AP Calculus AB

Limits and Continuity

AP Calculus AB — Unit 1

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Module 1

Foundations & Definitions

Overview, Objectives & Key DefinitionsTopic 1

Unit Overview

Limits are the foundation on which all of calculus is built. Before we can talk about instantaneous rates of change (derivatives) or accumulated change (integrals), we need a precise way to describe what a function approaches as its input gets arbitrarily close to some value. That idea — the limit — is what this unit develops.

By the end of this unit you should be able to reason about limits four ways: graphically, numerically, algebraically, and verbally, and connect them to the idea of continuity. On the AP exam, roughly 10–12% of the questions come from this unit, and the ideas reappear everywhere afterward.

Big picture: A limit answers the question, "Where is the function heading?" — regardless of whether it ever actually arrives.


Learning Objectives

After completing this unit you will be able to:

  • Distinguish average rate of change (slope of a secant line) from instantaneous rate of change (slope of a tangent line).
  • Express and evaluate limits using correct notation, including one-sided limits.
  • Estimate limits from graphs and tables.
  • Evaluate limits algebraically using limit laws, factoring, rationalization, and known trigonometric limits.
  • Apply the Squeeze Theorem.
  • Classify discontinuities (removable, jump, infinite) and remove removable ones.
  • Determine continuity at a point and over an interval.
  • Connect infinite limits to vertical asymptotes and limits at infinity to horizontal asymptotes.
  • Apply the Intermediate Value Theorem (IVT).

Prerequisite Concepts

You should be comfortable with:

  • Function notation and evaluation, domain and range.
  • Factoring (difference of squares, difference of cubes, common factors) and rationalizing with conjugates.
  • Piecewise functions and reading graphs.
  • Basic trigonometry (unit-circle values, identities like tan = sin/cos).
  • Rational functions and asymptote behavior from precalculus.

Key Definitions

  • Limit: We write lim_{x→c} f(x) = L if the values of f(x) get arbitrarily close to L as x gets arbitrarily close to c (from both sides), without requiring x = c.
  • One-sided limits: lim_{x→c⁻} f(x) (from the left) and lim_{x→c⁺} f(x) (from the right).
  • Existence of a two-sided limit: lim_{x→c} f(x) exists iff both one-sided limits exist and are equal.
  • Average rate of change of f on [a, b]: (f(b) − f(a)) / (b − a) — the slope of the secant line.
  • Instantaneous rate of change at x = a: lim_{h→0} (f(a + h) − f(a)) / h — the slope of the tangent line.
  • Continuity at x = a: f is continuous at a if all three hold: (1) f(a) is defined, (2) lim_{x→a} f(x) exists, (3) lim_{x→a} f(x) = f(a).
  • Removable discontinuity: a "hole" — the limit exists but does not equal f(a) (or f(a) is undefined).
  • Jump discontinuity: the one-sided limits both exist but are unequal.
  • Infinite discontinuity: the function grows without bound near the point (a vertical asymptote).

✎ Self-Check — 5 questions0 / 5
Q1.The position of a particle at time t is given by s(t) = t^2. What is the average velocity of the particle on the interval [1, 3]?
Explanation: Average velocity is the change in position divided by the change in time: (s(3) - s(1)) / (3 - 1) = (9 - 1) / 2 = 8 / 2 = 4.
Q2.If lim_{x→c} f(x) = L, which of the following MUST be true?
Explanation: The definition of a two-sided limit requires that the left-hand limit and the right-hand limit exist and are equal. The function need not be defined at c, nor does it need to be continuous.
Q3.Evaluate lim_{x→1} (x^2 + 3x - 2).
Explanation: Since the function is a polynomial, we can use direct substitution: (1)^2 + 3(1) - 2 = 1 + 3 - 2 = 2.
Q4.A table of values for f(x) near x = 5 shows f(4.9) = 2.9, f(4.99) = 2.99, f(5.1) = 3.1, and f(5.01) = 3.01. What is the best estimate for lim_{x→5} f(x)?
Explanation: The values from both the left (approaching 5 from values less than 5) and the right (approaching 5 from values greater than 5) suggest the function is approaching 3.
Q5.The average rate of change of a function f(x) over the interval [a, b] is given by:
Explanation: The average rate of change is the slope of the secant line connecting (a, f(a)) and (b, f(b)), calculated as (f(b) - f(a)) / (b - a).
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Module 2

Theorems & Formulas

Important Theorems & FormulasTopic 1

Important Theorems

Limit Laws (valid when the individual limits exist). If lim f = L and lim g = M:

  • Sum/Difference: lim (f ± g) = L ± M
  • Constant multiple: lim (k·f) = k·L
  • Product: lim (f·g) = L·M
  • Quotient: lim (f/g) = L/M, provided M ≠ 0
  • Power/Root: lim (f)^n = L^n

Squeeze (Sandwich) Theorem: If g(x) ≤ f(x) ≤ h(x) near c (except possibly at c) and lim_{x→c} g(x) = lim_{x→c} h(x) = L, then lim_{x→c} f(x) = L.

Intermediate Value Theorem (IVT): If f is continuous on [a, b] and N is any value between f(a) and f(b), then there exists at least one c in (a, b) with f(c) = N. (Guarantees existence, not the value or uniqueness.)

Continuity of common functions: Polynomials are continuous everywhere; rational, radical, trig, exponential, and log functions are continuous on their domains; sums, products, quotients (nonzero denominator), and compositions of continuous functions are continuous.


Important Formulas & Standard Limits

  • Direct substitution (when f is continuous at c): lim_{x→c} f(x) = f(c).
  • Fundamental trig limits:
    • lim_{x→0} (sin x)/x = 1
    • lim_{x→0} (1 − cos x)/x = 0
    • lim_{x→0} (tan x)/x = 1
  • Limits at infinity of rational functions (compare degrees of numerator N and denominator D):
    • deg N < deg D → limit is 0 (horizontal asymptote y = 0)
    • deg N = deg D → limit is the ratio of leading coefficients
    • deg N > deg D → no horizontal asymptote (limit is ±∞)

✎ Self-Check — 5 questions0 / 5
Q1.For a function f(x), if lim_{x→2^-} f(x) = 5 and lim_{x→2^+} f(x) = 3, which of the following is true?
Explanation: Since the left-hand limit (5) does not equal the right-hand limit (3), the two-sided limit at x = 2 does not exist.
Q2.Evaluate lim_{x→3} (x^2 - 9) / (x - 3).
Explanation: This is an indeterminate form 0/0. Factor the numerator: (x-3)(x+3) / (x-3). Canceling (x-3) leaves lim_{x→3} (x+3) = 3 + 3 = 6.
Q3.The function f(x) = (x^2 - 1) / (x - 1) has which type of discontinuity at x = 1?
Explanation: The function simplifies to f(x) = x + 1 for x ≠ 1. The limit exists at x = 1 (it is 2), but the function is undefined there, characterizing a removable discontinuity.
Q4.For a function to be continuous at x = a, which condition must be met?
Explanation: Continuity at a point requires the function to be defined, the limit to exist, and the limit to equal the function value.
Q5.Find the vertical asymptote(s) of f(x) = (x + 2) / (x^2 - 4).
Explanation: Simplify f(x) = (x + 2) / ((x - 2)(x + 2)). The factor (x + 2) cancels, leaving a removable discontinuity at x = -2. The factor (x - 2) remains in the denominator, creating a vertical asymptote at x = 2.
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Module 3

Concepts & Worked Examples

Concept Explanations & Solved ExamplesTopic 1

Concept Explanations

From average to instantaneous rate (1.1). The average rate of change over [a, b] is a secant slope you can compute directly. To get the instantaneous rate at a single point, shrink the interval: let the second point slide toward the first (h → 0). That limiting secant slope is the tangent slope — the derivative you'll define formally in Unit 2.

Why a limit ignores f(c) (1.2). The whole power of a limit is that it describes approach, not arrival. A function can have a limit of 3 at x = 4 while f(4) = 5, or while f(4) is undefined. This is exactly what lets us handle the 0/0 forms that derivatives produce.

Algebraic techniques (1.5–1.6). Try direct substitution first. If you get a real number, you're done. If you get 0/0 (indeterminate), do algebra: factor and cancel, rationalize with a conjugate, or use a known trig limit. If you get (nonzero)/0, the limit is infinite (check signs for ±∞).

Discontinuities and continuity (1.10–1.13). Test continuity with the three-part definition. A removable discontinuity can be "patched" by redefining f(c) to equal the limit. Jump and infinite discontinuities cannot be removed.

Asymptotes as limits (1.14–1.15). A vertical asymptote at x = c means a one-sided (or two-sided) limit is ±∞. A horizontal asymptote y = L means lim_{x→±∞} f(x) = L.


Worked Examples

Example 1 (Algebraic — factoring). Evaluate lim_{x→3} (x² − 9)/(x − 3). Direct substitution gives 0/0. Factor: (x − 3)(x + 3)/(x − 3) = x + 3. So the limit is 3 + 3 = 6.

Example 2 (Rationalizing). Evaluate lim_{x→0} (√(x + 1) − 1)/x. Multiply by the conjugate: ((x + 1) − 1) / (x(√(x+1) + 1)) = 1/(√(x+1) + 1). As x → 0 this is 1/(1 + 1) = 1/2.

Example 3 (Squeeze Theorem). Evaluate lim_{x→0} x² sin(1/x). Since −1 ≤ sin(1/x) ≤ 1, we have −x² ≤ x² sin(1/x) ≤ x². Both bounds → 0, so the limit is 0.

Example 4 (Continuity — solve for a constant). Find k so that f is continuous at x = 2, where f(x) = kx² for x ≤ 2 and f(x) = x + k for x > 2. Match the pieces: k(2)² = 2 + k → 4k = 2 + k → 3k = 2 → k = 2/3.

Example 5 (Limit at infinity). Evaluate lim_{x→∞} (3x² + 5)/(2x² − x). Equal degrees → ratio of leading coefficients = 3/2.

Example 6 (IVT). Show f(x) = x³ − x − 1 has a root in [1, 2]. f is continuous; f(1) = −1 < 0 and f(2) = 5 > 0. Since 0 lies between −1 and 5, IVT guarantees a c in (1, 2) with f(c) = 0.


✎ Self-Check — 5 questions0 / 5
Q1.Evaluate lim_{x→∞} (3x^2 + 5) / (2x^2 - x).
Explanation: For limits at infinity of rational functions where degrees are equal, the limit is the ratio of the leading coefficients: 3/2.
Q2.Evaluate lim_{x→0} (sqrt(x + 1) - 1) / x.
Explanation: Multiply numerator and denominator by the conjugate (sqrt(x+1)+1): ((x+1)-1) / (x(sqrt(x+1)+1)) = x / (x(sqrt(x+1)+1)) = 1 / (sqrt(x+1)+1). As x→0, this becomes 1 / (1+1) = 1/2.
Q3.Let f(x) = (x^2 - 4) / (x - 2) for x ≠ 2. What value should be assigned to f(2) to make the function continuous at x = 2?
Explanation: The function simplifies to f(x) = x + 2. To remove the discontinuity at x = 2, we must define f(2) = lim_{x→2} (x+2) = 4.
Q4.The slope of the line tangent to the curve y = f(x) at point (a, f(a)) is defined as:
Explanation: This is the definition of the derivative at a point, which represents the instantaneous rate of change or the slope of the tangent line at x = a.
Q5.A particle moves along a line with position s(t) = t^2 + 3t. What is the instantaneous rate of change of its position at t = 2?
Explanation: The instantaneous rate of change is the derivative s'(t) = 2t + 3. At t = 2, s'(2) = 2(2) + 3 = 7.
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Module 4

Graphs, Pitfalls & AP Exam Tips

Graphical Interpretation, Common Mistakes & TipsTopic 1

Graphical Interpretation

  • A two-sided limit exists when the left and right branches of the graph head to the same height — even if there's a hole or the point is plotted elsewhere.
  • A hole at (c, L) with the curve approaching L from both sides ⇒ lim = L (removable).
  • A jump (left branch heads to one height, right branch to another) ⇒ two-sided limit does not exist.
  • A vertical asymptote ⇒ the branch shoots to ±∞ (infinite limit).
  • A horizontal asymptote ⇒ the far-left/far-right tails flatten toward y = L.

Common Mistakes

  • Assuming lim_{x→c} f(x) = f(c) always. It only holds when f is continuous at c.
  • Stopping at 0/0 and writing "does not exist." 0/0 is indeterminate — do more algebra.
  • Confusing "limit does not exist" with "= ∞". An infinite limit is a specific behavior; a jump is a different reason a limit fails.
  • Forgetting to check both one-sided limits for piecewise functions and at asymptotes.
  • Sign errors on infinite limits. Near a vertical asymptote, test the sign from each side to decide +∞ vs −∞.
  • Misusing IVT. It requires continuity and only guarantees existence — never a specific value or uniqueness.
  • Canceling a factor and forgetting the hole. After canceling (x − c), the original function still has a removable discontinuity at x = c.

AP Exam Tips

  • Always try direct substitution first — it settles most non-calculator limits instantly.
  • Memorize the three fundamental trig limits; they appear on both multiple-choice and free-response.
  • For rational-function end behavior, compare degrees rather than doing long division under time pressure.
  • On free-response, when citing IVT or discussing continuity, state that the function is continuous — that justification earns points.
  • Read whether a question asks for the limit or the function value — they can differ.
  • Calculator-active tables: use x-values very close on both sides to estimate a limit.

Unit Summary

  • A limit describes where a function heads as x → c, independent of f(c).
  • A two-sided limit exists iff both one-sided limits exist and are equal.
  • Evaluate limits by substitution → algebra (factor/rationalize/trig) → interpret ∞.
  • Continuity requires the value, the limit, and their equality; classify breaks as removable, jump, or infinite.
  • Infinite limits ⇔ vertical asymptotes; limits at infinity ⇔ horizontal asymptotes.
  • The Squeeze Theorem handles oscillating bounded functions; the IVT guarantees solutions for continuous functions.

Next unit: Unit 2 turns the instantaneous-rate idea into the formal derivative.

✎ Self-Check — 5 questions0 / 5
Q1.Using the Squeeze Theorem, what is lim_{x→0} x^2 * sin(1/x)?
Explanation: Since -1 ≤ sin(1/x) ≤ 1, it follows that -x^2 ≤ x^2 * sin(1/x) ≤ x^2. As x approaches 0, both -x^2 and x^2 approach 0, so the limit is 0.
Q2.Given f(x) = x^3 - x - 1, which interval is guaranteed to contain a root by the Intermediate Value Theorem?
Explanation: f(1) = 1 - 1 - 1 = -1. f(2) = 8 - 2 - 1 = 5. Since f(x) is continuous and changes sign on [1, 2], there must be a root.
Q3.For a function f(x), what is lim_{x→4} f(x) if f(4) = 5 and the function has a hole at x = 4 with the curve approaching 3 from both sides?
Explanation: The limit as x approaches 4 is the value the function approaches, which is 3, regardless of the actual value f(4) = 5.
Q4.Which method is most appropriate for evaluating lim_{x→0} (sin x) / x?
Explanation: Direct substitution yields 0/0, an indeterminate form. Numerical estimation or knowledge of the standard limit lim_{x→0} (sin x)/x = 1 is required.
Q5.If f(x) is continuous on [0, 2] with f(0) = -3 and f(2) = 5, which of the following must be true?
Explanation: By the Intermediate Value Theorem, since f(x) is continuous and changes from -3 to 5, it must pass through 0 at some point c in (0, 2).

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