Continuity on an Interval
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 2.4: “Continuity” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Key idea
Continuity at a single point is the local statement. Continuity on an interval is the global statement: no breaks, no jumps, no vertical asymptotes anywhere along a stretch of the real line.
At interior points of an interval, the requirement is the two-sided limit equals the function value. At the endpoints of a closed interval, the requirement is relaxed to a one-sided condition: at the left endpoint $a$, the right-hand limit must equal $f(a)$; at the right endpoint $b$, the left-hand limit must equal $f(b)$. This relaxation is natural -- you cannot approach an endpoint from the outside when the function is only defined on the interval.
The Intermediate Value Theorem, which guarantees roots and crossing points, requires continuity on a closed interval. Getting the precise statement right matters for using that theorem correctly.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
Continuous on $(a, b)$. $f$ is continuous on the open interval $(a, b)$ if $f$ is continuous at every point in $(a, b)$.
Continuous on $[a, b]$. $f$ is continuous on the closed interval $[a, b]$ if:
- $f$ is continuous at every interior point $x \in (a, b)$, and
- $\lim_{x \to a^+} f(x) = f(a)$ (right-continuous at $a$), and
- $\lim_{x \to b^-} f(x) = f(b)$ (left-continuous at $b$).
Continuous from the right at $a$: $\lim_{x \to a^+} f(x) = f(a)$.
Continuous from the left at $b$: $\lim_{x \to b^-} f(x) = f(b)$.
Key Concepts
1. Interior vs. Endpoint Continuity
At an interior point $c \in (a, b)$, continuity requires: \[ \lim_{x \to c} f(x) = f(c) \quad (\text{two-sided limit equals function value}). \]
At the left endpoint $a$ of $[a, b]$, continuity requires: \[ \lim_{x \to a^+} f(x) = f(a) \quad (\text{right-continuous at }a). \]
At the right endpoint $b$ of $[a, b]$, continuity requires: \[ \lim_{x \to b^-} f(x) = f(b) \quad (\text{left-continuous at }b). \]
The reason for the one-sided requirement at endpoints: the function is only defined on one side of the endpoint (within the interval), so only a one-sided limit makes sense.
2. Standard Examples
Example 1. $f(x) = \sqrt{x}$ is continuous on $[0, \infty)$.
At every interior point $c > 0$: $\sqrt{x}$ is a composition of continuous functions, so it is continuous.
At the left endpoint $x = 0$: $\lim_{x \to 0^+} \sqrt{x} = 0 = \sqrt{0} = f(0)$. Right-continuous at 0.
So $\sqrt{x}$ is continuous on $[0, \infty)$. It is not defined for $x < 0$, so there is no question of two-sided continuity at 0.
Example 2. $f(x) = \sqrt{1 - x^2}$ is continuous on $[-1, 1]$.
The expression under the square root is $1 - x^2 \geq 0$ exactly when $-1 \leq x \leq 1$.
For interior $x \in (-1, 1)$: $1 - x^2 > 0$, so $f$ is a composition of continuous functions, hence continuous.
At the left endpoint $x = -1$: $\lim_{x \to -1^+} \sqrt{1 - x^2} = \sqrt{0} = 0 = f(-1)$. Right-continuous.
At the right endpoint $x = 1$: $\lim_{x \to 1^-} \sqrt{1-x^2} = 0 = f(1)$. Left-continuous.
So $f$ is continuous on $[-1, 1]$.
Example 3. The floor function $f(x) = \lfloor x \rfloor$ is NOT continuous on any closed interval containing an integer.
At each integer $n$: $\lim_{x \to n^-} f(x) = n - 1 \neq n = f(n)$. The left-hand limit fails, so $f$ is not left-continuous at $n$.
On any interval $(n, n+1)$ not containing an integer: $\lfloor x \rfloor = n$ is constant, hence continuous.
3. Describing the Maximal Domain of Continuity
Given a function, state the largest intervals on which it is continuous.
Example 4. Find the intervals of continuity for $f(x) = \dfrac{1}{x^2 - 4}$.
Denominator factors as $(x-2)(x+2)$, zero at $x = \pm 2$.
$f$ is a quotient of polynomials, continuous wherever denominator is nonzero. Intervals: $(-\infty, -2)$, $(-2, 2)$, and $(2, \infty)$.
Example 5. Find the intervals of continuity for $g(x) = \ln(x - 1)$.
$\ln$ is continuous on $(0, \infty)$. The inner function $x - 1 > 0$ when $x > 1$. So $g$ is continuous on $(1, \infty)$.
4. Continuity on $[a, b]$ and the Intermediate Value Theorem
The Intermediate Value Theorem (IVT) requires $f$ to be continuous on a closed interval $[a, b]$. The endpoint conditions matter: if $f$ has a jump at $a$ or $b$, the theorem may fail.
IVT statement (for reference). If $f$ is continuous on $[a, b]$ and $N$ is any number strictly between $f(a)$ and $f(b)$, then there exists $c \in (a, b)$ such that $f(c) = N$.
The geometric meaning: a continuous curve connecting $(a, f(a))$ to $(b, f(b))$ must cross every horizontal line $y = N$ between those two heights. You cannot draw the curve without lifting your pen.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Requiring two-sided limit at an endpoint | Claiming $\sqrt{x}$ is not continuous at 0 because $\lim_{x \to 0^-} \sqrt{x}$ fails | At an endpoint, only the one-sided limit from inside the interval is required |
| Confusing “continuous on $(a,b)$” with “continuous on $[a,b]$” | Applying IVT to an open-interval continuity result | IVT requires the closed interval $[a,b]$; check endpoint behavior separately |
| Listing $x$-values where $f$ is discontinuous instead of intervals of continuity | “Continuous at all $x \neq 2, -2$” instead of three intervals | State the maximal intervals explicitly: $(-\infty,-2)$, $(-2,2)$, $(2,\infty)$ |
Leveled Practice
Level 1 -- Reading Continuity Intervals
Problem 1. State the largest intervals of continuity for $h(x) = \dfrac{x+3}{x^2+x-2}$.
Show answer
Factor denominator: $(x+2)(x-1)$. Zero at $x = -2$ and $x = 1$.
Check numerator at each: $h(-2) = 1/0$ (undefined), $h(1) = 4/0$ (undefined).
Continuous on $(-\infty, -2)$, $(-2, 1)$, and $(1, \infty)$.
Problem 2. On which interval is $f(x) = \sqrt{4 - x}$ continuous?
Show answer
$4 - x \geq 0$ when $x \leq 4$. $f$ is continuous on $(-\infty, 4]$. (At $x = 4$: $\lim_{x \to 4^-} \sqrt{4-x} = 0 = f(4)$; left-continuous.)
Problem 3. Is the piecewise function $f(x) = \begin{cases} x^2 & x < 1 \\ 3 - x & x \geq 1 \end{cases}$ continuous on $(-\infty, \infty)$?
Show answer
Check at $x = 1$: left limit $= 1$, right limit $= 2$. Since $1 \neq 2$, $f$ is not continuous at $x = 1$.
$f$ is continuous on $(-\infty, 1)$ and on $[1, \infty)$ separately, but not on all of $(-\infty, \infty)$.
Level 2 -- Endpoint Analysis
Problem 4. Show that $f(x) = \sqrt{9 - x^2}$ is continuous on $[-3, 3]$.
Show answer
For $x \in (-3, 3)$: $9 - x^2 > 0$; $f$ is a composition of continuous functions, hence continuous.
At $x = -3$: $\lim_{x \to -3^+} \sqrt{9-x^2} = \sqrt{0} = 0 = f(-3)$. Right-continuous.
At $x = 3$: $\lim_{x \to 3^-} \sqrt{9-x^2} = 0 = f(3)$. Left-continuous.
So $f$ is continuous on $[-3, 3]$.
Level 3 -- Applying to IVT Setup
Problem 5. A student wants to apply the Intermediate Value Theorem to $f(x) = \tan x$ on $[0, \pi]$. Is this valid? Explain.
Show answer
No. $\tan x$ has a vertical asymptote at $x = \pi/2 \in (0, \pi)$, so $\tan x$ is not continuous on all of $[0, \pi]$. The IVT requires continuity on the entire closed interval. The student must restrict to $[0, \pi/2)$ or $(\pi/2, \pi]$ (open intervals that exclude the asymptote).
Problem 6. Show that $f(x) = x^3 - x - 1$ has a zero in the interval $[1, 2]$.
Show answer
$f$ is a polynomial, continuous on all of $\mathbb{R}$, and in particular on $[1, 2]$.
$f(1) = 1 - 1 - 1 = -1 < 0$.
$f(2) = 8 - 2 - 1 = 5 > 0$.
Since $f(1) < 0 < f(2)$ and $f$ is continuous on $[1, 2]$, by the IVT there exists $c \in (1, 2)$ such that $f(c) = 0$.
Common Misconceptions
“continuous on $(a, b)$” and “continuous on $[a, b]$” mean the same thing. On an open interval $(a, b)$, continuity requires only the two-sided limit condition at each interior point. On a closed interval $[a, b]$, continuity additionally requires right-continuity at $a$ (the right-hand limit equals $f(a)$) and left-continuity at $b$ (the left-hand limit equals $f(b)$). The distinction matters for theorems like the Extreme Value Theorem and IVT, which require the closed-interval condition.
a function continuous at every point of $(a, b)$ is automatically continuous on $[a, b]$. Continuity at every interior point says nothing about behavior at the endpoints. The function $f(x) = \sqrt{1 - x^2}$ is continuous on $(-1, 1)$ and at $x = \pm 1$ from the appropriate one-sided direction, so it is continuous on $[-1, 1]$. But $f(x) = \tan x$ is continuous on $(-\pi/2, \pi/2)$ yet cannot be extended continuously to the closed interval because of vertical asymptotes at the endpoints.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Continuity on an interval means you can draw the graph of $f$ over that stretch without lifting your pencil -- no breaks, no holes, no vertical runs off the page.
At interior points, this means the curve must approach the same height from both sides. At the endpoints of a closed interval, the curve only exists on one side, so only one direction of approach is required. A closed interval locks you into both endpoints; the IVT exploits this: if you start below zero and end above zero and never lift the pencil, you must have crossed zero somewhere in between.
Connections
Within Calculus I (MATH161)
- Intermediate Value Theorem: Requires continuity on $[a, b]$; the IVT is the first major theorem about continuous functions.
- Differentiability: The derivative is defined only where the function is continuous. Intervals of continuity are therefore the candidate intervals for differentiation.
- Integration (MATH161 Ch. 4): The definite integral $\int_a^b f(x)\,dx$ is guaranteed to exist when $f$ is continuous on $[a, b]$. The closed-interval hypothesis is the key.
Back to Calculus I Skills | Previous: Properties of Continuous Functions | Next: Intermediate Value Theorem