Properties of Continuous Functions

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Textbook Reference

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Key idea

If you know that two functions are continuous at a point, you can immediately conclude that their sum, difference, product, and (when the denominator is nonzero) quotient are also continuous there. This is not obvious from the definition; it follows directly from the limit laws.

The composite rule is especially important: if $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then $f \circ g$ is continuous at $a$. This means evaluating a limit of a composite is as simple as plugging in -- you do not need to decompose it. Almost every function you encounter in calculus is built from simple continuous pieces by sum, product, quotient, and composition, so these rules explain why direct substitution works so broadly.

Prerequisite Check

Before this lesson, make sure you can do all of the following:

• State the three-part definition of continuity at a point $a$
• Apply limit laws for sums, products, and quotients
• Evaluate a composite function $f(g(x))$ at a given $x$
• Identify when a rational function is undefined (denominator zero)

Quick Reference

Continuity laws. If $f$ and $g$ are both continuous at $a$, then so are:

1. $f + g$ and $f - g$
2. $c \cdot f$ for any constant $c$
3. $f \cdot g$
4. $f / g$, provided $g(a) \neq 0$

Composite rule. If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then $f \circ g$ is continuous at $a$: \[ \lim_{x \to a} f(g(x)) = f\!\left(\lim_{x \to a} g(x)\right) = f(g(a)). \]

The key consequence. You can evaluate the limit of a continuous composite by “pushing the limit inside”: $\lim_{x \to a} f(g(x)) = f(g(a))$ whenever both $f$ and $g$ meet the continuity hypotheses.

Key Concepts

1. The Arithmetic Laws for Continuity

These laws follow directly from the corresponding limit laws. At a point $a$ where both $f$ and $g$ are continuous:

\[ \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = f(a) + g(a). \]

So $f + g$ is continuous at $a$ (its limit equals its value at $a$). The same argument works for $f - g$, $f \cdot g$, and $f/g$.

Important note on division. The quotient $f/g$ is continuous at $a$ only if $g(a) \neq 0$. If $g(a) = 0$, the quotient may have a vertical asymptote or a removable discontinuity at $a$.

Example 1. If $f(x) = x^2 + 1$ and $g(x) = \cos x$, then $h(x) = (x^2 + 1)\cos x$ is continuous everywhere (product of two everywhere-continuous functions).

Example 2. $r(x) = \dfrac{x^2 + 3x}{x - 2}$ is continuous at every $x \neq 2$ (quotient of polynomials; denominator vanishes at $x = 2$).

2. Continuity of Composites

The composite rule. If $\lim_{x \to a} g(x) = L$ and $f$ is continuous at $L$, then \[ \lim_{x \to a} f(g(x)) = f(L). \]

When $g$ itself is continuous at $a$, $L = g(a)$, and the formula becomes $\lim_{x \to a} f(g(x)) = f(g(a))$.

This means the limit “passes through” the outer function $f$, because $f$ is continuous at the limiting value.

Example 3. Evaluate $\displaystyle\lim_{x \to 0} \sin(x^2 + 1)$.

The inner function $g(x) = x^2 + 1$ is continuous at $x = 0$ (polynomial). The outer function $f(t) = \sin t$ is continuous at $g(0) = 1$. So: \[ \lim_{x \to 0} \sin(x^2 + 1) = \sin(0 + 1) = \sin(1). \]

Example 4. Evaluate $\displaystyle\lim_{x \to 4} \sqrt{x^2 - 7}$.

$g(x) = x^2 - 7$ is continuous at $x = 4$; $g(4) = 9 > 0$. The square root function $f(t) = \sqrt{t}$ is continuous at $t = 9$ (positive). So: \[ \lim_{x \to 4} \sqrt{x^2 - 7} = \sqrt{4^2 - 7} = \sqrt{9} = 3. \]

3. Why the Hypothesis on the Outer Function Matters

The composite rule requires $f$ to be continuous at the limit of $g$. If $f$ has a discontinuity exactly at $g(a)$, the rule breaks down.

Example 5. Let $f(x) = \begin{cases} 1 & x > 0 \\ 0 & x \leq 0 \end{cases}$ and $g(x) = x\sin(1/x)$ for $x \neq 0$, $g(0) = 0$.

We know $\lim_{x \to 0} g(x) = 0 = g(0)$, so $g$ is continuous at 0.

But $f$ is discontinuous at 0 (jump: left limit is 0, right limit is 1).

Now consider $\lim_{x \to 0} f(g(x))$. Because $g(x)$ oscillates and passes through both positive and non-positive values near $x = 0$, $f(g(x))$ oscillates between 0 and 1 and does not settle. The composite rule fails because $f$ is not continuous at $g(0) = 0$.

4. Building Complex Continuous Functions

Every function built by finitely many arithmetic operations and compositions from continuous pieces is continuous wherever the result is defined.

Example 6. Show that $h(x) = \dfrac{\sin(x^2)}{\sqrt{x + 4}}$ is continuous on $(-4, \infty)$.

• $\sin$ is continuous everywhere.
• $x^2$ is a polynomial, continuous everywhere.
• By the composite rule, $\sin(x^2)$ is continuous everywhere.
• $x + 4$ is a polynomial, continuous everywhere, and positive for $x > -4$.
• $\sqrt{x + 4}$ is continuous for $x > -4$ (and defined there).
• The quotient is continuous wherever the denominator is nonzero: for $x > -4$.

So $h$ is continuous on $(-4, \infty)$.

Common Errors

Leveled Practice

Level 1 -- Applying the Laws

Problem 1. Explain why $h(x) = x^3 \cos x + 7$ is continuous everywhere, citing the relevant laws.

Problem 2. Where is $r(x) = \dfrac{x^2 - 1}{x^2 + x - 6}$ continuous?

Level 2 -- Composite Rule

Problem 3. Evaluate $\displaystyle\lim_{x \to \pi} \cos(x/2 + \pi)$.

Problem 4. Evaluate $\displaystyle\lim_{x \to 2} \sqrt{\frac{x^2 + 5}{x - 1}}$.

Level 3 -- Analysis

Problem 5. Let $f(x) = \dfrac{\sin x}{x}$ for $x \neq 0$ and $f(0) = 1$. Show that $f$ is continuous at $x = 0$ by verifying all three parts of the definition. Use the known limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

Problem 6. Suppose $f$ is continuous at 3 and $g$ is defined by $g(x) = f(x^2 - 5)$. At which value of $x$ does the composite rule guarantee $g$ is continuous? Explain.

Common Misconceptions

Mastery Checklist

You have mastered this skill when you can do all of the following without referring to notes:

• State the five continuity laws (sum, difference, constant multiple, product, quotient)
• State the composite rule and both its hypotheses
• Apply continuity laws to determine where a rational function is continuous
• Use the composite rule to evaluate limits of composed functions by direct substitution
• Verify continuity at a point using all three parts of the definition
• Construct a continuous function from simpler continuous pieces and justify continuity

Mental Model

The continuity laws say: nice operations preserve the “no surprises” property.

If two functions have no surprises at $a$ (they equal their limits there), then combining them by addition, multiplication, or division (when safe) produces a combined function that also has no surprises at $a$. The limit laws guarantee this.

The composite rule adds: passing a limit through an outer function is safe if the outer function has no surprises at the limiting value. The limit of the inner function becomes the input of the outer function, and continuity at that input is exactly what’s needed to push the limit all the way through.

This is why you can evaluate $\lim_{x \to 0} e^{\sin x}$ as $e^{\sin 0} = e^0 = 1$ without any further work: every function in the chain is continuous at every relevant point.

Connections

Within Calculus I (MATH161)

• Direct substitution: The breadth of direct substitution -- why it works for polynomials, rational functions, trig, exponential, and their combinations -- is entirely explained by the continuity laws plus the fact that each base family is continuous on its domain.
• Intermediate Value Theorem: This uses continuity on an interval; the laws here guarantee that specific composed or combined functions satisfy the hypothesis.
• Derivatives and differentiability: Differentiable functions are continuous, but not vice versa. The continuity laws built here apply to differentiable functions as a baseline.

Back to Calculus I Skills | Previous: Continuity at a Point | Next: Continuity on an Interval