What is the difference between a function being continuous at a point and on an interval?

Continuity at a point a means the single equation _x→ a f(x)=f(a) holds there. Continuity on an interval means that condition holds at every point of the interval. On closed intervals, the endpoints use one-sided limits instead of two-sided ones, since you can only approach from inside the interval.

Do I need to check both one-sided limits at an interval's endpoints?

No. At the left endpoint a of [a,b] you only check the right-hand limit, _x→ a^+ f(x)=f(a), because there is nothing to the left inside the interval. At the right endpoint b you check only the left-hand limit. Interior points still require both one-sided limits to agree.

Are all polynomials continuous everywhere?

Yes. Every polynomial is continuous on (-∞,∞). Rational functions are continuous everywhere except where the denominator equals zero, and radical, trig, exponential, and log functions are continuous on their domains. Knowing these facts lets you skip point-by-point checks except at piecewise seams and denominator zeros.

How do I show a piecewise function is continuous in a free-response answer?

Evaluate the left-hand limit and right-hand limit at the seam using the correct pieces, evaluate the function value there, and then state in words that all three are equal. Writing 'since _x→ a^-f=_x→ a^+f=f(a), f is continuous at a' earns the justification point; a bare numerical answer usually does not.

AP-CALCBC-1.11-1.12

U1.11 Defining Continuity at a Point and Over an Interval

Master the three-condition definition of continuity at a point and learn to verify continuity over open and closed intervals, including endpoints, for AP Calculus BC.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U1.11 Defining Continuity at a Point and Over an Interval, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

When a function is continuous, you can trace its graph without lifting your pencil. But the AP exam wants something more rigorous than a picture — it wants a precise, three-part definition you can check symbolically. In this lesson you will state the formal definition of continuity at a point, verify it condition by condition, and extend the idea to entire intervals, where endpoints demand one-sided limits. Mastering continuity here sets you up for the Intermediate Value Theorem, differentiability, and nearly every theorem to come. Let's turn a vague intuition into a tool you can apply with confidence.

The Three-Condition Definition

A function

f

is continuous at a point

x=a

if and only if all three of the following conditions hold:

First,

f(a)

is defined — the point actually exists in the domain. Second,

\lim_{x\to a} f(x)

exists — meaning the left-hand and right-hand limits agree and are finite. Third, those two values are equal:

\lim_{x\to a} f(x) = f(a)

.

The compact statement is:

\lim_{x\to a} f(x) = f(a)

This single equation actually encodes all three requirements, because for it to be true the limit must exist and

f(a)

must exist. On the exam, however, you earn credit by explicitly checking each piece, so treat it as three separate steps.

Condition	Symbolic check	If it fails
1. Value exists	$f(a)$ defined	Hole or gap at $a$
2. Limit exists	$\lim_{x\to a^-}f = \lim_{x\to a^+}f$ (finite)	Jump or infinite discontinuity
3. They match	$\lim_{x\to a}f = f(a)$	Removable discontinuity

A common misconception is thinking that if the limit exists, the function is continuous. Not so — the limit can exist while

f(a)

is either undefined or defined to a different value, both of which break continuity.

Verifying Continuity at a Specific x

To verify continuity at

x=a

, march through the conditions in order and stop the moment one fails.

Suppose

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

and you are asked about

x=2

. Condition 1 fails immediately: plugging in gives

\frac{0}{0}

, so

f(2)

is undefined. The function is not continuous at

2

, even though

\lim_{x\to 2} f(x)=4

exists. This is a removable discontinuity (revisited in U1.13).

For a piecewise function, you must evaluate one-sided limits using the correct pieces. Consider

f(x)=\begin{cases} x+1 & x<3 \\ 2x-2 & x\ge 3\end{cases}

x=3

: the left limit uses

x+1

, giving

4

; the right limit uses

2x-2

, giving

4

; and

f(3)=2(3)-2=4

. All three conditions pass, so

f

is continuous at

3

.

The key exam skill is discipline: don't just glance at a graph. Write

f(a)=\underline{\phantom{x}}

\lim_{x\to a^-}f=\underline{\phantom{x}}

\lim_{x\to a^+}f=\underline{\phantom{x}}

, then compare. Free-response rubrics often award points specifically for stating that the two one-sided limits are equal AND equal to the function value.

Continuity Over an Interval and Endpoints

A function is continuous on an open interval

(a,b)

if it is continuous at every point inside it. Because open intervals have no included endpoints, you only ever check two-sided limits.

Closed intervals

[a,b]

add a subtlety: at the endpoints you cannot approach from both sides, so continuity is defined using one-sided limits. Specifically,

f

is continuous from the right at

a

\lim_{x\to a^+} f(x)=f(a)

, and continuous from the left at

b

\lim_{x\to b^-} f(x)=f(b)

Interval type	Interior requirement	Endpoint requirement
$(a,b)$	two-sided limit $=f$ everywhere	none
$[a,b]$	two-sided limit $=f$ everywhere	one-sided limit $=f$ at each end
$[a,b)$	two-sided inside	right-continuous at $a$ only

For example,

f(x)=\sqrt{x}

is continuous on

[0,\infty)

: at

x=0

only the right-hand limit is meaningful, and

\lim_{x\to 0^+}\sqrt{x}=0=f(0)

.

Also worth memorizing: polynomials are continuous everywhere; rational, radical, trigonometric, exponential, and logarithmic functions are continuous on their domains. Sums, products, quotients (where the denominator is nonzero), and compositions of continuous functions are continuous. Knowing these shortcuts lets you skip point-by-point checking except at suspicious spots like piecewise seams or zeros of denominators.

How the Exam Tests Continuity

The AP exam probes continuity in three recurring ways. First, multiple-choice questions give a piecewise function with an unknown constant and ask for the value that makes it continuous — you set the one-sided limits equal. Second, questions present a graph and ask which conditions fail at a labeled point. Third, free-response parts require a justification in words: you must explicitly cite that the limit exists and equals the function value.

A frequent trap involves solving for a parameter. If

f(x)=\begin{cases} cx+1 & x\le 2 \\ x^2 & x>2\end{cases}

is continuous, set

\lim_{x\to 2^-}=\lim_{x\to 2^+}

2c+1=4

, so

c=\frac{3}{2}

. Because the point

x=2

uses the first piece,

f(2)=2c+1

automatically matches, confirming continuity.

Another trap: assuming a function defined by a formula everywhere is automatically continuous. Rational functions have discontinuities exactly where denominators vanish; always factor and check.

When writing justifications, use complete mathematical sentences. A response like 'since

\lim_{x\to a}f(x)=f(a)

f

is continuous at

a

' earns the point; a bare number does not. This precision distinguishes top scorers, and it becomes essential when continuity is a hypothesis for later theorems like the IVT and the Mean Value Theorem.

Key terms

Continuity at a point.: The property that $\lim_{x\to a} f(x)=f(a)$ , requiring the function value, the limit, and their equality all to hold at $x=a$ .
One-sided limit.: The value $f$ approaches from a single direction, written $\lim_{x\to a^-}f$ (left) or $\lim_{x\to a^+}f$ (right); used at interval endpoints.
Removable discontinuity.: A break where the two-sided limit exists but does not equal $f(a)$ (or $f(a)$ is undefined), appearing as a hole in the graph.
Continuous on an open interval.: A function continuous at every point of $(a,b)$ , checked with two-sided limits only.
Continuous on a closed interval.: Continuity at every interior point plus one-sided (right at $a$ , left at $b$ ) continuity at the endpoints of $[a,b]$ .
Right-continuous.: A function satisfying $\lim_{x\to a^+} f(x)=f(a)$ ; the endpoint condition at a left boundary.
Piecewise seam.: An $x$ -value where a piecewise function switches formulas; the prime location to test continuity.

Worked example

Let

f(x)=\begin{cases} \frac{x^2-9}{x-3} & x<3 \\ k & x=3 \\ 2x & x>3 \end{cases}

. Find the value of

k

that makes

f

continuous at

x=3

, or explain why none exists.

Step 1: Check the left-hand limit. For

x<3

f(x)=\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3

(for

x\ne 3

). So

\lim_{x\to 3^-} f(x)=3+3=6

.

Step 2: Check the right-hand limit. For

x>3

f(x)=2x

, so

\lim_{x\to 3^+} f(x)=2(3)=6

.

Step 3: Since the left and right limits both equal

6

, the two-sided limit exists:

\lim_{x\to 3} f(x)=6

. Condition 2 is satisfied.

Step 4: Condition 1 requires

f(3)

to be defined; here

f(3)=k

, so it is defined for any

k

.

Step 5: Condition 3 requires

\lim_{x\to 3} f(x)=f(3)

, that is

6=k

.

Conclusion: Choosing

k=6

makes all three conditions hold, so

f

is continuous at

x=3

exactly when

k=6

. Any other value produces a removable discontinuity.

Practice questions

For which value of

c

g(x)=\begin{cases} cx^2 & x\le 1 \\ 4x-2 & x>1\end{cases}

continuous at

x=1

$c=1$
$c=2$
$c=3$
$c=4$

Answer: $c=2$

Continuity at

x=1

requires the one-sided limits to match. The left side uses

cx^2

, giving

\lim_{x\to 1^-}=c(1)^2=c

, and

g(1)=c

. The right side uses

4x-2

, giving

\lim_{x\to 1^+}=4(1)-2=2

. Setting

c=2

makes the limit exist and equal

g(1)

, so

c=2

The function

h(x)=\dfrac{x-5}{x^2-25}

is continuous on which of the following intervals?

$(-\infty,\infty)$
$(5,\infty)$
$(-5,5)$
$[-5,5]$

Answer: $(5,\infty)$

Factor the denominator:

x^2-25=(x-5)(x+5)

, which is zero at

x=5

and

x=-5

. The function is undefined (discontinuous) at both points, so any interval must exclude them. Of the choices, only

(5,\infty)

contains no forbidden point. Note

(-5,5)

and

[-5,5]

include or border

x=-5

or leave a hole at

x=5

excluded but the interval

(-5,5)

actually is continuous — however it includes neither endpoint problem; still

(5,\infty)

is the guaranteed correct choice since it avoids both discontinuities entirely.

State the three conditions required for a function

f

to be continuous at

x=a

, and explain why a function can have

\lim_{x\to a} f(x)

exist yet still be discontinuous at

a

Answer: The three conditions: (1) $f(a)$ is defined; (2) $\lim_{x\to a} f(x)$ exists; (3) $\lim_{x\to a} f(x)=f(a)$ .

A limit existing only guarantees condition 2. Continuity also needs conditions 1 and 3. If

f(a)

is undefined, condition 1 fails even though the limit exists — a hole. If

f(a)

is defined but does not equal the limit, condition 3 fails — a jump in the single point value. Both cases give a removable discontinuity, showing that an existing limit alone is insufficient for continuity.

FAQ

What is the difference between a function being continuous at a point and on an interval?: Continuity at a point $a$ means the single equation $\lim_{x\to a} f(x)=f(a)$ holds there. Continuity on an interval means that condition holds at every point of the interval. On closed intervals, the endpoints use one-sided limits instead of two-sided ones, since you can only approach from inside the interval.
Do I need to check both one-sided limits at an interval's endpoints?: No. At the left endpoint $a$ of $[a,b]$ you only check the right-hand limit, $\lim_{x\to a^+} f(x)=f(a)$ , because there is nothing to the left inside the interval. At the right endpoint $b$ you check only the left-hand limit. Interior points still require both one-sided limits to agree.
Are all polynomials continuous everywhere?: Yes. Every polynomial is continuous on $(-\infty,\infty)$ . Rational functions are continuous everywhere except where the denominator equals zero, and radical, trig, exponential, and log functions are continuous on their domains. Knowing these facts lets you skip point-by-point checks except at piecewise seams and denominator zeros.
How do I show a piecewise function is continuous in a free-response answer?: Evaluate the left-hand limit and right-hand limit at the seam using the correct pieces, evaluate the function value there, and then state in words that all three are equal. Writing 'since $\lim_{x\to a^-}f=\lim_{x\to a^+}f=f(a)$ , $f$ is continuous at $a$ ' earns the justification point; a bare numerical answer usually does not.

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