Defining limits and using limit notation — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Intuitive and formal epsilon-delta definitions of one-sided, two-sided, infinite, and limits at infinity, standard limit notation conventions, and interpreting limits in graphical and contextual problems.

You should already know: Function notation and domain/range for common function types. Basic graphing of functions including asymptotes. Algebraic manipulation of polynomials and rational expressions.

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 limits and using limit notation?

Limits are the foundational building block of all calculus, describing the behavior of a function $f(x)$ as $x$ approaches a specific input, regardless of the value of $f$ at that input. The College Board AP Calculus CED allocates 10-12% of total exam weight to Unit 1: Limits and Continuity, with this topic specifically making up roughly a third of that unit. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections, and is embedded in questions across all units, since all core calculus concepts are defined using limits.

Limit notation is the standard language for communicating these behavior descriptions: ${\lim }_{x\to a}f(x)=L$ is read "the limit of $f$ of $x$ as $x$ approaches $a$ equals $L$", meaning $f(x)$ gets arbitrarily close to $L$ as $x$ gets arbitrarily close (but not equal) to $a$. Unlike evaluating a function at a point, limits describe approaching, not being at the point, which is critical for defining derivatives (where the difference quotient is never defined at the limit point) and integrals. Synonyms for the concept include limiting value, approached value, and long-run behavior (for limits at infinity).

2. One-Sided and Two-Sided Limits

A one-sided limit describes the behavior of $f(x)$ as $x$ approaches $a$ from only one side of the number line. A left-hand limit (approaching from values less than $a$) is written: $$\underset{x\to a^{-}}{\lim }f(x)=L$$ A right-hand limit (approaching from values greater than $a$) is written: $$\underset{x\to a^{+}}{\lim }f(x)=L$$

The core theorem for two-sided limits (limits where $x$ approaches $a$ from both sides) states that a two-sided limit ${\lim }_{x\to a}f(x)$ exists if and only if both one-sided limits exist and are equal. Formally: $$\underset{x\to a}{\lim }f(x)=L⟺\underset{x\to a^{-}}{\lim }f(x)=L\text{ and }\underset{x\to a^{+}}{\lim }f(x)=L$$ This rule is the most commonly tested concept from this topic, especially for piecewise functions, absolute value functions, and functions with jump discontinuities. If the one-sided limits are not equal, the two-sided limit does not exist, regardless of the value of $f(a)$.

Worked Example

Given the piecewise function $f(x)=\{\begin{array}{ll}2x+1& x<3\\ x^2-2x& x>3\end{array}$, find ${\lim }_{x\to 3}f(x)$ if it exists.

1. To find the two-sided limit, first calculate the left-hand limit as $x$ approaches 3 from the left. For $x<3$, $f(x)=2x+1$, so ${\lim }_{x\to 3^{-}}f(x)=2(3)+1=7$.
2. Next, calculate the right-hand limit as $x$ approaches 3 from the right. For $x>3$, $f(x)=x^2-2x$, so ${\lim }_{x\to 3^{+}}f(x)=(3{)}^2-2(3)=9-6=3$.
3. Compare the two one-sided limits: $7\ne 3$.
4. By the existence theorem for two-sided limits, the two-sided limit ${\lim }_{x\to 3}f(x)$ does not exist.

Exam tip: On multiple-choice questions asking if a two-sided limit exists for a piecewise function, always check both one-sided limits explicitly—don’t assume they match just because the function is defined at $x=a$ on both pieces.

3. Infinite Limits and Vertical Asymptotes

Infinite limits describe the behavior of $f(x)$ as $x$ approaches a finite value $a$, when $f(x)$ grows without bound (toward positive or negative infinity) instead of approaching a finite value $L$. The notation is: $$\underset{x\to a}{\lim }f(x)=\infty \text{or}\underset{x\to a}{\lim }f(x)=-\infty$$

A critical point to remember: this notation only describes the unbounded behavior of $f(x)$ near $a$. It does not mean the limit exists—an infinite limit is still a non-existent limit as a finite real number. When a function has an infinite limit at $x=a$, it has a vertical asymptote at $x=a$. Infinite limits almost always occur at points where the denominator of a rational function is zero and the numerator is non-zero, since dividing a finite number by an increasingly small number gives an increasingly large (in magnitude) result.

Worked Example

Find ${\lim }_{x\to 2^{-}}\frac{3x+1}{x-2}$, and state whether the function has a vertical asymptote at $x=2$.

1. First evaluate the numerator at $x=2$: $3(2)+1=7$, which is non-zero, so we expect an infinite limit at this point.
2. For the left-hand limit $x\to 2^{-}$, all $x$ near 2 are less than 2, so $x-2$ is a very small negative number.
3. We have a positive numerator (7) divided by a small negative number, which produces a large negative result.
4. Thus ${\lim }_{x\to 2^{-}}\frac{3x+1}{x-2}=-\infty$, and since an infinite limit exists at $x=2$, the function has a vertical asymptote at $x=2$.

Exam tip: When writing infinite limits on FRQ, saying "the limit equals infinity" is acceptable notation for describing unbounded behavior, but you must answer that the limit does not exist if explicitly asked whether a finite limit exists.

4. Formal Epsilon-Delta Definition of a Limit

The intuitive definition of a limit ("$f(x)$ gets close to $L$ when $x$ gets close to $a$") is imprecise, so the formal epsilon-delta definition gives a rigorous way to prove a limit equals a given value $L$. The definition states:

${\lim }_{x\to a}f(x)=L$ if for every $\epsilon >0$, there exists a $\delta >0$ such that if $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$.

Unpacked: $\epsilon$ is the maximum allowed error between $f(x)$ and $L$, and $\delta$ is how close $x$ must be to $a$ (not including $a$ itself, hence $0<|x-a|$) to satisfy the error bound. For a limit to exist, every positive $\epsilon$, no matter how small, has some corresponding $\delta$. On the AP Calculus BC exam, you will rarely be asked to write a full epsilon-delta proof, but you will often be asked to interpret the definition or find $\delta$ for a given $\epsilon$ on multiple-choice questions.

Worked Example

For ${\lim }_{x\to 4}(3x-5)=7$, find $\delta$ that satisfies the epsilon-delta condition for $\epsilon =0.1$.

1. The definition requires that $|(3x-5)-7|<0.1$ whenever $0<|x-4|<\delta$.
2. Simplify the left-hand inequality: $|3x-12|=3|x-4|<0.1$.
3. Divide both sides by 3 to isolate $|x-4|$: $|x-4|<\frac{0.1}{3}\approx 0.0333$.
4. Thus $\delta =0.0333$ (or any smaller positive number) satisfies the condition for $\epsilon =0.1$.

Exam tip: When finding $\delta$ for a linear function $f(x)=mx+b$, $\delta$ will always equal $\frac{\epsilon }{|m|}$—use this shortcut to save time on MCQ.

5. Common Pitfalls (and how to avoid them)

• Wrong move: After calculating only one one-sided limit for a piecewise function at the boundary, concluding the two-sided limit equals that value. Why: Students rush to answer and forget that two-sided limits require matching one-sided limits by definition. Correct move: Always explicitly calculate left-hand and right-hand limits, then compare them before concluding whether the two-sided limit exists.
• Wrong move: Stating that ${\lim }_{x\to a}f(x)=\infty$ means the limit exists. Why: Students confuse notation for describing unbounded behavior with the definition of an existing finite limit. Correct move: When asked if a finite limit exists, if the limit is infinite, answer that the limit does not exist, and note the notation only describes behavior.
• Wrong move: Evaluating $f(a)$ to find ${\lim }_{x\to a}f(x)$, and concluding the limit equals $f(a)$. Why: Students confuse function evaluation with limit behavior, especially when working with continuous functions. Correct move: Always check one-sided limits first, even if $f(a)$ is defined—discontinuities can cause the limit to differ from $f(a)$ or not exist at all.
• Wrong move: In epsilon-delta notation, dropping the $0<|x-a|$ condition and writing only $|x-a|<\delta$. Why: Students forget that the limit does not depend on the value of $f$ at $x=a$, only on values near $a$. Correct move: Always include the $0<$ inequality in epsilon-delta statements to exclude $x=a$ itself.
• Wrong move: Confusing ${\lim }_{x\to \infty }f(x)$ (limit at infinity) with ${\lim }_{x\to a}f(x)=\infty$ (infinite limit at finite $a$). Why: Both involve infinity in the notation, leading to mix-ups about the behavior being described. Correct move: Always check where the infinity is placed: infinity under the limit sign means $x$ grows without bound; infinity after the equals sign means $f(x)$ grows at a finite $a$.
• Wrong move: For absolute value functions, using the same expression for left and right limits at the critical point. Why: Students forget $|x-a|$ is a piecewise function that changes definition at $x=a$. Correct move: Rewrite $|x-a|$ into its piecewise form before calculating one-sided limits at $x=a$.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Given $g(x)=\frac{|x-4|}{x-4}$, what is ${\lim }_{x\to 4}g(x)$? A) $0$ B) $1$ C) $-1$ D) The limit does not exist

Worked Solution: First, rewrite the absolute value function as a piecewise function: $|x-4|=-(x-4)$ for $x<4$ and $|x-4|=x-4$ for $x>4$. Calculate the left-hand limit: ${\lim }_{x\to 4^{-}}g(x)=\frac{-(x-4)}{x-4}=-1$. Calculate the right-hand limit: ${\lim }_{x\to 4^{+}}g(x)=\frac{x-4}{x-4}=1$. Since the two one-sided limits are not equal, the two-sided limit does not exist. The correct answer is D.

Question 2 (Free Response)

Let $h(x)=\{\begin{array}{ll}{x}^2+k& x\le 2\\ 3x+2k& x>2\end{array}$, where $k$ is a constant. (a) Find ${\lim }_{x\to 2^{-}}h(x)$ and ${\lim }_{x\to 2^{+}}h(x)$ in terms of $k$. (b) Find the value of $k$ such that ${\lim }_{x\to 2}h(x)$ exists. Justify your answer. (c) Given your value of $k$ from part (b), what is ${\lim }_{x\to 2}h(x)$? If $k=1$ instead, would the limit exist? Explain.

Worked Solution: (a) For the left-hand limit ($x\le 2$): $$\underset{x\to 2^{-}}{\lim }h(x)=(2{)}^2+k=4+k$$ For the right-hand limit ($x>2$): $$\underset{x\to 2^{+}}{\lim }h(x)=3(2)+2k=6+2k$$

(b) A two-sided limit exists if and only if the one-sided limits are equal: $$4+k=6+2k⟹-k=2⟹k=-2$$ Justification: ${\lim }_{x\to 2}h(x)$ exists only when ${\lim }_{x\to 2^{-}}h(x)={\lim }_{x\to 2^{+}}h(x)$, which holds for $k=-2$.

(c) Substitute $k=-2$ into the one-sided limit: ${\lim }_{x\to 2}h(x)=4+(-2)=2$. If $k=1$, ${\lim }_{x\to 2^{-}}h(x)=5$ and ${\lim }_{x\to 2^{+}}h(x)=8$. Since the one-sided limits are not equal, the limit does not exist when $k=1$.

Question 3 (Application / Real-World Style)

A ecologist is modeling the population of deer on an island with a maximum carrying capacity of 1200 deer. The population at time $t$ years after the start of the study is given by $P(t)=\frac{1200t}{t+5}$ for $t\ge 0$. Write the correct limit notation for the long-term population of deer as time increases, find the value of the limit, and interpret the result in context.

Worked Solution: Long-term behavior means $t$ approaches infinity, so the limit notation is ${\lim }_{t\to \infty }P(t)$. Simplify the limit by dividing numerator and denominator by $t$: $$\underset{t\to \infty }{\lim }\frac{1200t}{t+5}=\underset{t\to \infty }{\lim }\frac{1200}{1+\frac{5}{t}}=\frac{1200}{1+0}=1200$$ In context, this means as time goes on, the deer population on the island will approach the carrying capacity of 1200 deer, getting arbitrarily close to this value over time.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational language for all of calculus. Immediately after mastering limit definitions and notation, you will move on to estimating limits from graphs and tables, then calculating limits using algebraic techniques like factoring and rationalizing. Without a solid understanding of what limits are and how to interpret their notation, you cannot correctly understand the definition of the derivative as the limit of a difference quotient, or the definition of the definite integral as the limit of a Riemann sum—both core concepts heavily tested on the AP Calculus BC exam. This topic also feeds into limits of sequences and series later in the course, which make up a large portion of the BC exam. Continue your study with these topics:

• Estimating limits from graphs and tables
• Algebraic techniques for calculating limits
• Continuity of functions
• Defining the derivative as a limit