MGU Mathematics Series — Trigonometry and Precalculus
Part V: Calculus Readiness and Mathematical Modeling
Limits · Continuity · Rates of Change · Calculus Preview
Chapter 13 of 14
MGU Mathematics Series  ·  Trigonometry and Precalculus  ·  Library Textbook Edition

Chapter 13:
Limits, Continuity,
and Rates of Change

An introduction — limits as approaching behavior, continuity as unbroken function graphs, average and instantaneous rates of change, secant and tangent lines, and area under a curve as a preview of integral calculus

■ Math Level: Precalculus / Calculus Bridge ■

■ 14 Sections  •  Part V of 5 ■

Follows Chapters 1–12: Functions, Trigonometry, Algebra, Analytic Geometry  ·  Prepares students for Calculus I

Purpose and Overview

This chapter serves as the bridge between precalculus and Calculus I. Every idea here grows out of something students have already studied: functions, graphs, domain, rate of change, and behavior at the edges of a domain. The goal is not to teach calculus here. The goal is to make calculus feel inevitable — like the natural next question after everything precalculus has built.

Calculus is the mathematics of change and accumulation. Precalculus is the foundation on which calculus stands. Students who understand functions deeply, who can read a graph carefully, who know what a rate of change means, and who have thought about what happens to a function's output as the input approaches a value — those students are genuinely ready for Calculus I.

This chapter introduces three central ideas of calculus as intuition, not formal theory: the limit, continuity, and the rate of change. Students will estimate limits from tables and graphs, interpret one-sided behavior, understand what continuity means and what breaks it, review average rates of change, see how secant lines approach tangent lines, get a first look at instantaneous rate of change as an idea, and preview the concept of area under a curve.

Opening Question

A car is traveling along a highway. Its speed is not constant — it accelerates, decelerates, and changes with traffic. You can measure the car's average speed over the last ten minutes: divide the total distance by ten minutes. But what is the car's speed at this exact instant — right now, at this precise moment?

You cannot measure an instant. An instant has zero length. But you can get closer and closer to an instant by shrinking the time interval: measure speed over five minutes, then two minutes, then thirty seconds, then ten seconds. As the interval shrinks toward zero, the average speed over that interval approaches a number. That number is the instantaneous speed. This is the idea of a limit. This is the beginning of calculus.

14Sections
15Objectives
16Key Terms
12+Worked Examples

Learning Objectives

  1. 1Explain what a limit describes in plain language and connect the idea to function behavior near a point.
  2. 2Estimate the limit of a function at a value using a table of inputs and outputs.
  3. 3Estimate the limit of a function at a value by reading a graph.
  4. 4Distinguish between one-sided limits from the left and from the right.
  5. 5Recognize when a limit exists and when it does not exist.
  6. 6Describe the behavior of a function as the input grows without bound or decreases without bound.
  7. 7State the definition of continuity in plain language and verify all three conditions.
  8. 8Identify the three types of discontinuity: removable, jump, and infinite.
  9. 9Determine whether a function is continuous at a point and on an interval.
  10. 10Compute and interpret an average rate of change.
  11. 11Describe the secant line through two points on a curve and connect it to average rate of change.
  12. 12Explain how the secant line approaches the tangent line as the interval shrinks toward zero.
  13. 13Describe the meaning of instantaneous rate of change as a limit of average rates of change.
  14. 14Explain the idea of area under a curve as an accumulation of infinitely thin strips.
  15. 15Recognize common misunderstandings about limits, continuity, and instantaneous behavior.

Sections at a Glance

Sections 13.1 – 13.6
Limits: What Functions Approach
Calculus Foundations · Limit Notation · Tables · Graphs · One-Sided Limits · Infinite Behavior

Why Calculus Needs Precalculus

Every major idea in calculus — derivatives, integrals, infinite series — depends directly on the function concepts studied in precalculus. Before students can understand what a derivative is, they must understand what a function is, how to evaluate it, how to read its graph, how to compute rates of change, and how to reason about its behavior near special values. Before students can understand integrals, they must be comfortable with area, accumulation, and summation.

Precalculus is not an obstacle before calculus. It is the architecture on which calculus is built.

Branch of CalculusCentral QuestionKey Tool
Differential CalculusAt any given instant, how fast is this function changing?The derivative
Integral CalculusHow much has accumulated over a period of time or distance?The integral
Common FoundationAs the input approaches a value, what does the output approach?The limit
Key Insight

Students have already studied average rates of change in Chapter 1. This chapter will show how that familiar idea leads naturally to instantaneous rate of change — and to calculus. The limit is the bridge that connects the two.

MGU Library Connection
  • See the Calculus I course pages for the formal definitions of derivative and integral
  • The ideas previewed in this chapter are developed fully in that course
  • Appendix R of this textbook: Calculus I Readiness Check

What a Limit Describes

Imagine approaching a building on foot. As you walk toward it, the building gets closer and closer. The concept of a limit is similar: you are asking what value a function's output approaches as the input gets closer and closer to a particular number.

Given a function f(x) and a value x = a, the central question is: as x gets closer and closer to a — from either side — what value does f(x) approach?

Limit Notation

The limit of f(x) as x approaches a equals L is written:

lim[x→a] f(x) = L

Read: "The limit of f of x as x approaches a equals L."

This is not asking for the value of f(a). It is asking what value f(x) gets closer and closer to as x approaches a. The input approaches a but may never equal a.

A Concrete Example Before Formulas

Consider the function f(x) = (x² – 1) / (x – 1). This function is undefined at x = 1 because the denominator equals zero there. But for x ≠ 1, the numerator factors:

f(x) = (x–1)(x+1) / (x–1) = x + 1 for x ≠ 1

So even though f(1) does not exist, the function behaves like y = x + 1 near x = 1. As x approaches 1, the output approaches 1 + 1 = 2. Therefore: lim[x→1] (x²–1)/(x–1) = 2.

Informal Definition

The limit of f(x) as x approaches a equals L if the values of f(x) can be made arbitrarily close to L by taking x sufficiently close to a (but not equal to a) on either side. A limit is about approaching, not arriving.

Example 13.2.1
Interpreting limit notation in plain English
  • ProblemState what lim[x→3] f(x) = 7 means in plain English.
  • ThinkLimit notation is a statement about approaching behavior. Translate each part separately.
  • SolveAs the input x gets closer and closer to 3 (from either side, without equaling 3), the output f(x) gets closer and closer to 7.
  • CheckThis statement says nothing about f(3) itself. The function could equal 7 at x = 3, equal something else entirely, or be undefined there.
Answer: As x approaches 3, the output of f approaches 7.

Estimating Limits from Tables

One of the most concrete ways to investigate a limit is by constructing a table of values. Students choose inputs that approach the target value from both sides and observe what the outputs approach.

Strategy: Estimating a Limit from a Table

1. Choose input values approaching a from the left (slightly less than a).

2. Choose input values approaching a from the right (slightly greater than a).

3. Compute the output f(x) for each input.

4. If outputs from both sides appear to approach the same value L, then lim[x→a] f(x) = L.

Example 13.3.1
Estimating a limit from a table
  • ProblemEstimate lim[x→2] (x² – 4) / (x – 2) using a table.
  • ThinkThis function is undefined at x = 2. Build a table with inputs approaching 2 from both sides.
  • Represent
    xf(x) = (x² – 4) / (x – 2)
    1.93.9
    1.993.99
    1.9993.999
    2.0014.001
    2.014.01
    2.14.1
  • SolveFrom the left, outputs approach 4. From the right, outputs also approach 4.
  • CheckAlgebraically: (x² – 4)/(x – 2) = (x–2)(x+2)/(x–2) = x + 2 for x ≠ 2. As x → 2, x + 2 → 4. ✓
Answer: lim[x→2] (x² – 4)/(x – 2) = 4
Example 13.3.2
A limit that does not exist — oscillation
  • ProblemEstimate lim[x→0] sin(1/x) using a table.
  • ThinkAs x → 0, the input to sine, 1/x, grows without bound. Sine oscillates between −1 and 1 infinitely many times.
  • Represent
    xsin(1/x)
    0.1sin(10) ≈ −0.544
    0.01sin(100) ≈ −0.506
    0.001sin(1000) ≈ 0.827
    0.0001sin(10000) ≈ −0.306
  • SolveThe outputs do not settle toward any single value. They oscillate unpredictably. No single value L is being approached.
Answer: lim[x→0] sin(1/x) does not exist.
Student Warning

A table can only suggest a limit. It cannot prove one. Use tables as estimation tools, not as definitive proofs. If the table values are not settling clearly toward a single number, suspect that the limit may not exist.

Estimating Limits from Graphs

A graph of a function gives visual evidence about limits. Students can trace a graph from either side of a target input value and observe whether the graph appears to approach a single output value from both sides. Crucially, look at where the graph is going — not at any dot that may or may not be marked at x = a.

Example 13.4.1
Reading a limit at a hole
  • ProblemEstimate lim[x→2] f(x) from the graph described below.
  • Represent
    Graph Description

    Draw the graph of y = x + 1 as a straight line with slope 1 and y-intercept 1. At the point (2, 3), place an open circle — indicating that the function is not defined at x = 2. There is no filled dot at x = 2. The graph is a straight line missing exactly one point.

  • SolveFrom the left, as x approaches 2, the graph rises toward height 3. From the right, as x approaches 2, the graph falls toward height 3. Both sides approach the same height.
  • CheckThe open circle confirms f(2) is undefined, but the limit still exists.
Answer: lim[x→2] f(x) = 3, even though f(2) is undefined.
Example 13.4.2
Reading a limit when the function value disagrees
  • ProblemEstimate lim[x→2] g(x) from the graph described below.
  • Represent
    Graph Description

    Draw the graph of y = x + 1 for all x ≠ 2, with an open circle at (2, 3). Also draw a filled dot at (2, 5), indicating that g(2) = 5. The isolated filled dot floats above the open circle of the line.

  • SolveFrom the left and right, the graph approaches height 3. The filled dot at 5 is an isolated point — the graph is not approaching 5 from either side.
Answer: lim[x→2] g(x) = 3, even though g(2) = 5.
Key Insight

The limit describes where the graph is heading, not where the dot is placed. These two things can be different, and recognizing that difference is the first step in understanding continuity.

One-Sided Limits as an Introduction

Sometimes a function approaches different values from the left and from the right. When this happens, the two-sided limit does not exist, but each one-sided limit may still be defined.

One-Sided Limit Notation

lim[x→a⁻] f(x) — the left-hand limit. As x approaches a from values less than a.

lim[x→a⁺] f(x) — the right-hand limit. As x approaches a from values greater than a.

The two-sided limit lim[x→a] f(x) = L exists if and only if both one-sided limits exist and are equal. If they differ, the two-sided limit does not exist.

Example 13.5.1
One-sided limits from a piecewise function
  • Problem Let f(x) = { x + 1 if x < 2 | 5 if x = 2 | 3x − 2 if x > 2 }.
    Find the left-hand limit, right-hand limit, and two-sided limit at x = 2.
  • ThinkUse the rule that applies for each approach direction. The value at x = 2 exactly (f(2) = 5) is irrelevant to the limits.
  • Solve Left-hand limit (x → 2⁻): use f(x) = x + 1 lim[x→2⁻] f(x) = 2 + 1 = 3 Right-hand limit (x → 2⁺): use f(x) = 3x − 2 lim[x→2⁺] f(x) = 3(2) − 2 = 4
  • CheckLeft limit = 3, right limit = 4. They are not equal. The two-sided limit does not exist.
Answer: Left-hand limit = 3, right-hand limit = 4. The two-sided limit does not exist.
Graph Description — Example 13.5.1

Draw three pieces: (1) A ray from the left following y = x + 1, approaching the open circle at (2, 3). (2) A filled dot at (2, 5), isolated above the graph. (3) A ray from the right following y = 3x − 2, approaching the open circle at (2, 4). The two open circles sit at different heights, showing clearly that the left and right approaches do not meet.

Infinite Behavior and End Behavior

Some limits describe what happens as the input grows without bound. Others describe what happens as a function's output grows without bound near a particular input. Both types appear in graphs students have already studied.

TypeNotationMeaningGraph Feature
Limit at infinitylim[x→∞] f(x) = LAs x grows without bound, f(x) approaches LHorizontal asymptote y = L
Limit at −∞lim[x→−∞] f(x) = LAs x decreases without bound, f(x) approaches LHorizontal asymptote y = L
Infinite limitlim[x→a] f(x) = ±∞f(x) grows without bound near x = aVertical asymptote x = a
Example 13.6.1
Limit at infinity for a rational function
  • ProblemEvaluate lim[x→∞] (3x + 1) / (x − 2).
  • ThinkFor large x, divide numerator and denominator by x to isolate the dominant behavior.
  • Solve (3x + 1) / (x − 2) = (3 + 1/x) / (1 − 2/x) As x → ∞: 1/x → 0 and 2/x → 0 lim = (3 + 0) / (1 − 0) = 3
  • CheckThe horizontal asymptote is y = 3, consistent with the degree of numerator and denominator being equal and the ratio of leading coefficients being 3/1.
Answer: lim[x→∞] (3x + 1)/(x − 2) = 3. Horizontal asymptote at y = 3.
Example 13.6.2
Infinite limit near a vertical asymptote
  • ProblemDescribe the behavior of f(x) = 1/(x − 3) as x → 3.
  • ThinkAs x approaches 3, the denominator approaches 0 while the numerator stays at 1. The sign of the denominator determines whether the output shoots to +∞ or −∞.
  • Solve From the right (x → 3⁺): denominator is small and positive 1 / (small positive) → +∞ From the left (x → 3⁻): denominator is small and negative 1 / (small negative) → −∞
  • CheckThe two one-sided limits are not equal (one is +∞, one is −∞), so the two-sided limit does not exist. There is a vertical asymptote at x = 3.
Answer: lim[x→3⁺] f(x) = +∞, lim[x→3⁻] f(x) = −∞. Two-sided limit does not exist.
Student Warning — Infinity Is Not a Number

When we write lim[x→a] f(x) = ∞, we mean the function grows without bound — this is a description of behavior, not a value the limit equals. Technically, such a limit does not exist as a finite number. ∞ is not a number you can substitute or compute with.

Sections 13.7 – 13.8
Continuity: Unbroken Function Behavior
Three Conditions for Continuity · Removable · Jump · Infinite Discontinuities

Continuity as Unbroken Behavior

A function is continuous if its graph can be drawn without lifting the pencil. That informal description captures something precise: a function is continuous at a point if the function has a value there, if the limit exists there, and if the value and the limit agree.

Three Conditions for Continuity at x = a

1. f(a) is defined.

2. lim[x→a] f(x) exists.

3. lim[x→a] f(x) = f(a).

If any one of these three conditions fails, f is not continuous at x = a — it has a discontinuity there.

Example 13.7.1
Checking continuity at a point — a hole
  • ProblemIs f(x) = (x² − 9)/(x − 3) continuous at x = 3?
  • ThinkCheck all three conditions in order.
  • Solve Condition 1: f(3) = (9 − 9)/(3 − 3) = 0/0 → Undefined. ✗ Since Condition 1 fails, f is not continuous at x = 3. (Note: the limit does exist — lim[x→3] f(x) = 6 — but that cannot rescue continuity if f(3) is undefined.)
Answer: Not continuous at x = 3. Removable discontinuity (hole) at (3, 6).
Example 13.7.2
Verifying continuity for a polynomial
  • ProblemIs g(x) = x² + 3x − 1 continuous at x = 2?
  • Solve Condition 1: g(2) = 4 + 6 − 1 = 9 ✓ (defined) Condition 2: lim[x→2] (x² + 3x − 1) = 9 ✓ (limit exists) Condition 3: limit (9) = function value (9) ✓
Answer: g is continuous at x = 2. All three conditions hold.
Shortcut for Familiar Functions

Polynomial, exponential, sine, and cosine functions are continuous everywhere on their domains. For these functions, the limit at any point always equals the function value — no need to verify each condition separately.

Discontinuities

When a function fails to be continuous at a point, it has a discontinuity there. There are three main types, each with a distinct graph appearance.

Type 1: Removable Discontinuity (Hole)

A removable discontinuity occurs when the limit exists, but either f(a) is undefined or f(a) does not equal the limit. This appears as a hole in the graph. The discontinuity is called "removable" because it can be repaired by redefining the function's value at that one point.

Graph Description — Removable Discontinuity

Draw the line y = x + 2. At the point (2, 4), place an open circle. The graph is a straight line missing exactly one point. Optionally, a filled dot at (2, 7) would indicate a redefined but still-discontinuous value.

Type 2: Jump Discontinuity

A jump discontinuity occurs when both one-sided limits exist but are not equal. The graph appears to jump from one height to another at x = a.

Graph Description — Jump Discontinuity

Draw a horizontal ray at height 2 for x < 1, ending in an open circle at (1, 2). Draw a horizontal ray at height 5 for x ≥ 1, starting with a filled dot at (1, 5). There is a visible vertical gap between the two pieces — the graph literally jumps from height 2 to height 5 at x = 1.

Type 3: Infinite Discontinuity

An infinite discontinuity occurs when f(x) grows without bound as x approaches a. The graph has a vertical asymptote at x = a.

Graph Description — Infinite Discontinuity

Draw the hyperbola y = 1/x. As x → 0⁺, the right branch rises toward +∞. As x → 0⁻, the left branch falls toward −∞. A dashed vertical line marks the y-axis (x = 0) as the asymptote. The function is not defined at x = 0.

Summary of Discontinuity Types

TypeLeft LimitRight LimitTwo-Sided Limitf(a) Defined?
Removable (hole)LLExists (= L)No, or f(a) ≠ L
JumpL₁L₂ ≠ L₁Does not existMay or may not be
Infinite±∞±∞Does not existNo
Practice 13.7–13.8Continuity Skill Check

For each function, determine whether it is continuous at the given point. Identify the type of discontinuity if any condition fails.

  1. f(x) = 3x² − 2x + 5 at x = −1
  2. g(x) = (x + 2)/(x − 4) at x = 4
  3. h(x) = (x² − 9)/(x − 3) at x = 3
  4. f(x) = {x+3 if x<1, 2x+2 if x≥1} at x = 1
  5. f(x) = sin(x) at x = π
  6. g(x) = 1/(x+2)² at x = −2
Sections 13.9 – 13.10
Average Rates of Change and Secant Lines
Average Rate of Change · Secant Line Slope · Shrinking Intervals

Average Rate of Change Review

Students have studied average rate of change throughout precalculus. This section reviews that idea carefully, because it is the direct algebraic ancestor of the derivative.

Average Rate of Change

The average rate of change of f on the interval [a, b] is:

[f(b) − f(a)] / (b − a)

This is the slope of the line connecting the two points (a, f(a)) and (b, f(b)) on the graph of f. It measures how much the output changes per unit of input, averaged across the entire interval.

Example 13.9.1
Computing average rate of change
  • ProblemFind the average rate of change of f(x) = x² on [1, 4].
  • ThinkThis is the slope of the line from (1, f(1)) to (4, f(4)).
  • Solve f(1) = 1, f(4) = 16 Average rate of change = (16 − 1) / (4 − 1) = 15 / 3 = 5
  • CheckOn average, the output increases by 5 for every 1-unit increase in input across [1, 4].
Answer: Average rate of change = 5.
Example 13.9.2
Average rate of change with context
  • ProblemA population of bacteria is modeled by P(t) = 200 · 2^t, where t is hours and P is the number of bacteria. Find the average rate of change from t = 1 to t = 3.
  • Solve P(1) = 200 · 2¹ = 400 P(3) = 200 · 2³ = 1600 Average rate = (1600 − 400) / (3 − 1) = 1200 / 2 = 600
  • CheckUnits: bacteria per hour. This is an average — the actual rate varies because the function is exponential, not linear.
Answer: The population grows at an average rate of 600 bacteria per hour between t = 1 and t = 3 hours.

Secant Lines

The line connecting two points on a curve is called a secant line. Its slope equals the average rate of change of the function between those two points.

Secant Line Slope

Given a function f and two distinct input values a and b, the secant line through (a, f(a)) and (b, f(b)) has slope:

m_sec = [f(b) − f(a)] / (b − a)

As the two points are moved closer together — as b is moved toward a — the secant line changes its angle. In the limit, it becomes the tangent line.

Example 13.10.1
Finding and interpreting a secant line
  • ProblemFind the equation of the secant line through (1, f(1)) and (3, f(3)) for f(x) = x².
  • Solve f(1) = 1, f(3) = 9 Slope: m = (9 − 1) / (3 − 1) = 8/2 = 4 Point-slope form using (1, 1): y − 1 = 4(x − 1) y = 4x − 3
  • Represent
    Graph Description

    Draw the parabola y = x². Mark the points (1, 1) and (3, 9). Draw the straight line y = 4x − 3 passing through both points. This is the secant line. It crosses the parabola at two locations and passes above the parabola between them.

Answer: The secant line is y = 4x − 3.

Shrinking the Interval — Watching Secant Slopes Converge

What happens if the second point moves closer to the first? The table below computes the secant slope for f(x) = x² with a = 1 and decreasing intervals.

b (second point)f(b)Secant Slope = [f(b) − 1] / (b − 1)
394
243
1.52.252.5
1.11.212.1
1.011.02012.01
1.0011.0020012.001
What the Table Shows

As b approaches 1, the secant slope approaches 2. This suggests that the slope of the curve at x = 1 — the instantaneous rate of change — is 2. This is the idea of the tangent line.

Sections 13.11 – 13.13
Calculus Previews: Tangent Lines, Derivatives, and Area
Tangent Lines · Instantaneous Rate of Change · Riemann Sums · Definite Integral Preview

Tangent Lines as a Preview

A tangent line at a point on a curve is the line that just touches the curve at that point and has the same direction (slope) as the curve at that exact location. Unlike a secant line, which crosses the curve at two points, the tangent line touches the curve at one point and represents the curve's local direction there.

The Tangent Line as a Limit of Secant Lines

As b approaches a, the secant line through (a, f(a)) and (b, f(b)) rotates and approaches a limiting position: the tangent line at x = a. Its slope is:

m_tangent = lim[b→a] [f(b) − f(a)] / (b − a)

In Calculus I, this limit is called the derivative f′(a).

Example 13.11.1
Slope of the tangent line via a limit
  • ProblemFind the slope of the tangent line to f(x) = x² at x = 1 using a limit.
  • ThinkCompute [f(b) − f(1)] / (b − 1) and take b → 1.
  • Solve [f(b) − f(1)] / (b − 1) = (b² − 1) / (b − 1) = (b − 1)(b + 1) / (b − 1) = b + 1 for b ≠ 1 lim[b→1] (b + 1) = 2
  • Represent
    Graph Description

    Draw the parabola y = x². Mark the point (1, 1). Draw the tangent line y = 2x − 1 with slope 2. The tangent line touches the parabola exactly at (1, 1) and runs alongside the curve locally, neither crossing it steeply nor flattening away.

  • CheckIn Calculus I, the derivative of f(x) = x² is f′(x) = 2x. At x = 1, f′(1) = 2. ✓ This chapter derived that result using limits alone.
Answer: Slope of tangent = 2. Tangent line: y = 2x − 1.

Instantaneous Rate of Change as a Preview

The instantaneous rate of change of a function at a single point describes how fast the function is changing at that exact input — not over an interval, but at a single moment. Limits resolve the apparent paradox of measuring change at a single point.

Instantaneous Rate of Change (Limit Definition — Preview)

The instantaneous rate of change of f at x = a is:

lim[h→0] [f(a + h) − f(a)] / h

Here, h is the width of the interval. As h → 0, the two points (a, f(a)) and (a+h, f(a+h)) merge into one, and the secant slope approaches the tangent slope. This limit is the definition of f′(a) in Calculus I.

Example 13.12.1
Instantaneous rate of change via the difference quotient
  • ProblemFind the instantaneous rate of change of f(x) = x² + 1 at x = 2.
  • ThinkCompute [f(2 + h) − f(2)] / h, simplify, and take the limit as h → 0.
  • Solve f(2) = 4 + 1 = 5 f(2 + h) = (2 + h)² + 1 = 4 + 4h + h² + 1 = 5 + 4h + h² [f(2+h) − f(2)] / h = [(5 + 4h + h²) − 5] / h = (4h + h²) / h = h(4 + h) / h = 4 + h lim[h→0] (4 + h) = 4
  • CheckIn Calculus I, the derivative of f(x) = x² + 1 is f′(x) = 2x. At x = 2, f′(2) = 4. ✓
Answer: The instantaneous rate of change of f(x) = x² + 1 at x = 2 is 4. At exactly x = 2, the function is increasing at a rate of 4 output units per input unit.
FieldInstantaneous Rate of Change Means
PhysicsInstantaneous velocity — the derivative of position with respect to time
BiologyInstantaneous population growth rate
EconomicsMarginal cost — the derivative of the cost function
MedicineRate of change of drug concentration in the bloodstream
EngineeringRate of temperature change, rate of current change in a circuit

Area Under a Curve as a Preview

Integral calculus studies accumulation: how much of something accumulates as a variable changes. One of the most geometric ways to understand accumulation is through the idea of area under a curve.

Given a function f(x) ≥ 0 on an interval [a, b], what is the total area of the region bounded above by the graph, below by the x-axis, on the left by x = a, and on the right by x = b? For straight-line graphs, this question is straightforward. For curves, it requires limits.

Riemann Sums — Approximating Area with Rectangles

Divide [a, b] into n equal subintervals of width Δx = (b − a)/n. On each subinterval, build a rectangle whose height equals the function's value at a representative input. The area approximation is:

Area ≈ Σ f(xₖ) · Δx (sum from k = 1 to n)

As n → ∞ and Δx → 0, the rectangles fill the region precisely. The exact area is the limit of these Riemann sums, written in Calculus I as:

∫[a to b] f(x) dx

Example 13.13.1
Approximating area with rectangles
  • ProblemApproximate the area under f(x) = x² on [0, 2] using four equal rectangles with right endpoints.
  • ThinkDivide [0, 2] into four subintervals of width Δx = 2/4 = 0.5. Use the right endpoint of each subinterval as the rectangle height.
  • Solve
    SubintervalRight Endpointf(x) = x²Rectangle Area
    [0, 0.5]x = 0.50.250.25 × 0.5 = 0.125
    [0.5, 1]x = 111 × 0.5 = 0.5
    [1, 1.5]x = 1.52.252.25 × 0.5 = 1.125
    [1.5, 2]x = 244 × 0.5 = 2.0
    Total ≈ 0.125 + 0.5 + 1.125 + 2.0 = 3.75
  • CheckThe exact area (from Calculus I: ∫[0 to 2] x² dx = 8/3 ≈ 2.67). The approximation overestimates because right endpoints overshoot the curve on this increasing function. More rectangles reduce the error.
  • Represent
    Graph Description

    Draw the parabola y = x² from x = 0 to x = 2. Draw four rectangles on [0, 0.5], [0.5, 1], [1, 1.5], and [1.5, 2], each using the right endpoint height. The rectangles extend above the curve on each subinterval, showing the overestimate. As more rectangles are used, the staircase pattern of rectangle tops approaches the smooth curve.

Answer: Approximate area ≈ 3.75 (overestimate). Exact area = 8/3 ≈ 2.67.
The Fundamental Connection

In Calculus I, students will discover that differentiation and integration are inverse processes. The derivative of the area function (with respect to the upper limit) returns the original function. This relationship is called the Fundamental Theorem of Calculus — one of the most important results in the history of mathematics.

Common Limit and Continuity Misunderstandings

Misunderstanding 1 — Confusing the Limit with the Function Value
lim[x→a] f(x) = f(a) (automatic assumption)

The limit describes approaching behavior, not the value at the point. They may differ. Check all three continuity conditions separately — never assume they agree without verification.

Misunderstanding 2 — Believing a Limit Must Be Achieved
"If lim[x→a] f(x) = L, then f reaches the value L at x = a."

A limit is about approaching, not arriving. The function may approach L without ever equaling L at x = a — for example, when there is a hole at (a, L).

Misunderstanding 3 — Treating ∞ as a Number
lim[x→a] f(x) = ∞, so the limit equals infinity.

Writing lim[x→a] f(x) = ∞ describes unbounded behavior, not a finite value the limit equals. Technically, the limit does not exist in this case. ∞ is not a number.

Misunderstanding 4 — Thinking Limits Only Matter at Undefined Points
"Limits only apply when the function is undefined."

Limits apply everywhere. Even perfectly continuous functions have limits at every point — they just happen to equal the function values. Limits are universal; continuity is the special case where limit and value agree.

Misunderstanding 5 — Confusing Average Rate with Instantaneous Rate
"Average rate of change over [1, 4] is 5, so the instantaneous rate at x = 1 is 5."

Average rate of change is the slope of the secant — it describes behavior over an interval. Instantaneous rate is the slope of the tangent — it describes behavior at one exact point. They are related through a limit, but not equal in general.

Misunderstanding 6 — Recording a Riemann Sum as the Exact Area
Riemann sum with 4 rectangles = 3.75 = exact area.

A Riemann sum is an approximation. The exact area is the limit as the number of rectangles grows without bound. The approximation improves as rectangles become thinner, but it equals the exact area only in the limit.

Chapter Summary

Key Terms Review

limit
The value L that f(x) approaches as x approaches a; written lim[x→a] f(x) = L. About approaching behavior, not the function value at the point.
left-hand limit
The limit as x approaches a from values less than a. Notation: lim[x→a⁻] f(x).
right-hand limit
The limit as x approaches a from values greater than a. Notation: lim[x→a⁺] f(x).
limit does not exist
Occurs when left and right limits differ, the output oscillates without settling, or the output grows without bound.
end behavior
How f(x) behaves as x → ∞ or x → −∞. Described by limits at infinity and connected to horizontal asymptotes.
infinite limit
A limit in which f(x) grows without bound as x approaches a. Associated with a vertical asymptote. Not a finite number.
continuity at a point
f(a) is defined, the limit at a exists, and they are equal. All three conditions must hold.
removable discontinuity
A hole in the graph. The limit exists, but f(a) is undefined or disagrees with it. Can be repaired by redefining the function at one point.
jump discontinuity
Both one-sided limits exist but are not equal. The graph jumps at x = a.
infinite discontinuity
The function grows without bound near x = a. The graph has a vertical asymptote.
average rate of change
[f(b) − f(a)] / (b − a). The slope of the secant line through two points on the graph of f.
secant line
The line through two points on a curve. Its slope equals the average rate of change between those points.
tangent line
The line touching a curve at one point, matching its local direction. Its slope is the limit of secant slopes as the interval shrinks to zero.
instantaneous rate of change
The limit of the average rate of change as the interval shrinks to zero. Equals the slope of the tangent line. The derivative in Calculus I.
Riemann sum
An approximation of the area under a curve using rectangles. Becomes more precise as the number of rectangles increases.
definite integral
The limit of Riemann sums as the number of rectangles grows without bound. The exact area under a curve on [a, b]. Written ∫[a to b] f(x) dx in Calculus I.

Concept Review Questions

  1. In your own words, explain what a limit describes. How is a limit different from the value of the function at the point?
  2. What must be true for a two-sided limit to exist? What is the relationship between the one-sided limits and the two-sided limit?
  3. State the three conditions required for a function to be continuous at a point.
  4. Describe each type of discontinuity — removable, jump, and infinite — in your own words and describe what each looks like on a graph.
  5. What is the relationship between the average rate of change and the slope of the secant line?
  6. How does the tangent line arise from the secant line? What happens as the two points on a curve get closer together?
  7. What is the instantaneous rate of change? How is it defined using a limit?
  8. What is a Riemann sum, and how does it approximate the area under a curve?
  9. Explain in your own words why the limit is the central idea connecting average rates of change to instantaneous rates of change.
  10. Explain why precalculus is necessary preparation for calculus, using at least three specific ideas from this chapter.

Limit Estimation Practice

Estimate each limit using a table or graph description. Show your reasoning.

  1. Use a table to estimate lim[x→1] (x³ − 1)/(x − 1). Try x = 0.9, 0.99, 0.999, 1.001, 1.01, 1.1.
  2. Use a table to estimate lim[x→0] sin(x)/x (x in radians). Try x = ±0.1, ±0.01, ±0.001.
  3. A graph shows a curve approaching height −2 from both sides as x → 4, with an open circle at (4, −2) and a filled dot at (4, 1). State: (a) lim[x→4] f(x), (b) f(4), (c) whether f is continuous at x = 4.
  4. A graph shows: curve approaching height 3 as x → 2⁻, and height 6 as x → 2⁺. State: (a) the left-hand limit, (b) the right-hand limit, (c) the two-sided limit.
  5. Evaluate: lim[x→∞] (5x² − 3)/(2x² + 1).
  6. Evaluate: lim[x→−∞] (4x − 7)/(x + 2).
  7. Evaluate: lim[x→5⁺] 1/(x − 5).
  8. Evaluate: lim[x→0] (x² + 3x)/x.
  9. Given f(x) = (x² − 16)/(x − 4), estimate lim[x→4] f(x) and state whether f is continuous at x = 4. Justify.
  10. Given h(x) = {2x+1 if x<3, x²−2 if x≥3}: Find (a) lim[x→3⁻], (b) lim[x→3⁺], (c) lim[x→3], (d) h(3), (e) is h continuous at x = 3?
PracticeContinuity

Determine whether each function is continuous at the given point. State which continuity condition fails, if any, and identify the type of discontinuity.

  1. f(x) = 3x² − 2x + 5 at x = −1
  2. g(x) = (x+2)/(x−4) at x = 4
  3. h(x) = (x²−9)/(x−3) at x = 3
  4. f(x) = {x+3 for x<1, 2x+2 for x≥1} at x = 1
  5. f(x) = sin(x) at x = π
  6. g(x) = 1/(x+2)² at x = −2
  7. f(x) = {x²−1 for x≠2, 5 for x=2} at x = 2
PracticeRates of Change
  1. Average rate of change of f(x) = 3x − 5 on [−2, 4]
  2. Average rate of change of f(x) = x² − x on [1, 5]
  3. Average rate of change of f(x) = √x on [4, 9]
  4. Average rate of change of f(x) = 2^x on [0, 3]
  5. Secant slope through (2, f(2)) and (5, f(5)) for f(x) = x² + 1
  6. Equation of secant line through (1, f(1)) and (3, f(3)) for f(x) = x² − 2

For Problems 7–8: For f(x) = x², complete the table of secant slopes from x = 3 to x = 3 + h: h = 1, 0.5, 0.1, 0.01, 0.001. Then use the limit definition to find the instantaneous rate of change of f(x) = x² + 4x at x = 1.

Error Analysis

Each problem contains a student's work with an error. Identify the error and provide the correct solution.

Problem 1
lim[x→3] (x² − 9) = 0, therefore lim[x→3] (x² − 9)/(x − 3) = 0/0 = 0

What is wrong? The form 0/0 is indeterminate — it does not equal 0. What is the correct limit?

Problem 2
"f(x) = |x| has a sharp corner at x = 0, so it is discontinuous there."

What is wrong? Check all three continuity conditions for f(x) = |x| at x = 0. Is this function actually continuous there?

Problem 3
lim[x→∞] (x² + 1)/(3x + 2) = ∞/∞ = 1

Two errors here: ∞/∞ is indeterminate, and the limit is not 1. What is the correct approach and answer?

Problem 4
"f(x) = 1/x is discontinuous only at x = 0, so it is continuous on [−1, 1]."

What is wrong? An interval containing a discontinuity cannot be called continuous.

Problem 5
"Average rate of change of f(x) = x² from x = 2 to x = 4 is 6, so instantaneous rate at x = 2 is 6."

What is the error? What would the instantaneous rate of change at x = 2 be? (Use the limit definition.)

Chapter 13 Checkpoint

Chapter 13 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 13
Part A — Limits
  1. Define a limit in your own words. What does lim[x→a] f(x) = L mean?
  2. Evaluate: lim[x→4] (x² − 16)/(x − 4).
  3. Given the table below, estimate lim[x→2] f(x): inputs 1.9, 1.99, 1.999, 2.001, 2.01, 2.1 give outputs 4.61, 4.9601, 4.996001, 5.004001, 5.0401, 5.41.
  4. Let g(x) = {3x − 1 if x < 2, x² + 1 if x ≥ 2}. Find: (a) lim[x→2⁻] g(x), (b) lim[x→2⁺] g(x), (c) lim[x→2] g(x).
  5. Evaluate: lim[x→∞] (6x² − 1)/(2x² + 5).
Part B — Continuity
  1. State all three conditions required for f to be continuous at x = a.
  2. Determine whether f(x) = (x² − 25)/(x − 5) is continuous at x = 5. Identify the type of discontinuity if any condition fails.
  3. Is h(x) = cos(x) continuous at x = π? Justify using all three conditions.
  4. A function has a jump discontinuity at x = 3. Describe what the graph looks like near x = 3.
Part C — Rates of Change
  1. Find the average rate of change of f(x) = x² − 4x + 1 on [1, 5].
  2. Find the equation of the secant line through (2, f(2)) and (4, f(4)) for f(x) = x².
  3. Use the limit definition to find the instantaneous rate of change of f(x) = 3x² at x = 2. Show each step.
Part D — Calculus Preview
  1. Approximate the area under f(x) = x + 1 on [0, 4] using four equal rectangles with left endpoints. Compare to the exact area (a trapezoid).
  2. In two to three sentences, describe the connection between secant lines, tangent lines, and the derivative.
  3. Explain why lim[h→0] [f(a+h) − f(a)]/h gives the instantaneous rate of change of f at x = a.
Bridge to Calculus I
Chapter 14 and Calculus I Readiness

Chapter 13 introduced the foundational ideas of calculus — limits, continuity, and rates of change — as intuition built on precalculus. Chapter 14 will close the book by bringing together all function families, modeling techniques, and mathematical reasoning skills developed throughout the course in a capstone review.

In Calculus I, the derivative will be formalized: derivative rules (Power Rule, Product Rule, Chain Rule) will make computation far faster, while the conceptual foundation built here explains why those rules are valid. The Intermediate Value Theorem will use continuity to prove the existence of solutions. The Fundamental Theorem of Calculus will reveal that differentiation and integration — the two great branches of calculus — are inverse operations.

Students who have worked through this textbook arrive at Calculus I knowing functions deeply, reading graphs fluently, working with trigonometric, exponential, logarithmic, and polynomial functions, and understanding the limit as the bridge between algebra and calculus. That readiness is what this book was designed to build.

MGU Library Connection
  • Calculus I textbook — formal derivative rules, Intermediate Value Theorem, Fundamental Theorem of Calculus
  • Appendix P of this textbook: Limit and Continuity Preview Reference
  • Appendix R of this textbook: Calculus I Readiness Check
  • MGU Dictionary entries: limit, continuity, derivative, integral, Riemann sum
  • Physics I — instantaneous velocity and acceleration as derivatives of position
  • Finance — marginal cost, marginal revenue as derivatives of cost and revenue functions