Defining continuity at a point — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The three-part definition of continuity at a point, one-sided continuity, testing continuity on piecewise functions, classifying discontinuities, and solving for unknown constants to make a function continuous at a point.

You should already know: How to evaluate one-sided and two-sided limits algebraically. How to simplify and evaluate piecewise functions. The definition of domain for common function types.

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 BC 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?

Defining continuity at a point is the core foundational concept that connects limits to the graphical and algebraic behavior of functions in AP Calculus. As part of Unit 1: Limits and Continuity, which accounts for 10–12% of the total AP Calculus BC exam score, questions about continuity at a point appear in both multiple-choice (MCQ) and free-response (FRQ) sections, and the definition is required for nearly every subsequent topic in the course. Intuitively, 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 from the page. Formally, it requires three non-negotiable conditions that tie the value of the function at $a$ to the value the function approaches near $a$. Synonyms you may encounter on the exam include “continuous at $x=a$”, “has no break at $a$”, and “satisfies the continuity condition at $a$”. This concept resolves the ambiguity of undefined points and mismatched function values by using limits to formalize what a “break” in a graph actually means.

2. The Three-Part Formal Definition of Continuity at a Point

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

$$\text{A function }f(x)\text{ is continuous at }x=a\text{ if and only if:}$$ $$1.f(a)\text{ is defined (}a\text{ is in the domain of }f\text{)}$$ $$2.\underset{x\to a}{\lim }f(x)\text{ exists (left-hand limit = right-hand limit)}$$ $$3.\underset{x\to a}{\lim }f(x)=f(a)$$

Intuition for each condition: The first condition eliminates points where the function does not exist, such as holes or vertical asymptotes. The second condition ensures the function approaches a single consistent value from both sides of $a$, eliminating jumps where left and right limits disagree. The third condition connects the limit (the value the function approaches) to the actual function value at $a$, catching the common case where a hole exists (limit exists) but the function is defined to a different value at that point.

Worked Example

Let $f(x)=\frac{x^2-9}{x-3}$. Does $f(x)$ satisfy all three conditions for continuity at $x=3$?

1. Check condition 1: Substitute $x=3$ to get $f(3)=\frac{0}{0}$, which is undefined. Condition 1 fails immediately.
2. For completeness, check condition 2: Simplify the function to get ${\lim }_{x\to 3}\frac{(x-3)(x+3)}{x-3}={\lim }_{x\to 3}(x+3)=6$, so the limit exists and condition 2 holds.
3. Since $f(3)$ is undefined, ${\lim }_{x\to 3}f(x)$ cannot equal an undefined value, so condition 3 also fails.
4. Conclusion: $f(x)$ is discontinuous at $x=3$.

Exam tip: On MCQ questions asking if a function is continuous at a point, check conditions in order. You can stop as soon as you find a failed condition, saving time on exam day.

3. One-Sided Continuity and Piecewise Functions

Piecewise functions are the most common context for AP exam questions testing continuity at a point, because the function’s rule changes at the boundary between pieces. To analyze boundaries correctly, we use the concept of one-sided continuity: A function is left-continuous at $x=a$ if ${\lim }_{x\to a^{-}}f(x)=f(a)$, and right-continuous at $x=a$ if ${\lim }_{x\to a^{+}}f(x)=f(a)$. For full continuity at an interior boundary point $x=a$, the function must be both left-continuous and right-continuous, which matches the three-part definition: if both one-sided limits equal $f(a)$, then the two-sided limit equals $f(a)$ by definition. For endpoints of the domain, continuity is only defined as one-sided continuity matching the side that is inside the domain. For example, $f(x)=\sqrt{x}$ with domain $x\ge 0$ is continuous at $x=0$ because it is right-continuous there, even though the left-hand limit does not exist.

A very common exam question asks you to solve for an unknown constant that makes a piecewise function continuous at its boundary point. The process always uses matching one-sided limits to the function value at the boundary.

Worked Example

Given the piecewise function $$f(x)=\{\begin{array}{ll}2x+k& x<1\\ x^2-3& x\ge 1\end{array}$$ Find the value of $k$ that makes $f(x)$ continuous at $x=1$.

1. Confirm condition 1: $f(1)$ is defined by the right-hand piece: $f(1)=1^2-3=-2$.
2. Evaluate one-sided limits: Left-hand limit (uses the $x<1$ piece): ${\lim }_{x\to 1^{-}}(2x+k)=2(1)+k=2+k$. Right-hand limit (uses the $x>1$ piece): ${\lim }_{x\to 1^{+}}(x^2-3)=1-3=-2$.
3. For the two-sided limit to exist, left and right limits must be equal: $2+k=-2⟹k=-4$.
4. Check condition 3: ${\lim }_{x\to 1}f(x)=-2=f(1)$, so all three conditions are satisfied. Conclusion: $k=-4$ gives continuity at $x=1$.

Exam tip: Always label which piece goes with which one-sided limit before you start calculating. A majority of student errors on these problems come from mixing up the left and right pieces.

4. Classifying Discontinuities at a Point

When a function fails the continuity definition at $x=a$, it has a discontinuity at that point, and AP questions regularly ask to classify the type of discontinuity. There are three main types tested on the exam, split into removable and non-removable discontinuities:

1. Removable discontinuity: Occurs when ${\lim }_{x\to a}f(x)$ exists (left limit equals right limit), but either $f(a)$ is undefined, or ${\lim }_{x\to a}f(x)\ne f(a)$. It is called removable because you can redefine $f(a)$ to equal the limit, which makes the function continuous at $a$. Removable discontinuities appear as holes on the graph.
2. Jump discontinuity: Occurs when the left-hand and right-hand limits both exist, but are not equal. The function “jumps” from one value to another at $x=a$, and no redefinition of $f(a)$ can make it continuous. This is non-removable.
3. Infinite discontinuity: Occurs when one or both one-sided limits approach $\pm \infty$, which almost always happens at a vertical asymptote. This is also a non-removable discontinuity.

Worked Example

Classify the discontinuity of $$g(x)=\{\begin{array}{ll}\frac{x^2-4x+3}{x-1}& x\ne 1\\ 0& x=1\end{array}$$ at $x=1$.

1. Check condition 1: $g(1)=0$, so it is defined, condition 1 holds.
2. Simplify to find the limit: ${\lim }_{x\to 1}\frac{(x-1)(x-3)}{x-1}={\lim }_{x\to 1}(x-3)=-2$, so the limit exists, condition 2 holds.
3. Check condition 3: ${\lim }_{x\to 1}g(x)=-2\ne 0=g(1)$, so condition 3 fails, meaning $g(x)$ is discontinuous at $x=1$.
4. Classify: Since the two-sided limit exists, but does not equal $g(1)$, this is a removable discontinuity. We could remove the discontinuity by redefining $g(1)=-2$.

Exam tip: If the question asks whether a discontinuity is removable, remember the rule: only discontinuities with a finite, existing two-sided limit are removable. All other discontinuities are non-removable.

5. Common Pitfalls (and how to avoid them)

• Wrong move: After confirming that ${\lim }_{x\to a}f(x)$ exists, stopping and concluding the function is continuous at $x=a$, without checking that the limit equals $f(a)$. Why: Students often confuse the existence of the limit with continuity, forgetting that the function value can be different at the point. Correct move: Always check all three conditions explicitly, especially the third condition that matches the limit to the function value.
• Wrong move: When finding $k$ to make a piecewise function continuous at boundary $x=a$, using the boundary piece's expression to compute the left-hand limit instead of the $x<a$ piece. Why: Students mix up which piece corresponds to which side of the boundary. Correct move: Write "left limit uses $x<a$, right limit uses $x>a$" at the top of your work before starting calculations.
• Wrong move: Classifying a vertical asymptote as a removable discontinuity because the function is undefined there. Why: Students associate "undefined at $a$" with removable discontinuity, forgetting that removable requires the limit to exist. Correct move: If one or both one-sided limits are infinite, always classify as an infinite (non-removable) discontinuity regardless of whether the point is defined.
• Wrong move: Claiming a function is discontinuous at a point not in its domain, for example calling $f(x)=1/x$ discontinuous at $x=0$ without context. Why: Students learn discontinuities occur where the function is undefined, but continuity is only defined for points in the domain. Correct move: Only discuss continuity or discontinuity at points in the domain of the function; if the point is not in the domain, state it is not in the domain instead of incorrectly calling it discontinuous unless explicitly asked.
• Wrong move: For endpoints of the domain, requiring both left and right limits to exist for continuity. Why: Students memorize the definition for interior points and incorrectly apply it to endpoints. Correct move: For a left endpoint $a$ of the domain, continuity only requires right-continuity; for a right endpoint, only left-continuity.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Let $f(x)=\{\begin{array}{ll}\frac{3x^2-12}{x-2}& x\ne 2\\ 12& x=2\end{array}$. Which of the following statements is true? A) $f$ is continuous at $x=2$ because ${\lim }_{x\to 2}f(x)$ exists B) $f$ is continuous at $x=2$ because all three conditions for continuity are satisfied C) $f$ has a removable discontinuity at $x=2$ because ${\lim }_{x\to 2}f(x)\ne f(2)$ D) $f$ has an infinite discontinuity at $x=2$ because the denominator is 0 when $x=2$

Worked Solution: First, $f(2)=12$, so it is defined, satisfying condition 1. Next, simplify the limit: ${\lim }_{x\to 2}\frac{3(x^2-4)}{x-2}={\lim }_{x\to 2}3(x+2)=12$, so the limit exists, satisfying condition 2. Finally, the limit equals $f(2)$, satisfying condition 3. All three conditions are met, so $f$ is continuous. Option A is incorrect because it does not check the third condition, options C and D are incorrect because no discontinuity exists. Correct answer: $\boxed{B}$

Question 2 (Free Response)

Let $h(x)=\{\begin{array}{ll}(kx-1)(x+2)& x\le 0\\ \ln (x+4)+m& x>0\end{array}$ (a) State the three conditions that must be satisfied for $h(x)$ to be continuous at $x=0$. (b) Given that $h(x)$ is continuous at $x=0$, and ${\lim }_{x\to 0^{-}}\frac{h(x)-h(0)}{x}=3$, find the values of constants $k$ and $m$. (c) Classify the behavior of $h(x)$ at $x=-4$. Justify your answer.

Worked Solution: (a) The three conditions for continuity at $x=0$ are:

1. $h(0)$ is defined (0 is in the domain of $h$)
2. ${\lim }_{x\to 0}h(x)$ exists (left-hand limit equals right-hand limit)
3. ${\lim }_{x\to 0}h(x)=h(0)$

(b) First, calculate $h(0)$ from the left piece: $h(0)=(0-1)(0+2)=-2$. Next, calculate the right-hand limit: ${\lim }_{x\to 0^{+}}h(x)=\ln (0+4)+m=2\ln 2+m$. For continuity, set equal to $h(0)$: $2\ln 2+m=-2⟹m=-2-2\ln 2$. Now evaluate the difference quotient: $$\underset{x\to 0^{-}}{\lim }\frac{(kx-1)(x+2)-(-2)}{x}=\underset{x\to 0^{-}}{\lim }\frac{k{x}^2+(2k-1)x}{x}=\underset{x\to 0^{-}}{\lim }(kx+2k-1)=2k-1$$ Set equal to 3: $2k-1=3⟹k=2$. Final values: $k=2$, $m=-2-2\ln 2$.

(c) $x=-4$ is not in the domain of $h(x)$, because the right piece $\ln (x+4)$ requires $x>-4$. The right-hand limit as $x\to -4^{+}$ is ${\lim }_{x\to -4^{+}}\ln (x+4)+m=-\infty$, so $x=-4$ is a vertical asymptote with an infinite (non-removable) discontinuity.

Question 3 (Application / Real-World Style)

The concentration $C(t)$ (in mg/L) of a drug in a patient's bloodstream $t$ hours after an initial oral dose, with a planned adjustment to the dosage at $t=2$ hours, is given by: $$C(t)=\{\begin{array}{ll}\frac{8t}{t+2}& 0\le t<2\\ 12-\frac{4}{t}+A& t\ge 2\end{array}$$ Where $A$ is a constant adjustment to the concentration after $t=2$. Find the value of $A$ that makes $C(t)$ continuous at $t=2$, and interpret the result in context.

Worked Solution: First, calculate the left-hand limit as $t$ approaches 2 from below: ${\lim }_{t\to 2^{-}}\frac{8t}{t+2}=\frac{16}{4}=4$ mg/L. Next, calculate $C(2)$ from the right-hand rule: $C(2)=12-\frac{4}{2}+A=10+A$. For continuity at $t=2$, set the limit equal to $C(2)$: $4=10+A⟹A=-6$. Interpretation: A value of $A=-6$ means there is no instantaneous jump in the patient's drug concentration at $t=2$ hours: the concentration immediately after the dosage adjustment matches the concentration immediately before the adjustment.

7. Quick Reference Cheatsheet

8. What's Next

Mastering the definition of continuity at a point is an absolute prerequisite for all upcoming calculus topics, starting with continuity over an interval, which you will study next. You will immediately apply this definition to confirm the hypotheses of key theorems like the Intermediate Value Theorem and the Extreme Value Theorem, both of which require continuity on a closed interval to apply. Continuity is also a necessary condition for differentiability: every differentiable function is continuous at a point, so you will use this definition to test differentiability at boundary points of piecewise functions later in Unit 2. This concept also underpins integration, since all continuous functions are integrable.

• Continuity over an interval
• Intermediate Value Theorem
• Definition of the derivative
• Differentiability and continuity