Limits and Continuity — AP Calculus AB Calc AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Limit definition and notation, algebraic and graphical limit computation, indeterminate forms with introductory L'Hôpital's rule, continuity and the Intermediate Value Theorem (IVT), and asymptotes with end behaviour analysis.

You should already know: Strong precalculus (functions, trig, algebra).

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 papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Limits and Continuity?

Limits describe the value a function approaches as its input gets arbitrarily close to a given point, even if the function is undefined at that point. Continuity extends this framework to describe functions where the approached value exactly matches the actual function output at every point in their domain. This unit makes up 10-12% of your AP Calculus AB multiple-choice score and is frequently embedded in free-response questions testing differentiation, integration, and function analysis.

2. Limit definition and notation

The intuitive definition of a limit, tested most often on AP Calculus AB, states that the limit of $f(x)$ as $x$ approaches $a$, written ${\lim }_{x\to a}f(x)=L$, means $f(x)$ gets arbitrarily close to $L$ as $x$ draws near to $a$ (from both left and right sides, without ever touching $a$).

One-sided limits describe behaviour from a single direction:

• ${\lim }_{x\to a^{-}}f(x)$: Left-hand limit, the value $f(x)$ approaches as $x$ moves toward $a$ from values smaller than $a$
• ${\lim }_{x\to a^{+}}f(x)$: Right-hand limit, the value $f(x)$ approaches as $x$ moves toward $a$ from values larger than $a$

A key rule for two-sided limits: ${\lim }_{x\to a}f(x)$ exists if and only if ${\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)=L$. Formal epsilon-delta limit proofs are rarely tested on AB, but you should recognize the definition wording for multiple-choice questions.

Worked Example: For $f(x)=\frac{x^2-9}{x-3}$, $f(3)$ is undefined (division by zero), but evaluating near $x=3$ gives $f(2.99)=5.99$ and $f(3.01)=6.01$, so ${\lim }_{x\to 3}f(x)=6$.

3. Computing limits — algebra and graphs

You will be asked to compute limits from both graphs and algebraic expressions on the exam:

Graphical limit computation

To find ${\lim }_{x\to a}f(x)$ from a graph:

1. Trace the graph from the left of $x=a$ to identify the y-value it approaches
2. Trace the graph from the right of $x=a$ to identify the right-hand limit
3. If the two values match, that is the two-sided limit; ignore the value of $f(a)$ even if there is a hole, jump, or isolated point at $x=a$

Algebraic limit computation

1. Direct substitution first: If $f(x)$ is continuous at $a$ (polynomials, rational functions with non-zero denominators at $a$, trig, exponential, and log functions in their domains), substitute $x=a$ directly to get the limit.
2. Indeterminate form workarounds: If direct substitution gives $\frac{0}{0}$, use factoring, rationalizing, or combining fractions to cancel problematic terms before substituting.

Worked Example 1: Compute ${\lim }_{x\to 4}\frac{x^2-16}{x-4}$ Direct substitution gives $\frac{0}{0}$. Factor the numerator to get $\frac{(x-4)(x+4)}{x-4}$, cancel the $(x-4)$ term (valid because $x\ne 4$ when taking the limit), so the limit simplifies to ${\lim }_{x\to 4}x+4=8$.

Worked Example 2: Compute ${\lim }_{x\to 0}\frac{\sqrt{x+9}-3}{x}$ Direct substitution gives $\frac{0}{0}$. Rationalize the numerator by multiplying numerator and denominator by $\sqrt{x+9}+3$: $$\frac{(\sqrt{x+9}-3)(\sqrt{x+9}+3)}{x(\sqrt{x+9}+3)}=\frac{(x+9)-9}{x(\sqrt{x+9}+3)}=\frac{x}{x(\sqrt{x+9}+3)}=\frac{1}{\sqrt{x+9}+3}$$ Substitute $x=0$ to get $\frac{1}{3+3}=\frac{1}{6}$.

4. Indeterminate forms and L'Hôpital's rule introduction

Indeterminate forms are expressions that do not yield a clear limit when you substitute directly. The two most common forms tested on AP Calculus AB are $\frac{0}{0}$ and $\frac{\pm \infty }{\pm \infty }$.

L'Hôpital's Rule (LR) is a shortcut for evaluating these forms, stated as: If ${\lim }_{x\to a}f(x)=0$ and ${\lim }_{x\to a}g(x)=0$, or ${\lim }_{x\to a}f(x)=\pm \infty$ and ${\lim }_{x\to a}g(x)=\pm \infty$, then: $$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$$ provided the second limit exists or is $\pm \infty$.

Critical exam notes: You cannot use L'Hôpital's Rule for non-indeterminate forms, and you differentiate the numerator and denominator separately (do not use the quotient rule). You may apply LR multiple times if the resulting limit is still indeterminate.

Worked Example: Compute ${\lim }_{x\to 0}\frac{\sin (2x)}{x}$ Direct substitution gives $\frac{0}{0}$, so LR applies. The derivative of $\sin (2x)$ is $2\cos (2x)$, and the derivative of $x$ is 1, so the limit becomes ${\lim }_{x\to 0}2\cos (2x)=2(1)=2$.

5. Continuity and the IVT

Continuity definition

A function $f(x)$ is continuous at $x=a$ if and only if all three conditions are met:

1. $f(a)$ is defined
2. ${\lim }_{x\to a}f(x)$ exists
3. ${\lim }_{x\to a}f(x)=f(a)$

Discontinuities fall into three categories:

• Removable (hole): Condition 3 fails, the limit exists but does not match $f(a)$ or $f(a)$ is undefined
• Jump discontinuity: Condition 2 fails, left and right limits exist but are not equal
• Infinite discontinuity: Condition 2 fails, one or both one-sided limits are $\pm \infty$

A function is continuous on a closed interval $[a,b]$ if it is continuous at every point in the open interval $(a,b)$, right-continuous at $a$, and left-continuous at $b$.

Intermediate Value Theorem (IVT)

The IVT states: If $f$ is continuous on the closed interval $[a,b]$, and $N$ is any number between $f(a)$ and $f(b)$, then there exists at least one value $c\in (a,b)$ such that $f(c)=N$. The most common exam use case is proving a function has a root (zero) on an interval: if $f(a)$ and $f(b)$ have opposite signs, there is at least one root between $a$ and $b$.

Worked Example: Prove $f(x)=e^x+x-2$ has a root on $[0,1]$

1. $f(x)$ is the sum of a continuous exponential and linear function, so it is continuous on $[0,1]$
2. $f(0)=1+0-2=-1$, $f(1)=e+1-2\approx 1.718$
3. 0 lies between $-1$ and $1.718$, so by the IVT, there exists a $c\in (0,1)$ where $f(c)=0$. Exam tip: You will lose points on free-response questions if you do not explicitly state the continuity precondition for the IVT.

6. Asymptotes and end behaviour

Asymptotes describe the behaviour of functions near discontinuities and as inputs grow very large or very small. Three types are tested on AP Calculus AB:

1. Vertical Asymptote (VA): $x=a$ is a VA if ${\lim }_{x\to a^{+}}f(x)=\pm \infty$ or ${\lim }_{x\to a^{-}}f(x)=\pm \infty$. For rational functions, VA occur at points where the denominator is zero and the numerator is non-zero (if both are zero, the point is a removable hole, not a VA).
2. Horizontal Asymptote (HA): $y=L$ is a HA if ${\lim }_{x\to \infty }f(x)=L$ or ${\lim }_{x\to -\infty }f(x)=L$, describing end behaviour. For rational functions:

• If degree of numerator < degree of denominator: HA at $y=0$
• If degrees are equal: HA is the ratio of leading coefficients
• If degree of numerator is higher by 1: Slant asymptote (rarely tested on AB)
• If degree of numerator is higher by 2+: No HA

3. Slant Asymptote: Linear asymptote for rational functions where numerator degree is exactly 1 higher than denominator degree, found via polynomial long division.

Worked Example: Find all asymptotes of $f(x)=\frac{3x^2-7x+2}{x^2-4}$ Factor numerator: $(3x-1)(x-2)$, denominator: $(x-2)(x+2)$. Cancel the $(x-2)$ term, so there is a hole at $x=2$. Remaining denominator zero at $x=-2$, so VA at $x=-2$. Degrees of numerator and denominator are equal, leading coefficients 3 and 1, so HA at $y=3$.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Assuming ${\lim }_{x\to a}f(x)=f(a)$ for all functions. Why: Students get used to direct substitution for polynomials and apply it to discontinuous functions. Correct move: Check if the function is continuous at $a$, or compute left/right limits for piecewise functions or points of discontinuity.
• Wrong move: Applying L'Hôpital's Rule to non-indeterminate forms (e.g. limits that evaluate to $\frac{3}{0}$ or $\frac{\infty }{0}$). Why: Students treat LR as a universal limit shortcut without checking preconditions. Correct move: Confirm direct substitution gives exactly $\frac{0}{0}$ or $\frac{\pm \infty }{\pm \infty }$ before using LR.
• Wrong move: Forgetting to state the continuity condition when applying the IVT on free response. Why: Students focus on the sign change and skip the required theorem precondition. Correct move: Always explicitly state "f is continuous on [a,b]" as your first step when using IVT.
• **Wrong move: Identifying common zeros of numerator and denominator of rational functions as vertical asymptotes. Why: Students associate all denominator zeros with VA without checking the numerator. Correct move: Factor and cancel common terms first; canceled terms correspond to holes, remaining denominator zeros are VA.
• Wrong move: Only computing one side of the limit for piecewise functions or absolute value expressions. Why: Students assume left and right limits are always equal. Correct move: Compute both one-sided limits for piecewise functions, absolute values, or graphs with jumps before concluding the two-sided limit value.

8. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

What is the value of ${\lim }_{x\to 1}\frac{x^3-1}{x-1}$? A) 0 B) 1 C) 3 D) Does not exist

Worked Solution: Direct substitution gives $\frac{0}{0}$, indeterminate. Factor the numerator using difference of cubes: $x^3-1=(x-1)(x^2+x+1)$. Cancel the $(x-1)$ term to get ${\lim }_{x\to 1}{x}^2+x+1=1+1+1=3$. You can also use L'Hôpital's Rule: derivative of numerator is $3x^2$, derivative of denominator is 1, so limit is $3(1{)}^2=3$. Correct answer: C.

Question 2 (Free Response Part A)

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

Worked Solution: For continuity at $x=3$, left-hand limit = right-hand limit = $f(3)$.

1. Left-hand limit: ${\lim }_{x\to 3^{-}}2x+c=6+c$
2. Right-hand limit: ${\lim }_{x\to 3^{+}}c{x}^2-1=9c-1$
3. Set equal: $6+c=9c-1\to 7=8c\to c=\frac{7}{8}$
4. Verify $f(3)=9(\frac{7}{8})-1=\frac{63}{8}-\frac{8}{8}=\frac{55}{8}$, which matches the limit value. Correct answer: $c=\frac{7}{8}$.

Question 3 (Free Response Part B)

Use the Intermediate Value Theorem to show that $g(x)=x^5+2x-7$ has at least one root on the interval $[1,2]$.

Worked Solution:

1. $g(x)$ is a polynomial, which is continuous for all real $x$, so it is continuous on $[1,2]$.
2. Compute $g(1)=1+2-7=-4$, $g(2)=32+4-7=29$.
3. 0 lies between $-4$ and $29$, so by the Intermediate Value Theorem, there exists at least one $c\in (1,2)$ such that $g(c)=0$, meaning $g(x)$ has a root on $[1,2]$.

9. Quick Reference Cheatsheet

10. What's Next

Limits and continuity are the foundational building blocks for all remaining content on the AP Calculus AB exam. The limit definition of the derivative directly extends this topic, allowing you to calculate instantaneous rates of change for any function, and continuity is a required precondition for the Fundamental Theorem of Calculus, which connects differentiation and integration. You will also regularly use limit computation to evaluate improper integrals, identify critical points of functions, and analyze the end behaviour of derivatives and antiderivatives in multi-part free-response questions.

If you struggle with any of the concepts covered in this guide, or want to practice more exam-style questions tailored to your weak spots, you can ask Ollie for personalized help at any time on the homepage. The next unit in the AP Calculus AB syllabus covers differentiation: definition and basic derivative rules, which builds directly on the limit definitions you just learned.