What is AP Calculus AB/BC unit 1?
Calculus begins with a deceptively simple question: what value does a function approach as its input gets close to some number? Unit 1 answers that question rigorously using limits, then uses limits to define continuity and to prove the Intermediate Value Theorem.
Unit 1 is about limits and continuity. You learn to express, estimate, and evaluate limits using graphs, tables, algebra, and the Squeeze Theorem, then use the formal definition of continuity to classify discontinuities, confirm continuity over intervals, remove removable discontinuities, and apply the IVT.
Limits: what they are and how to find them
A limit lim[x→c] f(x) = R describes what output a function approaches as x gets close to c, independent of the actual value f(c). You can estimate limits from graphs and tables, or evaluate them exactly using limit laws, algebraic manipulation such as factoring and conjugate rationalization, and the Squeeze Theorem for cases like lim[x→0] sin(x)/x = 1.
Continuity: definition, types, and intervals
A function is continuous at x = c when f(c) exists, lim[x→c] f(x) exists, and those two values are equal. Discontinuities are classified as removable (hole), jump (unequal one-sided limits), or infinite (vertical asymptote). Polynomials, rationals, exponentials, logarithms, and trig functions are continuous on their entire domains.
Limits involving infinity
Infinite limits describe unbounded behavior near a vertical asymptote: lim[x→a] f(x) = ±∞. Limits at infinity describe end behavior: if lim[x→∞] f(x) = L, then y = L is a horizontal asymptote. For rational functions, compare degrees of numerator and denominator to determine the horizontal asymptote or confirm none exists.
Why limits matter beyond Unit 1The derivative in Unit 2 is defined as lim[h→0] (f(a+h) - f(a))/h, a limit of an average rate of change. The definite integral in Unit 6 is a limit of Riemann sums. For BC students, infinite series in Unit 10 rely on limits of partial sums. Every major concept in calculus is built on the limit idea introduced here.
Unit 1 review notes
1.1
Instantaneous vs. average rate of change
Calculus asks whether change can happen at a single instant. The average rate of change over an interval [a, a+h] is Δy/Δx = (f(a+h) - f(a))/h, which is the slope of a secant line. As h shrinks toward zero, the secant line approaches the tangent line, and the limit of that ratio defines the instantaneous rate of change at x = a. This is the conceptual origin of the derivative.
- Average rate of change: (f(a+h) - f(a))/h over an interval; slope of the secant line through two points on the graph.
- Instantaneous rate of change: The limit of the average rate of change as the interval width h approaches zero; slope of the tangent line at a point.
- Secant line: A line connecting two points on a curve; its slope equals the average rate of change between those points.
- Why average rate is undefined at a point: Dividing by Δx = 0 is undefined, so a limit is required to define change at a single instant.
Given f(x) = x², write the average rate of change from x = 2 to x = 2+h and describe what happens as h → 0.
1.2
Defining and estimating limits
The limit lim[x→c] f(x) = R means f(x) can be made arbitrarily close to R by taking x sufficiently close to c, without requiring x to equal c. Limits can be expressed graphically (reading where the graph heads), numerically (building a table of values approaching c from both sides), or analytically (using notation and algebra). A two-sided limit exists only when the left-hand limit and right-hand limit both exist and are equal. A limit fails to exist when the function is unbounded near c, oscillates (like sin(1/x) near 0), or has unequal one-sided limits.
- Limit notation: lim[x→c] f(x) = R; left-hand: lim[x→c⁻] f(x); right-hand: lim[x→c⁺] f(x).
- Two-sided limit: Exists when lim[x→c⁻] f(x) = lim[x→c⁺] f(x); that shared value is the limit.
- Limit does not exist (DNE): Occurs when one-sided limits differ, the function is unbounded, or the function oscillates without settling near c.
- Graphical estimation: Trace the graph from both sides toward x = c; the y-value the graph approaches (not the plotted point) is the limit.
- Numerical estimation: Build a table with x-values approaching c from both sides; if outputs converge to the same value, that is the estimated limit.
A graph shows an open circle at (3, 5) and a filled dot at (3, 2). What is lim[x→3] f(x)? What is f(3)?
| Representation | How to estimate the limit | Key caution |
|---|
| Graph | Read the y-value both sides approach as x → c | Open vs. filled circles show f(c), not the limit |
| Table | Check that outputs from left and right converge to the same value | Outputs may converge slowly; use values very close to c |
| Analytic | Use limit notation and algebraic or theorem-based evaluation | Direct substitution may give 0/0; further work needed |
1.5
Evaluating limits algebraically
Limit laws let you break complex limits into simpler pieces: the limit of a sum, difference, product, or quotient equals the corresponding operation on the individual limits (provided the limits exist and denominators are nonzero). Direct substitution works whenever the function is continuous at c. When substitution gives the indeterminate form 0/0, use algebraic manipulation: factor and cancel common factors in rational functions, multiply by a conjugate to simplify radical expressions, or use alternate trigonometric forms. Topic 1.7 focuses on choosing the right technique for a given limit.
- Direct substitution: Plug x = c directly into f(x); valid when f is continuous at c and produces a defined value.
- Indeterminate form 0/0: Signals that algebraic manipulation is needed before the limit can be evaluated; not the final answer.
- Factor and cancel: Factor numerator and denominator of a rational function, cancel the common factor causing 0/0, then substitute.
- Conjugate rationalization: Multiply numerator and denominator by the conjugate of a radical expression to eliminate the radical and allow cancellation.
- Limit laws: Sum, difference, product, quotient, and power rules that let you evaluate limits of combined functions from the limits of their parts.
Evaluate lim[x→3] (x² - 9)/(x - 3) by factoring. Then identify which limit law justifies evaluating lim[x→2] [f(x) + g(x)] when both individual limits exist.
1.8
Squeeze Theorem and connecting representations
The Squeeze Theorem states that if g(x) ≤ f(x) ≤ h(x) near c and lim[x→c] g(x) = lim[x→c] h(x) = L, then lim[x→c] f(x) = L. The two most important results are lim[x→0] sin(x)/x = 1 and lim[x→0] (1 - cos x)/x = 0, both proved by squeezing. Topic 1.9 asks you to move fluidly between graphical, numerical, and analytic representations of the same limit, recognizing that all three describe the same underlying behavior.
- Squeeze Theorem: If a function is bounded above and below by two functions that share the same limit at c, the middle function has that same limit.
- lim[x→0] sin(x)/x = 1: A foundational trigonometric limit proved via the Squeeze Theorem; used in derivative proofs for sine and cosine.
- lim[x→0] (1 - cos x)/x = 0: A companion trigonometric limit also proved by squeezing; appears in the derivative of cosine.
- Connecting representations: The same limit can be read from a graph, estimated from a table, or computed analytically; all three should give consistent results.
Use the Squeeze Theorem to find lim[x→0] x·sin(1/x), given that -|x| ≤ x·sin(1/x) ≤ |x|.
1.10
Types of discontinuities and continuity at a point
A function is continuous at x = c when all three conditions hold: f(c) exists, lim[x→c] f(x) exists, and lim[x→c] f(x) = f(c). Failing any one condition produces a discontinuity. Removable discontinuities (holes) occur when the limit exists but either f(c) is undefined or f(c) does not equal the limit. Jump discontinuities occur when the one-sided limits exist but are unequal. Infinite discontinuities occur when the limit is ±∞, corresponding to a vertical asymptote.
- Three conditions for continuity at c: f(c) exists; lim[x→c] f(x) exists; lim[x→c] f(x) = f(c). All three must hold.
- Removable discontinuity: The limit exists at c but f(c) is undefined or differs from the limit; appears as a hole on the graph.
- Jump discontinuity: Left-hand and right-hand limits both exist but are not equal; the graph jumps between two values.
- Infinite discontinuity: The function grows without bound near x = c; the limit is ±∞ and a vertical asymptote exists at x = c.
For f(x) = (x² - 4)/(x - 2), identify the type of discontinuity at x = 2 and state which of the three continuity conditions fails.
| Type | One-sided limits | f(c) | Removable? |
|---|
| Removable (hole) | Both exist and are equal | Undefined or wrong value | Yes, redefine f(c) |
| Jump | Both exist but are unequal | May or may not exist | No |
| Infinite (vertical asymptote) | One or both are ±∞ | Undefined | No |
1.12
Continuity over an interval and removing discontinuities
A function is continuous on an interval when it is continuous at every point in that interval, with appropriate one-sided conditions at closed endpoints. Polynomials are continuous everywhere. Rational functions are continuous wherever the denominator is nonzero. Exponential, logarithmic, and trigonometric functions are continuous on their domains. To remove a removable discontinuity, redefine f(c) to equal the limit at that point. For piecewise functions, continuity at a boundary requires the expressions on both sides to produce the same value at the boundary point, which often means solving for an unknown parameter.
- Continuous on an interval: Continuous at every interior point; for closed intervals, also continuous from the right at the left endpoint and from the left at the right endpoint.
- Domain-based continuity: Polynomials: all reals. Rationals: all reals except where denominator = 0. Logarithms: x > 0. Trig: depends on function (tan has gaps at π/2 + kπ).
- Removing a discontinuity: If lim[x→c] f(x) = L exists, define or redefine f(c) = L to make the function continuous at c.
- Piecewise continuity at a boundary: Set the left-side expression equal to the right-side expression at the boundary point and solve for any unknown constants.
Find the value of k that makes f(x) = {kx + 1, x < 2; x² - 1, x ≥ 2} continuous at x = 2.
1.14
Infinite limits, limits at infinity, and asymptotes
Infinite limits describe behavior near a vertical asymptote: if lim[x→a⁺] f(x) = +∞, the graph rises without bound just to the right of x = a. Check both one-sided limits because the signs can differ. Limits at infinity describe end behavior: lim[x→∞] f(x) = L means the graph levels off toward y = L, a horizontal asymptote. For rational functions, compare the degrees of numerator and denominator: equal degrees give a horizontal asymptote at the ratio of leading coefficients; lower numerator degree gives y = 0; higher numerator degree means no horizontal asymptote. Relative magnitudes of growth rates matter: exponential functions grow faster than power functions, which grow faster than logarithms.
- Vertical asymptote: x = a is a vertical asymptote when lim[x→a] f(x) = ±∞; occurs at zeros of the denominator that do not cancel.
- Infinite limit sign analysis: Substitute a value just to the left or right of a to determine whether the function approaches +∞ or -∞ from each side.
- Horizontal asymptote: y = L when lim[x→∞] f(x) = L or lim[x→-∞] f(x) = L; describes long-run end behavior.
- Degree comparison for rational functions: Numerator degree < denominator: y = 0. Equal degrees: y = ratio of leading coefficients. Numerator degree > denominator: no horizontal asymptote.
- End behavior: What a function's output approaches as x → +∞ or x → -∞; described precisely using limits at infinity.
Find all vertical and horizontal asymptotes of f(x) = (3x² + 1)/(x² - 4) using limits.
| Asymptote type | Limit form | What it describes |
|---|
| Vertical | lim[x→a] f(x) = ±∞ | Unbounded behavior near x = a |
| Horizontal | lim[x→∞] f(x) = L | End behavior as x grows large |
| None (rational) | Numerator degree > denominator degree | No horizontal asymptote; may have slant asymptote |
1.16
Intermediate Value Theorem (IVT)
The IVT states: if f is continuous on the closed interval [a, b] and d is any number strictly between f(a) and f(b), then there exists at least one c in (a, b) such that f(c) = d. The most common application is proving a root exists: if f(a) and f(b) have opposite signs and f is continuous on [a, b], then f(c) = 0 for some c in (a, b). The IVT guarantees existence but does not find c or say c is unique. Always state continuity on the closed interval before invoking the theorem.
- IVT conditions: f must be continuous on the closed interval [a, b], and the target value d must satisfy f(a) < d < f(b) or f(b) < d < f(a).
- IVT conclusion: There exists at least one c in the open interval (a, b) such that f(c) = d.
- Sign-change root criterion: If f(a) and f(b) have opposite signs and f is continuous on [a, b], the IVT guarantees a root in (a, b).
- IVT does not guarantee uniqueness: The theorem says at least one such c exists; there may be more than one.
- Continuity requirement: A jump or infinite discontinuity on [a, b] can invalidate the IVT conclusion; always verify continuity first.
Show that f(x) = x³ - x - 1 has a root on [1, 2] by applying the IVT. State each condition explicitly.
Practice AP Calculus AB/BC unit 1 questions
Try AP-style multiple-choice questions and written prompts after you review the notes.
QuestionA function r(x) satisfies: limx→4−r(x)=7, limx→4+r(x)=7, and r(4)=5. Which statement correctly interprets these conditions?
limx→4r(x)=7 because both one-sided limits equal 7
limx→4r(x)=5 because the function value is r(4)=5
limx→4r(x) does not exist because r(4)=7
r(4)=7 because the one-sided limits both equal 7
QuestionA student evaluates limx→2x2−4x3−8 and gets 00. Which algebraic rearrangement correctly identifies the limit?
Factor numerator as (x−2)(x2+2x+4) and denominator as (x−2)(x+2); cancel (x−2) to get x+2x2+2x+4, yielding limit 3.
Divide both numerator and denominator by x2 to obtain 1−4/x2x−8/x2, which approaches 12=2.
Recognize that both numerator and denominator equal 0 at x=2, so apply L'Hôpital's rule to get limx→22x3x2=3.
Substitute x=2 into the original expression to confirm the limit is undefined, since 00 is indeterminate.
3. The function f is defined by f(x)=x−2x2−4 for x=2 and f(2)=5. Let g(x)=x−23x2−12 for x=2 and g(2)=12.
1. The following functions are defined for this question: f(x)=x2−4 g(x)=x−2
k(x)=4
Water flows through a filter. For x=2, the flow rate (in liters per minute) is modeled by the function F defined by F(x)=x−2x2−4, where x is the setting on the filter’s control dial. The graph of F for 0≤x≤6 is shown in Figure 1.
f(x)=x2−4
g(x)=x−2
Figure 1. Graph of F on 0 ≤ x ≤ 6, showing a removable discontinuity at x = 2