Defining continuity at a point

AP Calculus AB· AP Calculus AB CED — Limits and Continuity· 14 min read

1. Definition of Continuity at an Interior Point★★☆☆☆⏱ 4 min

Informally, a function is continuous at a point $x = a$ if you can draw the graph of the function through $x = a$ without lifting your pencil off the paper. Continuity is a foundational property that underpins nearly all advanced results in calculus, from the Intermediate Value Theorem to derivatives and the Fundamental Theorem of Calculus.

📘 Definition

Continuity at an interior point $x=a$

$f$ is continuous at $x = a$

Three independent conditions must all hold: 1. $f (a)$ is defined ( $a$ is in the domain of $f$ ), 2. $lim_{x \to a} f (x)$ exists (the two one-sided limits are equal and finite), 3. $lim_{x \to a} f (x) = f (a)$ . All three conditions are captured by the single equality:

Example:

Verified for any point inside the function's domain

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

Each condition serves a distinct purpose: the first confirms we have a function value to compare, the second confirms the graph approaches the same value from both sides, and the third confirms the point actually lies on the approaching curve.

📐 Worked Example

Let $f (x) = {\frac{x ^{2} - 4}{x - 2} 3 x \neq = 2 x = 2$ . Does $f$ satisfy all three conditions for continuity at $x = 2$ ?

1
Check if $f (2)$ is defined: per the piecewise definition, $f (2) = 3$ , so condition 1 holds.
2
Evaluate the two-sided limit by factoring the numerator:
3
$\frac{(x-2)(x+2)}{x-2} = x+2 \quad \text{for } x \neq 2$
4
So $lim_{x \to 2} f (x) = lim_{x \to 2} (x + 2) = 4$ . Left and right limits are equal, so the limit exists, condition 2 holds.
5
Compare the limit to the function value: $lim_{x \to 2} f (x) = 4 \neq = 3 = f (2)$ , so condition 3 fails. Conclusion: $f$ is not continuous at $x = 2$ .

Exam tip:

On AP FRQs, always explicitly check all three conditions when justifying continuity. Skipping any condition will cost you a point, even if your final conclusion is correct.

2. One-Sided Continuity for Domain Endpoints★★☆☆☆⏱ 3 min

For functions defined on closed intervals $[a, b]$ , a two-sided limit cannot exist at the endpoints, since the function is not defined outside the interval. For this reason, we use one-sided continuity to define continuity at endpoints.

📘 Definition

One-sided continuity

A function is right-continuous at $x = a$ if $lim_{x \to a^{+}} f (x) = f (a)$ , used for left endpoints of domains. A function is left-continuous at $x = b$ if $lim_{x \to b^{-}} f (x) = f (b)$ , used for right endpoints of closed domains.

A function is continuous on the full closed interval $[a, b]$ if it is continuous at all interior points, right-continuous at the left endpoint, and left-continuous at the right endpoint.

📐 Worked Example

Let $f (x) = 9 - x^{2}$ , which has domain $[- 3, 3]$ . Is $f$ continuous at the left endpoint $x = - 3$ ? Justify your answer.

1
Confirm $f (- 3)$ is defined: $f (- 3) = 9 - (- 3)^{2} = 0 = 0$ , so the function value exists.
2
Since $x = - 3$ is the left endpoint of the domain, only the right-hand limit needs to be checked (no values of $x < - 3$ are in the domain).
3
Evaluate the right-hand limit:
4
$\lim_{x \to -3^+} \sqrt{9 - x^2} = \sqrt{ \lim_{x \to -3^+} (9 - x^2) } = \sqrt{9 - 9} = 0$
5
Compare: $lim_{x \to - 3^{+}} f (x) = 0 = f (- 3)$ , so $f$ satisfies the definition of continuity at an endpoint. Conclusion: $f$ is continuous at $x = - 3$ .

Exam tip:

Never check for a two-sided limit at a domain endpoint. AP exam graders will penalize you for referencing a non-existent limit outside the function's domain.

3. Checking Continuity at Piecewise Function Boundaries★★★☆☆⏱ 4 min

The most common AP exam question on this topic asks you to check continuity at the boundary point where the definition of a piecewise function changes. Each individual piece of a piecewise function is almost always continuous (polynomials, rationals, trig functions are continuous on their domains), so the only possible discontinuity is at the boundary between pieces.

Find $f (a)$ from the piece that includes $x = a$ .
Calculate the left-hand limit $lim_{x \to a^{-}} f (x)$ using the piece defined for $x < a$ .
Calculate the right-hand limit $lim_{x \to a^{+}} f (x)$ using the piece defined for $x > a$ .
Confirm both one-sided limits are equal (so the two-sided limit exists), then confirm the limit equals $f (a)$ .

📐 Worked Example

Let $g (x) = {2 x + 1 x^{2} + 2 x < 1 x \geq 1$ . Determine if $g$ is continuous at $x = 1$ , the boundary between the two pieces. Justify your answer.

1
Check $g (1)$ : from the second piece ( $x \geq 1$ ), $g (1) = 1^{2} + 2 = 3$ , so $g (1)$ is defined.
2
Evaluate the left-hand limit (use the first piece for $x < 1$ ):
3
$\lim_{x \to 1^-} (2x + 1) = 2(1) + 1 = 3$
4
Evaluate the right-hand limit (use the second piece for $x > 1$ ):
5
$\lim_{x \to 1^+} (x^2 + 2) = 1^2 + 2 = 3$
6
Since left and right limits are equal, $lim_{x \to 1} g (x) = 3$ , which equals $g (1) = 3$ . All three conditions hold. Conclusion: $g (x)$ is continuous at $x = 1$ .

Exam tip:

Always match the direction of the limit to the inequality of the piece. Swapping left and right pieces is the most common mistake on these problems.

4. AP-Style Worked Practice Problems★★★☆☆⏱ 3 min

📐 Worked Example

Let $h (x) = ⎩ ⎨ ⎧ \frac{sin ( 2 x )}{x} k x \neq = 0 x = 0$ . For what value of $k$ is $h (x)$ continuous at $x = 0$ ? A) $0$ B) $1$ C) $2$ D) $\frac{1}{2}$

1
For continuity at $x = 0$ , we require $lim_{x \to 0} h (x) = h (0) = k$ . Using the standard trigonometric limit $lim_{x \to 0} \frac{s i n ( a x )}{x} = a$ , we get:
2
$\lim_{x \to 0} \frac{\sin(2x)}{x} = 2$
3
For all three conditions to hold, $k$ must equal 2: $h (0) = 2$ is defined, the limit exists, and $lim_{x \to 0} h (x) = h (0)$ . The correct answer is C.

📐 Worked Example

Let $f (x) = {x^{3} + a x + 4 b x^{2} - 3 x x < 2 x \geq 2$ . (a) Find all values of $a$ and $b$ such that $f (x)$ is continuous at $x = 2$ if $a = b$ . (b) Suppose instead that $a = 1$ , what value of $b$ makes $f$ continuous at $x = 2$ ? (c) Is $f$ continuous at $x = 0$ for any values of $a$ and $b$ ? Justify your answer.

1
Part (a): First, calculate $f (2)$ from the second piece:
2
$f(2) = b(2)^2 - 3(2) = 4b - 6$
3
Next, calculate the left limit as $x \to 2^{-}$ :
4
$\lim_{x \to 2^-} (x^3 + ax + 4) = 8 + 2a + 4 = 12 + 2a$
5
Set the limit equal to $f (2)$ for continuity: $12 + 2 a = 4 b - 6$ . Substitute $a = b$ : $12 + 2 a = 4 a - 6 ⟹ 18 = 2 a ⟹ a = 9$ , so $b = 9$ .
6
Part (b): For $a = 1$ , the left limit becomes $12 + 2 (1) = 14$ . Set equal to $4 b - 6$ : $14 = 4 b - 6 ⟹ 20 = 4 b ⟹ b = 5$ .
7
Part (c): $x = 0$ is an interior point of the interval $x < 2$ , where $f (x)$ is a polynomial. All polynomials are continuous at all real numbers, so $lim_{x \to 0} f (x) = 0 + 0 + 4 = 4 = f (0)$ . Therefore, $f$ is continuous at $x = 0$ for all values of $a$ and $b$ .

📐 Worked Example

A coffee shop models the price of a large coffee over a 10-day period with the function $P (d) = {4.50 4.50 + 0.25 (d - 4) 0 \leq d < 4 4 \leq d \leq 10$ , where $P (d)$ is the price in dollars on day $d$ , for $0 \leq d \leq 10$ . Is the price function continuous at $d = 4$ , the day the shop starts a gradual price increase? Justify your answer, and explain what continuity means in this context.

1
Check $P (4)$ : using the second piece, $P (4) = 4.50 + 0.25 (4 - 4) = 4.50$ , so $P (4)$ is defined.
2
Evaluate the left-hand limit:
3
$\lim_{d \to 4^-} P(d) = 4.50$
4
Evaluate the right-hand limit:
5
$\lim_{d \to 4^+} \left(4.50 + 0.25(d-4)\right) = 4.50 + 0 = 4.50$
6
Compare: $lim_{d \to 4} P (d) = 4.50 = P (4)$ , so all three conditions are satisfied. Conclusion: The price function is continuous at $d = 4$ . In context, this means there is no abrupt jump in the price of coffee on day 4; the price increases gradually starting from the original $4.50 price after day 4, rather than jumping immediately to a higher price.

5. Common Pitfalls

Wrong move:

For $f (x) = \frac{x ^{2} - 9}{x - 3}$ for $x \neq = 3$ , $f (3) = 6$ , concluding $f$ is discontinuous at $x = 3$ because the original expression is undefined at $x = 3$ .

Why:

Students confuse the unsimplified algebraic expression with the given function's definition, forgetting the function explicitly defines $f (3)$ .

Correct move:

Always use the given function definition to check if $f (a)$ is defined, not the unsimplified expression.

Wrong move:

Checking only that $lim_{x \to a} f (x)$ exists, then concluding continuity at $a$ .

Why:

Students forget that $f (a)$ could be undefined, or could be a different value than the limit.

Correct move:

Always explicitly check all three conditions in order when justifying continuity on an FRQ.

Wrong move:

Claiming a function is discontinuous at an endpoint $x = a$ because the two-sided limit doesn't exist.

Why:

Students confuse two-sided continuity for interior points with the definition of continuity at endpoints.

Correct move:

For any endpoint of the function's domain, only check the one-sided limit that exists (right for left endpoint, left for right endpoint) and compare it to the function value.

Wrong move:

For a piecewise function with boundary $x = a$ , calculating the left limit using the right piece and vice versa.

Why:

Students rush and misread the inequality signs defining the pieces.

Correct move:

Underline the inequality for each piece before evaluating one-sided limits, and match the direction of the limit to the inequality.

Wrong move:

Concluding $f$ is continuous at interior point $x = a$ because $lim_{x \to a^{-}} f (x) = f (a)$ , without checking the right-hand limit.

Why:

Students mix up one-sided continuity for endpoints with the requirements for interior points.

Correct move:

For any interior point, always confirm that both one-sided limits exist, are equal, and equal $f (a)$ .

6. Quick Reference Cheatsheet

Category	Formula/Rule	Notes
Continuity at interior point $x = a$	$lim_{x \to a} f (x) = f (a)$	Requires 3 conditions: $f (a)$ defined, $lim_{x \to a} f (x)$ exists, equality holds
Right-continuous at $x = a$	$lim_{x \to a^{+}} f (x) = f (a)$	Used for left endpoints of domains, only requires the right-hand limit
Left-continuous at $x = b$	$lim_{x \to b^{-}} f (x) = f (b)$	Used for right endpoints of domains, only requires the left-hand limit
Continuity at endpoint	One-sided limit equals function value	No two-sided limit required, since the function is undefined outside the interval
Piecewise boundary check	Left limit from left piece, right limit from right piece	Match the limit direction to the inequality defining the piece
Polynomial continuity	All polynomials are continuous at all real points	Any interior point in a polynomial piece is automatically continuous
Rational function continuity	Continuous at all points where denominator $\neq = 0$	Discontinuous at points where the denominator equals zero
Radical function continuity	Continuous at all points in domain	Check one-sided continuity at endpoints of the domain

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Check continuity of piecewise function
2022 · FRQ
Find constant for continuity

What's Next

Defining continuity at a point is the foundational prerequisite for almost all of the rest of calculus. Next, you will extend this definition to continuity over an interval, which is required to apply the Intermediate Value Theorem (IVT), one of the most frequently tested theorems on the AP Calculus AB exam. You will also use this definition to classify different types of discontinuities and solve for constants that make piecewise functions continuous, a common problem on both MCQ and FRQ sections.