Defining continuity at a point — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: The three-part formal definition of continuity at a point, one-sided continuity for domain endpoints, checking continuity for piecewise, rational, and radical functions, and justifying continuity for AP Calculus AB exam questions.

You should already know: 1. How to evaluate one-sided and two-sided limits, 2. How to evaluate function values at a given point, 3. How to interpret piecewise function definitions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Defining continuity at a point?

Continuity at a point is the core foundational property of functions that underpins almost all advanced results in calculus, from the Intermediate Value Theorem to the definition of the derivative and the Fundamental Theorem of Calculus. Per the AP Calculus AB Course and Exam Description (CED), this topic is part of Unit 1: Limits and Continuity, which counts for 10–12% of the total AP exam weight, and questions on this topic appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a lead-in to derivative or integral questions. 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. Formally, it requires three specific conditions to all be true, which we break down in the sections below. Notation-wise, we write "$f$ is continuous at $x=a$" to indicate the property, and the equivalent limit notation is ${\lim }_{x\to a}f(x)=f(a)$, which condenses the three conditions into a single equality. Synonyms include "no break at $a$" or "connected at $a$", but only the formal definition earns full credit on justifications.

2. The Three-Part Formal Definition of Continuity at $x=a$

The formal definition requires three independent conditions, all of which must hold for $f$ to be continuous at an interior point $x=a$ (a point that is not an endpoint of the function's domain). If even one condition fails, $f$ is discontinuous at $x=a$. The three conditions are:

1. $f(a)$ is defined (meaning $a$ is in the domain of $f$; there is a well-defined function value at $x=a$),
2. ${\lim }_{x\to a}f(x)$ exists (meaning the two-sided limit exists, so ${\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)$, and both are finite),
3. ${\lim }_{x\to a}f(x)=f(a)$ (the limit value equals the actual function value at $a$).

All three conditions are captured in the single equality: $$\underset{x\to a}{\lim }f(x)=f(a)$$

Intuition: The first condition makes sure we have a function value to compare to the limit. The second makes sure the left and right sides of the graph approach the same finite value near $a$. The third makes sure that the point $f(a)$ actually sits where the graph is approaching, not somewhere else (like a hole with the point moved above or below). This definition applies to all types of functions.

Worked Example

Let $f(x)=\{\begin{array}{ll}\frac{x^2-4}{x-2}& x\ne 2\\ 3& x=2\end{array}$. Does $f$ satisfy all three conditions for continuity at $x=2$?

1. Check if $f(2)$ is defined: By the piecewise definition, $f(2)=3$, so condition 1 holds.
2. Evaluate the two-sided limit: Factor the numerator to get $\frac{(x-2)(x+2)}{x-2}=x+2$ for $x\ne 2$, so ${\lim }_{x\to 2}(x+2)=4$. Left and right limits both equal 4, so the limit exists, condition 2 holds.
3. Compare the limit to the function value: ${\lim }_{x\to 2}f(x)=4\ne 3=f(2)$. Condition 3 fails.
4. Conclusion: $f$ is not continuous at $x=2$.

Exam tip: On FRQs, always explicitly name and check all three conditions in order if asked to justify continuity—skipping any condition will cost you a point, even if your final conclusion is correct.

3. One-Sided Continuity for Endpoints

When working with functions defined on closed intervals $[a,b]$, we cannot take a two-sided limit at the endpoints $a$ and $b$, because the function is not defined outside the interval. For this reason, we use the concept of one-sided continuity to define continuity at endpoints. A function $f$ is right-continuous at $x=a$ if: $$\underset{x\to a^{+}}{\lim }f(x)=f(a)$$ Similarly, $f$ is left-continuous at $x=b$ if: $$\underset{x\to b^{-}}{\lim }f(x)=f(b)$$ A function defined on a closed interval $[a,b]$ is continuous on the full interval if it is continuous at every interior point, right-continuous at the left endpoint $a$, and left-continuous at the right endpoint $b$. This definition is commonly tested with radical functions (which have restricted domains) and piecewise functions defined on closed intervals.

Worked Example

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

1. Confirm $f(-3)$ is defined: $f(-3)=\sqrt{9-(-3{)}^2}=\sqrt{0}=0$, so the function value exists.
2. Since $x=-3$ is the left endpoint of the domain, only the right-hand limit exists (there are no $x<-3$ in the domain, so no left-hand limit to check).
3. Evaluate the right-hand limit: ${\lim }_{x\to -3^{+}}\sqrt{9-x^2}=\sqrt{{\lim }_{x\to -3^{+}}(9-x^2)}=\sqrt{9-9}=0$.
4. Compare: ${\lim }_{x\to -3^{+}}f(x)=0=f(-3)$, so $f$ is right-continuous at $x=-3$, which meets the definition of continuity at an endpoint. Conclusion: Yes, $f$ is continuous at $x=-3$.

Exam tip: Never try to check a two-sided limit at an endpoint of a function's domain—AP exam graders will mark you down for incorrectly referencing a non-existent limit outside the domain.

4. Checking Continuity of Piecewise Functions at Boundary Points

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 a continuous function (like a polynomial, trigonometric function, or rational function that is continuous on its interval of the piece), so the only possible point of discontinuity is at the boundary between pieces. The standard process for checking continuity at an interior boundary point $x=a$ is:

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

Worked Example

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

1. Check if $g(1)$ is defined: From the second piece ($x\ge 1$), $g(1)=(1{)}^2+2=3$, so $g(1)$ is defined.
2. Evaluate the left-hand limit (for $x<1$, use the first piece): ${\lim }_{x\to 1^{-}}(2x+1)=2(1)+1=3$.
3. Evaluate the right-hand limit (for $x>1$, use the second piece): ${\lim }_{x\to 1^{+}}(x^2+2)=1^2+2=3$.
4. 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 the pieces for left/right limits is the most common mistake on these problems, leading to an incorrect conclusion.

5. Common Pitfalls (and how to avoid them)

• Wrong move: For $f(x)=\frac{x^2-9}{x-3}$ for $x\ne 3$, $f(3)=6$, 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. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Let $h(x)=\{\begin{array}{ll}\frac{\sin (2x)}{x}& x\ne 0\\ k& x=0\end{array}$. For what value of $k$ is $h(x)$ continuous at $x=0$? A) $0$ B) $1$ C) $2$ D) $\frac{1}{2}$

Worked Solution: For $h(x)$ to be continuous at $x=0$, we need ${\lim }_{x\to 0}h(x)=h(0)=k$. Using the standard trigonometric limit result ${\lim }_{x\to 0}\frac{\sin (ax)}{x}=a$, we get ${\lim }_{x\to 0}\frac{\sin (2x)}{x}=2$. This means $k$ must equal 2 to satisfy ${\lim }_{x\to 0}h(x)=h(0)$: $h(0)=2$ is defined, the limit exists, and the equality holds. The correct answer is C.

Question 2 (Free Response)

Let $f(x)=\{\begin{array}{ll}{x}^3+ax+4& x<2\\ b{x}^2-3x& x\ge 2\end{array}$. (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.

Worked Solution: (a) First, calculate $f(2)$ from the second piece: $f(2)=b(2{)}^2-3(2)=4b-6$. Next, calculate the left limit as $x\to 2^{-}$: ${\lim }_{x\to 2^{-}}(x^3+ax+4)=8+2a+4=12+2a$. For continuity, set the limit equal to $f(2)$: $12+2a=4b-6$. Substitute $a=b$: $12+2a=4a-6⟹18=2a⟹a=9$, so $b=9$. (b) For $a=1$, the left limit becomes $12+2(1)=14$. Set equal to $4b-6$: $14=4b-6⟹20=4b⟹b=5$. (c) $x=0$ is an interior point of the interval $x<2$, where $f(x)=x^3+ax+4$, 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 $b$.

Question 3 (Application / Real-World Style)

A coffee shop models the price of a large coffee over a 10-day period with the function $P(d)=\{\begin{array}{ll}4.50& 0\le d<4\\ 4.50+0.25(d-4)& 4\le d\le 10\end{array}$, where $P(d)$ is the price in dollars on day $d$, for $0\le d\le 10$. 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.

Worked Solution:

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 limit: ${\lim }_{d\to 4^{-}}P(d)=4.50$.
3. Evaluate the right limit: ${\lim }_{d\to 4^{+}}(4.50+0.25(d-4))=4.50+0=4.50$.
4. 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.

7. Quick Reference Cheatsheet

8. 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. More broadly, the derivative of a function at a point is only defined if the function is continuous at that point, so without mastering this definition, you will not be able to correctly answer questions about the relationship between differentiability and continuity.

• Types of discontinuities
• Continuity over an interval
• Intermediate Value Theorem (IVT)
• Differentiability implies continuity