Unit 1: Limits and Continuity

Unit 1 builds the foundation of calculus. It covers average and instantaneous rates of change, the definition and notation of limits, techniques for evaluating limits, types of discontinuities, continuity, asymptotes, and the Intermediate Value Theorem.

1.1 Introducing Calculus: Can Change Occur at an Instant?

Calculus starts with a question about change. The average rate of change of a function over an interval is the change in output divided by the change in input, [f(b) − f(a)] / (b − a). On a graph it is the slope of the secant line through the two points (a, f(a)) and (b, f(b)). For f(x) = x², the average rate of change over [1, 2] is (4 − 1) / (2 − 1) = 3.

The instantaneous rate of change is the rate at a single point. It is defined as the limit of average rates of change as the interval shrinks down to that point. On a graph it is the slope of the tangent line, and it is what the derivative measures. For f(x) = x² at x = 1, shrink the interval to [1, 1 + h]: the average rate is [(1 + h)² − 1] / h = (2h + h²) / h = 2 + h, and as h approaches 0 this approaches 2. So the instantaneous rate at x = 1 is 2.

1.2 Defining Limits and Using Limit Notation

The limit of a function, written lim_x→c f(x) = L, means that 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." On the exam, correct limit notation is required when you justify a claim about a limit, so write the lim_x→c part every time.

The function does not need to be defined at c for the limit to exist. The limit describes what the function approaches, not what happens at the point itself. A limit value and a function value are two different numbers that happen to agree when a function is continuous.

1.3 Estimating Limit Values from Graphs

A one-sided limit looks at x approaching c from one side only. From the right it is written lim_x→c⁺ f(x), and from the left lim_x→c⁻ f(x). On a graph, trace the curve toward x = c from each direction separately and read the y-value each side approaches. The two-sided limit lim_x→c f(x) exists only if both one-sided limits exist and are equal to each other.

This is the tool for jump discontinuities. If the left branch approaches y = 1 and the right branch approaches y = 4 at x = 2, then lim_x→2⁻ f(x) = 1 and lim_x→2⁺ f(x) = 4, and because they differ, lim_x→2 f(x) does not exist. The value f(2), wherever the dot is drawn, does not change either one-sided limit.

1.5 Determining Limits Using Algebraic Properties of Limits

The algebraic properties of limits, also called the limit laws, let you break a limit apart. The limit of a sum, difference, product, or quotient equals the corresponding combination of the individual limits, as long as each limit exists and the denominator limit is not zero. Constant multiples and powers work the same way.

The first move is almost always direct substitution, which the limit laws justify for polynomials and for rational functions wherever the denominator is not zero. For example, lim_x→2 (x² + 3x) / (x + 4) = (4 + 6) / (2 + 4) = 10/6 = 5/3.

If substitution gives 0/0, the expression is indeterminate and needs algebra before you can judge it. Factoring and canceling is the most common fix. For lim_x→3 (x² − 9) / (x − 3), substitution gives 0/0, but factoring gives (x − 3)(x + 3) / (x − 3), which equals x + 3 for x ≠ 3, so the limit is 6.

1.8 Determining Limits Using the Squeeze Theorem

The Squeeze Theorem handles functions you cannot evaluate directly. 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 outer functions pin the middle one down to the same limit.

The standard example is lim_x→0 x · sin(1/x). The sine factor oscillates between −1 and 1 forever, so direct substitution fails, but −|x| ≤ x · sin(1/x) ≤ |x| for all x, and both bounds approach 0 as x → 0. The squeeze forces the limit to 0. A related result worth memorizing is lim_x→0 sin x / x = 1, which is proved the same way with a geometric squeeze and appears constantly in later units.

1.10 Exploring Types of Discontinuities

A discontinuity is a break in the graph, and there are three types. A removable discontinuity is a hole: the limit exists at x = c, but f(c) is undefined or does not equal the limit. Filling in the hole at the limit value would repair it. A jump discontinuity is where the left- and right-hand limits exist but are not equal, so the graph jumps 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 has a three-part definition. 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). In words, you can draw through the point without lifting the pencil, and the value you approach matches the value that is there.

Each discontinuity type fails the definition in a characteristic way. A removable discontinuity passes the limit condition but fails the equality, because f(c) is missing or wrong. A jump discontinuity fails the limit condition itself, because the two sides disagree. An infinite discontinuity fails both, since 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 needs to equal the function value, since there is no other side to check.

This is why the familiar function families are easy to work with. Polynomials are continuous everywhere. Rational functions are continuous everywhere they are defined, which is everywhere except where the denominator is zero. Sums, differences, products, and quotients of continuous functions stay 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 − 2)², the values blow up to positive infinity from both sides, so lim_x→2 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 is enough. The sign can differ by side, so check each. For f(x) = 1/x, lim_x→0⁺ 1/x = ∞ while lim_x→0⁻ 1/x = −∞, and x = 0 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 is the description of what happens in both directions, x → ∞ and x → −∞, often summarized by the horizontal asymptotes.

For rational functions, compare degrees. If the degree on top 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→∞ (3x² + 1) / (2x² − 5) = lim_x→∞ (3 + 1/x²) / (2 − 5/x²) = 3/2, so y = 3/2 is a horizontal asymptote. If the degree on top 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 that 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 typical justification: f(x) = x³ − x − 2 is a polynomial, so it is continuous everywhere, in particular on [1, 2]. Since f(1) = 1 − 1 − 2 = −2 and f(2) = 8 − 2 − 2 = 4, and 0 lies between −2 and 4, the IVT guarantees some c in (1, 2) with f(c) = 0. The theorem promises existence, not the location of c.

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 − 3)(x + 3) / (x − 3) = x + 3 for x ≠ 3, and the limit is 6. 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 is vague hand-waving that avoids doing the algebra.

2. C. The one-sided limits are 1 from the left and 4 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(2), which equals 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 − 2)(x + 2) / (x − 2) = x + 2 for x ≠ 2, so the limit is 4, but f(2) = 5. 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(2). C is wrong because a jump requires disagreeing one-sided limits, and here both sides approach 4. D confuses f(2) = 5 with the limit, which is 4.

4. A. Both bounds approach 0 as x → 0: lim_x→0 x² = 0 and lim_x→0 (−x²) = 0. The Squeeze Theorem forces g(x) to 0 as well. B ignores the theorem entirely. C may come from misreading x² near 0 as approaching 1. D is the trap of thinking the bounds are not enough; matching bounds are exactly what the theorem needs.

5. B. From the left, x − 2 is a small negative number, so 1 / (x − 2) is a large negative number: the limit is −∞. A ignores the side and reports the right-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 x³: (4 − 1/x²) / (2 + 5/x³), which approaches 4/2 = 2. 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 x³ terms control the behavior.

7. B. The IVT applies: f is continuous on [0, 3] and 0 lies between f(0) = −1 and f(3) = 5, so some c in (0, 3) satisfies f(c) = 0. A and C both claim the theorem locates the zero; it only guarantees existence, never the position. D confuses continuity with differentiability, which the IVT never requires.

8. A. A horizontal asymptote y = 3 as x → ∞ is exactly the statement that lim_x→∞ f(x) = 3. B confuses the asymptote's y-value with a function value at x = 3. 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

Say each answer out loud before you look back, and mark the ones you cannot finish. Anything you cannot say out loud yet belongs in your flashcard deck. In Rycal, add those items to the Limits and Continuity deck and let spaced review bring them back over the next few days.

• 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→3 (x² − 9) / (x − 3) from scratch, showing the algebra.
• State the Squeeze Theorem and use it to find lim_x→0 x · sin(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 AB 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.