)))$, work from inside out:

Innermost: What does $k(x)$ require? → Domain $D_1$
Middle: What does $g(u)$ require of its input? What values does $k(x)$ produce? → Restrict to $D_2$
Outermost: What does $f(v)$ require? What values does $g(k(x))$ produce? → Final domain $D_3$

The answer is the intersection: $D_1 \cap D_2 \cap D_3$

Example: $\sqrt{\ln(x^2)}$

$x^2$ defined for all $x$ ✓
$\ln u$ requires $u > 0$, so $x^2 > 0$ → $x \neq 0$
$\sqrt{v}$ requires $v \geq 0$, so $\ln(x^2) \geq 0$ → $x^2 \geq 1$ → $\vert x\vert \geq 1$
Final domain: $\vert x\vert \geq 1$, i.e., $(-\infty, -1] \cup [1, \infty)$

Practice Problems

Level 1 Identifying Continuity of Sum

If $f(x) = x^2$ and $g(x) = \sin x$, explain why $h(x) = x^2 + \sin x$ is continuous for all real numbers.

Thought Process

$f(x) = x^2$ is a polynomial → continuous on all of $\mathbb{R}$
$g(x) = \sin x$ is a trig function → continuous on all of $\mathbb{R}$
The sum of continuous functions is continuous

Show Answer

$f(x) = x^2$ is a polynomial, so it’s continuous on $\mathbb{R}$
$g(x) = \sin x$ is continuous on $\mathbb{R}$
By the sum rule: $h(x) = f(x) + g(x)$ is continuous on $\mathbb{R}$

Therefore $h(x) = x^2 + \sin x$ is continuous for all real numbers.

Level 2 CCI: Understanding Why Rules Work

A student says: “Since $\sin x$ and $\frac{1}{x}$ are both continuous on their domains, $\sin x \cdot \frac{1}{x}$ must be continuous for all $x \neq 0$.”

Is this reasoning correct?

(A) Yes (the product rule guarantees this)

(B) No (you need both functions defined at the SAME points)

(C) No (the product rule only works for polynomials)

(D) It depends on whether the functions are differentiable

Thought Process

The product rule says: if $f$ and $g$ are both continuous at $a$, then $fg$ is continuous at $a$.

For this to apply:

Both $f$ and $g$ must be defined at the point
Both must be continuous at that point

Here, $\sin x$ is continuous everywhere, and $\frac{1}{x}$ is continuous where defined ($x \neq 0$).

At any $a \neq 0$: both are continuous at $a$, so the product is continuous at $a$.

Show Answer

Answer: (A) Yes (the product rule guarantees this)

The student’s reasoning is correct:

$\sin x$ is continuous at every point
$\frac{1}{x}$ is continuous at every $x \neq 0$
At any $a \neq 0$, both functions are continuous at $a$
By the product rule, $\sin x \cdot \frac{1}{x} = \frac{\sin x}{x}$ is continuous at $a$

Why (B) is wrong: The functions ARE defined at the same points we’re checking ($x \neq 0$). The student correctly restricted to this domain.

Key insight: “Continuous on its domain” means we only check points where the function is defined. The product rule applies at each such point.

Level 2 Composition Continuity

Show that $F(x) = e^{\cos x}$ is continuous on $\mathbb{R}$.

Thought Process

Identify the composition:

Inner function: $g(x) = \cos x$
Outer function: $f(u) = e^u$

Check each is continuous on the relevant domain, then apply composition rule.

Show Answer

Write $F(x) = f(g(x))$ where:

$g(x) = \cos x$ (inner function)
$f(u) = e^u$ (outer function)

Continuity check:

$g(x) = \cos x$ is continuous on all of $\mathbb{R}$
For any $a \in \mathbb{R}$, $g(a) = \cos a$ is some real number
$f(u) = e^u$ is continuous for all $u \in \mathbb{R}$, so it’s continuous at $g(a)$

By the composition rule: $F = f \circ g$ is continuous at every $a \in \mathbb{R}$.

Therefore $F(x) = e^{\cos x}$ is continuous on $\mathbb{R}$.

Level 3 Quotient with Restricted Domain

Determine where $G(x) = \dfrac{\sin x}{x^2 - 4}$ is continuous.

Thought Process

For a quotient $\frac{f}{g}$:

Both $f$ and $g$ must be continuous
The quotient is continuous where $g(x) \neq 0$

Find where $x^2 - 4 = 0$ and exclude those points.

Show Answer

Numerator: $\sin x$ is continuous on $\mathbb{R}$

Denominator: $x^2 - 4$ is a polynomial, continuous on $\mathbb{R}$

Quotient rule: $G(x)$ is continuous where the denominator is nonzero.

Find zeros of denominator: $$x^2 - 4 = 0 \implies (x-2)(x+2) = 0 \implies x = 2 \text{ or } x = -2$$

Conclusion: $G(x) = \dfrac{\sin x}{x^2 - 4}$ is continuous on: $$\mathbb{R} \setminus \{-2, 2\} = (-\infty, -2) \cup (-2, 2) \cup (2, \infty)$$

Level 4 Nested Composition Domain

Determine the domain on which $H(x) = \sqrt{\ln(x)}$ is continuous.

Thought Process

Work from the inside out:

$\ln(x)$ requires $x > 0$
$\sqrt{u}$ requires $u \geq 0$
So we need $\ln(x) \geq 0$, which means $x \geq 1$

Combine: need $x > 0$ AND $x \geq 1$, so $x \geq 1$.

Show Answer

Step 1: Domain of $\ln(x)$

Requires $x > 0$

Step 2: Domain of $\sqrt{u}$

Requires $u \geq 0$
Here $u = \ln(x)$, so we need $\ln(x) \geq 0$
$\ln(x) \geq 0$ when $x \geq 1$

Step 3: Combine conditions

Need both $x > 0$ and $x \geq 1$
The intersection is $x \geq 1$, i.e., $[1, \infty)$

Step 4: Verify continuity

On $[1, \infty)$: $\ln(x)$ is continuous (continuous on its domain)
$\sqrt{u}$ is continuous for $u \geq 0$
By composition rule: $H(x) = \sqrt{\ln(x)}$ is continuous on $[1, \infty)$

Answer: $H(x)$ is continuous on $[1, \infty)$

Level 5 Proof Using Continuity Rules

Prove that if $f$ is continuous at $a$ and $f(a) > 0$, then there exists an interval $(a - \delta, a + \delta)$ on which $f(x) > 0$.

Thought Process

This is asking us to show that “positivity is preserved near $a$.”

Use the definition of continuity via limits:

$\lim_{x \to a} f(x) = f(a) > 0$
By the definition of limit, for $\epsilon = \frac{f(a)}{2} > 0$, there exists $\delta > 0$ such that...
Show that within this $\delta$-neighborhood, $f(x)$ stays positive

Show Answer

Given: $f$ is continuous at $a$ and $f(a) > 0$.

To prove: There exists $\delta > 0$ such that $f(x) > 0$ for all $x \in (a - \delta, a + \delta)$.

Proof:

Since $f$ is continuous at $a$: $$\lim_{x \to a} f(x) = f(a)$$

Let $\epsilon = \dfrac{f(a)}{2} > 0$ (since $f(a) > 0$).

By the definition of limit, there exists $\delta > 0$ such that: $$\vert x - a\vert < \delta \implies \vert f(x) - f(a)\vert < \epsilon = \frac{f(a)}{2}$$

The inequality $\vert f(x) - f(a)\vert < \frac{f(a)}{2}$ means: $$-\frac{f(a)}{2} < f(x) - f(a) < \frac{f(a)}{2}$$

Adding $f(a)$ to all parts: $$\frac{f(a)}{2} < f(x) < \frac{3f(a)}{2}$$

In particular: $f(x) > \frac{f(a)}{2} > 0$ for all $x \in (a - \delta, a + \delta)$.

Conclusion: The interval $(a - \delta, a + \delta)$ has the desired property. $\square$

Common Misconceptions

Common misconception

the composition of two continuous functions is always continuous. The composition $f \circ g$ is continuous at $a$ provided $g$ is continuous at $a$ and $f$ is continuous at $g(a)$. If $f$ has a discontinuity precisely at the value $g(a)$, the composition fails at $a$. The rule is not “compose two continuous functions and get a continuous function everywhere”; it is a pointwise condition that must be verified.

Common misconception

a quotient of two continuous functions is continuous everywhere both are defined. The quotient $f/g$ is continuous where $g(x) \neq 0$, not merely where both $f$ and $g$ are individually continuous. At a zero of $g$, the quotient may have a removable discontinuity, a jump, or a vertical asymptote, depending on the behavior of $f$ at that point. The domain restriction “$g(x) \neq 0$” must be checked explicitly.

Mastery Checklist

Novice (Level 1-2):

I can state the arithmetic rules (sum, product, quotient) for combining continuous functions
I can identify the “building block” functions (polynomials, trig, exp, log)
I can evaluate limits by direct substitution when I recognize continuity

Competent (Level 3):

I can apply the composition rule to determine continuity of $f(g(x))$
I can identify the domain of continuity for quotients (where denominator ≠ 0)
I can work “inside-out” to find the domain of nested compositions

Proficient (Level 4-5):

I can determine the complete domain of continuity for complex functions like $\sqrt{\ln(x)}$
I can prove properties (like positivity preservation) using continuity definitions
I can explain WHY the continuity rules follow from limit laws

Mental Model

The “Lego Blocks” Analogy:

The basic continuous functions (polynomials, trig, exp, log) are like Lego blocks. You can snap them together using addition, multiplication, division, and composition. As long as you follow the rules (no dividing by zero, no square roots of negatives), you automatically get a continuous structure.

When you see a complex function like $\ln(\sin(x^2 + 1))$, don’t panic. Just trace the Lego blocks:

$x^2 + 1$: polynomial ✓
$\sin(\cdot)$: trig ✓
$\ln(\cdot)$: log (but need input $> 0$) ✓ when $\sin(x^2+1) > 0$

Connections

Looking back:

Continuity at a point provides the definition we’re extending
Limit laws (from Section 5) justify the arithmetic rules for continuity

Looking ahead:

Intermediate Value Theorem applies to continuous functions built this way
The Chain Rule (Chapter 2) mirrors the composition rule: derivatives of compositions involve derivatives of the pieces
These rules will be used constantly when differentiating complex functions

Real-world connections:

Physical quantities like position, temperature, and pressure are often modeled by continuous functions
Combining physical models (sums, products) preserves continuity

Real-World Example: Gravitational Force

🌍 Is Gravity Continuous? (Stewart Exercise 46)

The gravitational force exerted by Earth on a unit mass at distance $r$ from Earth’s center is:

$$F(r) = \begin{cases} \dfrac{GMr}{R^3} & \text{if } r < R \text{ (inside Earth)} \\[10pt] \dfrac{GM}{r^2} & \text{if } r \geq R \text{ (outside Earth)} \end{cases}$$

where $M$ = Earth’s mass, $R$ = Earth’s radius, $G$ = gravitational constant.

Is $F$ continuous?

At the boundary $r = R$:

From inside: $\lim_{r \to R^-} F(r) = \dfrac{GMR}{R^3} = \dfrac{GM}{R^2}$

From outside: $\lim_{r \to R^+} F(r) = \dfrac{GM}{R^2}$

Value at boundary: $F(R) = \dfrac{GM}{R^2}$

Both limits equal the function value! ✓

So $F$ is continuous at $r = R$, meaning gravity transitions smoothly as you pass through Earth’s surface.

Physical insight: Nature “chose” the formulas so that they match at the boundary, with no abrupt changes in gravitational force.

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Types of Discontinuities	Skills Index	Intermediate Value Theorem

Last updated: 2026-01-22