Unit 1: Limits and Continuity

Unit 1 is where calculus begins. It introduces rates of change, the limit as the central idea, the tools for evaluating limits, the meaning of continuity, and the Intermediate Value Theorem. Everything in this unit reappears later: the derivative is a limit, improper integrals are limits, and infinite series are limits of partial sums.

1.1 Introducing Calculus: Can Change Occur at an Instant?

The unit opens with two ways to measure change. The average rate of change of a function over an interval is the change in the output divided by the change in the input, written [f(b) − f(a)] / (b − a). On a graph, it is the slope of the secant line joining the points (a, f(a)) and (b, f(b)). For f(x) = x³, the average rate of change over [0, 2] is (8 − 0) / (2 − 0) = 4.

The instantaneous rate of change is the rate at one point. It is defined as the limit of average rates of change as the interval collapses down to that point. On a graph it is the slope of the tangent line, and it is what the derivative computes. For f(x) = x³ at x = 2, shrink the interval to [2, 2 + h]: the average rate is [(2 + h)³ − 8] / h = (12h + 6h² + h³) / h = 12 + 6h + h², which approaches 12 as h approaches 0. So the instantaneous rate at x = 2 is 12.

1.2 Defining Limits and Using Limit Notation

The limit of a function, written lim_x→c f(x) = L, means f(x) can be made arbitrarily close to L by taking x sufficiently close to c, with x not equal to c. Read the notation as "the limit of f of x as x approaches c equals L." The exam requires correct limit notation in justifications, so write the lim_x→c part every time you claim something about a limit.

The function need not be defined at c for the limit to exist. The limit says what the function is approaching, not what is happening at the point itself. The limit value and the function value are separate numbers that coincide exactly when the function is continuous there.

1.3 Estimating Limit Values from Graphs

A one-sided limit approaches c from one direction only. From the right it is lim_x→c⁺ f(x); from the left, lim_x→c⁻ f(x). On a graph, follow the curve toward x = c from each side separately and read the y-value each side is heading to. The two-sided limit lim_x→c f(x) exists only when both one-sided limits exist and match.

One-sided limits are the natural tool for jump discontinuities. Suppose the left branch of a graph approaches y = 2 at x = 3 and the right branch approaches y = 5. Then lim_x→3⁻ f(x) = 2 and lim_x→3⁺ f(x) = 5, and because the two differ, lim_x→3 f(x) does not exist. The plotted value f(3), whatever dot is drawn, changes neither one-sided limit.

1.5 Determining Limits Using Algebraic Properties of Limits

The algebraic properties of limits, or limit laws, let you split a limit apart. The limit of a sum, difference, product, quotient, or power is the matching combination of the individual limits, as long as each limit exists and the denominator's limit is not zero. Constant multiples and powers follow the same pattern.

Direct substitution is almost always the first move, and the limit laws justify it for polynomials and for rational functions wherever the denominator is nonzero. For example, lim_x→1 (x² + 2x) / (x + 3) = (1 + 2) / (1 + 3) = 3/4.

When substitution gives 0/0, the expression is indeterminate and needs algebra first. Factoring and canceling is the usual fix. For lim_x→−2 (x² − 4) / (x + 2), substitution gives 0/0, but factoring gives (x − 2)(x + 2) / (x + 2), which equals x − 2 for x ≠ −2, so the limit is −4.

1.8 Determining Limits Using the Squeeze Theorem

The Squeeze Theorem handles functions that resist direct evaluation. If g(x) ≤ f(x) ≤ h(x) for x near c, and lim_x→c g(x) = lim_x→c h(x) = L, then lim_x→c f(x) = L. The two outer functions trap the middle one at the same limit.

The classic application is lim_x→0 x² · cos(1/x). The cosine factor oscillates between −1 and 1 forever, so substitution tells you nothing, but −x² ≤ x² · cos(1/x) ≤ x² for all x, and both bounding functions approach 0 as x → 0. The squeeze pins the limit at 0. The same idea proves lim_x→0 sin x / x = 1, a result that returns constantly in differentiation and in series.

1.10 Exploring Types of Discontinuities

A discontinuity is a break in the graph, and there are three kinds. A removable discontinuity is a hole: the limit exists at x = c, but f(c) is undefined or does not match the limit. Redefining f(c) to equal the limit would repair it. A jump discontinuity is where the left- and right-hand limits exist but disagree, so the graph steps from one value to another. A discontinuity due to a vertical asymptote is where the function grows without bound as x approaches c, an infinite limit.

1.11 Defining Continuity at a Point

Continuity at a point rests on three conditions. A function f is continuous at x = c when f(c) is defined, lim_x→c f(x) exists, and the two are equal: lim_x→c f(x) = f(c). Informally, you can draw through the point without lifting the pencil, and the value you approach matches the value sitting there.

Each discontinuity type breaks the definition in its own way. A removable discontinuity satisfies the limit condition but fails the equality, because f(c) is missing or wrong. A jump discontinuity fails the limit condition outright, since the two sides disagree. An infinite discontinuity fails both, because the function is undefined at c and the limit does not exist as a real number.

1.12 Confirming Continuity over an Interval

Continuity over an interval means the function is continuous at every point of the interval. For interior points, that is the three-condition test from 1.11. At an endpoint of a closed interval, only the relevant one-sided limit must equal the function value, since there is no second side to check.

This is why the familiar function families are straightforward. Polynomials are continuous everywhere. Rational functions are continuous everywhere they are defined, which is everywhere the denominator is nonzero. Sums, differences, products, and quotients of continuous functions remain continuous wherever the result is defined, so most expressions on the exam are continuous on their domains by construction.

1.14 Connecting Infinite Limits and Vertical Asymptotes

An infinite limit describes a function growing without bound near a point, written lim_x→c f(x) = ∞ or −∞. Strictly speaking the limit does not exist as a real number, but the infinity notation records the behavior, and the exam expects you to use it. For f(x) = 1 / (x − 1)², the values blow up to positive infinity from both sides, so lim_x→1 f(x) = ∞.

A vertical asymptote is the line x = c, and it appears exactly when f has an infinite limit as x approaches c from at least one side. One side suffices. The sign can differ by side, so check each one. For f(x) = 1/(x − 1), lim_x→1⁺ 1/(x − 1) = ∞ while lim_x→1⁻ 1/(x − 1) = −∞, and x = 1 is a vertical asymptote either way.

1.15 Connecting Limits at Infinity and Horizontal Asymptotes

A limit at infinity, lim_x→∞ f(x) = L, means f(x) approaches L as x grows without bound. A horizontal asymptote is the line y = L, and it appears when lim_x→∞ f(x) = L or lim_x→−∞ f(x) = L. End behavior describes what happens in both directions, x → ∞ and x → −∞, often summarized by the horizontal asymptotes.

For rational functions, compare degrees. If the top degree is smaller, the limit is 0. If the degrees are equal, divide by the highest power and read the ratio of leading coefficients: lim_x→∞ (5x² + 2) / (x² − 3x) = lim_x→∞ (5 + 2/x²) / (1 − 3/x) = 5/1 = 5, so y = 5 is a horizontal asymptote. If the top degree is larger, the function grows without bound and there is no horizontal asymptote.

1.16 Working with the Intermediate Value Theorem (IVT)

The Intermediate Value Theorem says a continuous function cannot skip values. If f is continuous on the closed interval [a, b], and N is any value between f(a) and f(b), then there is at least one c in the open interval (a, b) with f(c) = N. The hypotheses matter: the interval must be closed, the function must be continuous on it, and N must lie between the endpoint values.

A standard justification: k(x) = x³ − 3x + 1 is a polynomial, hence continuous everywhere, in particular on [0, 1]. Since k(0) = 0 − 0 + 1 = 1 and k(1) = 1 − 3 + 1 = −1, and 0 lies between 1 and −1, the IVT guarantees some c in (0, 1) with k(c) = 0. The theorem promises that such a c exists, not where it is.

Confusions That Cost Points

Practice Questions

Original questions written for this guide in the style of the AP exam. Answers and explanations are on the next page, so complete the questions before checking them.

Answer Key

1. B. Direct substitution gives 0/0, so factor: (x − 2)(x + 2) / (x + 2) = x − 2 for x ≠ −2, and the limit is −4. A stops at the indeterminate form 0/0 and treats it as an answer. C confuses "indeterminate" with "does not exist"; the algebra reveals a finite limit. D changes the sign of the correct value.

2. C. The one-sided limits are 2 from the left and 5 from the right. They disagree, so the two-sided limit does not exist. A reports only the left-hand limit. B reports the function value f(3), which matches the right-hand limit here but is not the two-sided limit. D confuses the x-value where the jump happens with a y-value.

3. B. Factoring gives (x − 1)(x + 1) / (x − 1) = x + 1 for x ≠ 1, so the limit is 2, but f(1) = 3. The limit exists while the function value disagrees, which is the definition of a removable discontinuity. A ignores the mismatch between the limit and f(1). C is wrong because a jump requires disagreeing one-sided limits, and here both sides approach 2. D is wrong because the limit is finite; a vertical asymptote needs an infinite limit.

4. A. Both bounds approach 0 as x → 0: lim_x→0 2x² = 0 and lim_x→0 (−2x²) = 0. The Squeeze Theorem forces h(x) to 0 as well. B ignores the theorem entirely. C may come from misreading 2x² near 0 as approaching 2. D is the trap of thinking the bounds are not enough; matching bounds are exactly what the theorem needs.

5. A. From the right, x − 1 is a small positive number, so 1 / (x − 1) is a large positive number: the limit is ∞. B ignores the side and reports the left-hand behavior instead. C confuses an infinite limit with a limit at infinity of 1/x. D is tempting because infinite limits do not exist as real numbers, but the question asks for the one-sided behavior, which is precisely ∞.

6. A. The degrees are equal, so divide by x²: (5 + 2/x²) / (1 − 3/x), which approaches 5/1 = 5. B would be correct only if the denominator degree were larger. C treats the leading term as dominating without checking that the degrees match. D takes the ratio of the constant terms, which is meaningless here since the x² terms control the behavior.

7. B. The IVT applies: f is continuous on [1, 4] and 3 lies between f(1) = −2 and f(4) = 7, so some c in (1, 4) satisfies f(c) = 3. A and C both claim the theorem locates the value; it only guarantees existence, never the position. D confuses continuity with differentiability, which the IVT never requires.

8. A. A horizontal asymptote y = −2 as x → −∞ is exactly the statement that lim_x→−∞ f(x) = −2. B confuses the asymptote's y-value with a function value at x = −2. C is the classic trap: a graph may cross its horizontal asymptote at finite x values; the asymptote only governs far-out behavior. D confuses a horizontal asymptote with a vertical one.

One-Page Recall Check

• State the average rate of change formula and explain what the secant line represents.
• Explain how the instantaneous rate of change is defined as a limit of average rates.
• Write lim_x→c f(x) = L in words, and explain why f(c) need not exist.
• State the condition for a two-sided limit to exist in terms of one-sided limits.
• List the limit laws and the two conditions needed to use the quotient law.
• Evaluate lim_x→−2 (x² − 4) / (x + 2) from scratch, showing the algebra.
• State the Squeeze Theorem and use it to find lim_x→0 x² · cos(1/x).
• Describe each discontinuity type and name which continuity condition it violates.
• State the three conditions for continuity at a point.
• Explain what changes at an endpoint when checking continuity over a closed interval.
• Explain how a vertical asymptote follows from an infinite limit on at least one side.
• Give the degree rules for limits at infinity of rational functions.
• State the Intermediate Value Theorem with all of its hypotheses.

Key terms for this unit

Average rate of change, Instantaneous rate of change, Limit of a function, Limit notation, One-sided limit, Algebraic properties of limits (limit laws), Squeeze Theorem, Removable discontinuity, Jump discontinuity, Discontinuity due to a vertical asymptote, Continuity at a point, Continuity over an interval, Infinite limit, Vertical asymptote, Limit at infinity, Horizontal asymptote, End behavior, Intermediate Value Theorem (IVT).

About this guide. Written for Rycal and aligned to the College Board AP Calculus BC course framework, Unit 1. All questions and explanations are original Rycal writing. Rycal is independent and is not affiliated with or endorsed by the College Board.