MGU Mathematics Series — Calculus
Part I: Limits and the Foundations of Calculus
Limits · Continuity · Asymptotes · Intermediate Value Theorem
Chapter 2 of 14
MGU Mathematics Series  ·  Calculus  ·  Library Textbook Edition

Chapter 2:
Limits and
Continuity

The language that makes calculus rigorous — what functions approach, where they are continuous, and how limiting behavior reveals the foundations of every derivative and integral

■ Math Level: Calculus I ■

■ 14 Sections  •  Part I of 5 ■

Follows Chapter 1: What Calculus Studies  ·  Prepares students for Chapter 3: The Derivative

Chapter 2 — Front Matter

Purpose and Overview

Chapter 1 introduced the central ideas of calculus: that calculus studies change, accumulation, motion, and limiting behavior. It showed that the derivative would answer questions about instantaneous rate of change and that the integral would answer questions about accumulated quantity. Both of those ideas rest on a foundation that has not yet been fully built.

That foundation is the concept of a limit.

This chapter builds that foundation carefully. Before computing a derivative or evaluating an integral, a student must understand what it means for a function to approach a value, what it means for a limit to exist, and what it means for a function to be continuous. These ideas are not merely preliminary formalities — they are the precise language that makes calculus work. Every major definition in calculus, the derivative, the definite integral, the Fundamental Theorem, is stated in terms of limits.

Chapter 2 introduces limits visually and numerically before moving to algebraic techniques. Students will estimate limits from tables and graphs, evaluate limits using algebra, understand one-sided limits, explore limits involving infinity, and study continuity as a property of functions on intervals. The chapter closes with an introduction to the Intermediate Value Theorem, a foundational result that depends directly on continuity.

Opening Question

Consider the function f(x) = (x² − 4) / (x − 2). If you try to evaluate this at x = 2, the denominator becomes zero — the expression is undefined. But look at the output values as x approaches 2 from both sides. They approach 4. The function is undefined at x = 2, but its values are heading toward a clear destination. This raises the central question of limits: What value does a function approach as its input approaches a given point — even if the function is undefined at that point? The answer is the limit, and it is the key to everything that follows.

14Sections
15Objectives
22Key Terms
20+Worked Examples

Learning Objectives

  1. 1Explain in plain language what a limit describes and how it differs from a function value.
  2. 2Estimate limits from numerical tables by observing approach behavior from both sides.
  3. 3Estimate limits from graphs, including cases where the limit does not exist.
  4. 4Evaluate one-sided limits using limit notation.
  5. 5Evaluate two-sided limits and determine whether they exist.
  6. 6Evaluate limits using direct substitution and the limit laws.
  7. 7Resolve indeterminate 0/0 forms using factoring, cancellation, and rationalization.
  8. 8Analyze limits involving rational functions near points of discontinuity.
  9. 9Identify and interpret infinite limits and vertical asymptotes.
  10. 10Identify and interpret limits at infinity and horizontal asymptotes.
  11. 11Define continuity at a point using all three required conditions.
  12. 12Classify types of discontinuities: removable, jump, and infinite.
  13. 13Determine whether a function is continuous on an interval.
  14. 14State and apply the Intermediate Value Theorem.
  15. 15Recognize and avoid the most common limit and continuity mistakes.
Chapter 2 — Contents

Sections at a Glance

Sections 2.1 – 2.5
What a Limit Is: Approach, Notation, and One-Sided Behavior
Limit Description · Tables · Graphs · One-Sided Limits · Two-Sided Limits
Section 2.1

What a Limit Describes

A limit asks a specific question about a function: as the input approaches some value, what output does the function approach? The word "approach" is essential. A limit is not asking for the function's value at a point. It is asking about the function's behavior as the input gets arbitrarily close to that point. The function may or may not even be defined at the point itself, and that fact does not change whether the limit exists.

This distinction — between the value of a function at a point and the limit of the function as the input approaches that point — is one of the most important conceptual separations in all of calculus.

Limit Notation

The statement lim(x→a) f(x) = L is read: "the limit of f(x) as x approaches a equals L."

This means: as x gets closer and closer to a — but does not equal a — the value f(x) gets closer and closer to L. The variable x may approach from either direction.

Three Situations for Limits

SituationDescriptionExample
Function defined; limit equals valueThe function behaves smoothly near a; the limit simply equals f(a). This is continuity.lim(x→3) (x+1) = 4 = f(3)
Function undefined; limit still existsThere is a hole at x = a, but the function approaches a clear value from both sides.lim(x→2) (x²−4)/(x−2) = 4, even though f(2) is undefined
Limit does not existThe function oscillates, jumps to different values from left and right, or grows without bound.lim(x→0) sin(1/x) — no single value is approached
Critical Distinction

A limit is not the same as a function's value at a point. The statement lim(x→a) f(x) = L says nothing about what f(a) is. The function might be undefined at a, defined with f(a) = L, or defined with f(a) ≠ L. All three are possible. The limit depends only on approach behavior — not on what the function equals at the target point.

Practice 2.1Concept Check
  1. In your own words, explain what it means to say that a function "approaches" a value. Why is "approach" different from "equals"?
  2. A function f is undefined at x = 5. Can lim(x→5) f(x) still exist? Explain your reasoning and give a hypothetical example.
  3. A function g satisfies g(3) = 7 but lim(x→3) g(x) = 10. Draw a rough sketch of what this graph might look like. What kind of discontinuity is present?
Section 2.2

Estimating Limits from Tables

Before developing algebraic techniques, it is valuable to observe limit behavior numerically. A table shows approach behavior directly: choose inputs close to the target value from both sides and observe what the outputs do.

Method: Numerical Estimation

1. Choose input values close to a but not equal to a, from both sides.

2. Compute or observe the corresponding outputs.

3. If outputs approach a single number from both sides, that number is the estimated limit.

Example 2.2.1
Estimating a Limit from a Table
  • ProblemEstimate lim(x→3) (x² − 9)/(x − 3).
  • ThinkThe function is undefined at x = 3 (denominator is zero). Evaluate for values of x close to 3 from both sides.
  • Represent
    x2.92.992.9993.0013.013.1
    f(x)5.95.995.9996.0016.016.1
  • AnalyzeFrom both the left and the right, the outputs are approaching 6.
  • InterpretThe estimated limit is 6. This result will be confirmed algebraically in Section 2.7.
Answer: lim(x→3) (x² − 9)/(x − 3) = 6
Example 2.2.2
A Limit That Does Not Appear to Exist
  • ProblemEstimate lim(x→0) sin(1/x).
  • ThinkThis function oscillates rapidly near x = 0. Build a table to observe the behavior.
  • Represent
    x0.10.010.0010.0001
    sin(1/x)≈ −0.544≈ −0.506≈ 0.827≈ −0.306
  • AnalyzeThe values do not settle toward any particular number. The function oscillates increasingly rapidly as x approaches 0.
  • InterpretThe limit does not exist. The function oscillates between −1 and 1 infinitely many times near x = 0 with no single value being approached.
Answer: lim(x→0) sin(1/x) does not exist (oscillation).
Graph Instruction — Section 2.2

Draw the graph of y = sin(1/x) near x = 0, showing the function oscillating between −1 and 1 with increasing frequency as x approaches 0. Label the horizontal bounds y = 1 and y = −1. Note that no single y-value is being approached. This is an example of a limit that does not exist due to oscillation — distinct from limits that do not exist because of jumps or infinite behavior.

Limitation of Numerical Methods

A table can only show finitely many values. The function might behave differently for inputs even closer to a. Rounding errors can give misleading results. Use tables to build intuition, then confirm results algebraically when possible.

Practice 2.2Numerical Estimation
  1. Complete a table with x values 1.9, 1.99, 1.999, 2.001, 2.01, 2.1 and estimate lim(x→2) (x² − 4)/(x − 2).
  2. Use a calculator to estimate lim(x→0) (sin x)/x using x = ±0.1, ±0.01, ±0.001 (radians). What value does the function approach?
  3. Use a table to estimate lim(x→0) (eˣ − 1)/x. What does your table suggest about this limit?
Section 2.3

Estimating Limits from Graphs

Graphs are among the most powerful tools for understanding limits. A graph shows approach behavior visually and can quickly reveal whether a limit exists, whether it is finite or infinite, and where the function behaves unusually. The key reading skill: trace along the curve toward the target point from both sides and observe which y-value the curve is heading toward. The open or closed circle at that point does not affect the limit — only the approach behavior does.

Example 2.3.1
A Removable Discontinuity
  • GraphA smooth curve resembling y = x + 2 everywhere except at x = 3, where an open circle sits at (3, 5) and a filled circle sits at (3, 1).
  • AnalyzeTracing from the left: the curve heads toward y = 5 as x → 3. Tracing from the right: also heading toward y = 5. The open circle shows the curve doesn't meet that value there; the filled circle shows the actual defined value f(3) = 1.
  • Interpretlim(x→3) f(x) = 5 but f(3) = 1. The limit exists even though the function value differs from it.
lim(x→3) f(x) = 5  ·  f(3) = 1  ·  Removable discontinuity at x = 3
Example 2.3.2
A Jump Discontinuity
  • GraphA piecewise function: the curve equals y = 2 for x < 1 (with an open circle at (1, 2)) and y = 4 for x ≥ 1 (with a filled circle at (1, 4)). The graph jumps from 2 to 4 at x = 1.
  • AnalyzeTracing from the left: approaches 2. Tracing from the right: approaches 4. The two traces lead to different y-values.
  • InterpretBecause the left and right approaches disagree, lim(x→1) f(x) does not exist. The function has a jump discontinuity at x = 1.
lim(x→1) f(x) does not exist — jump discontinuity.
Graph Instruction — Section 2.3

Draw three separate labeled graphs side by side. Graph A: a smooth curve with a single open circle (hole) — the function is undefined at that point but the curve approaches the hole from both sides. Graph B: a piecewise function with a visible jump — the left and right branches approach different heights at x = a. Graph C: a smooth curve passing through a filled circle — the function is defined and continuous at that point. Ask students: in which graph does the limit exist? In which does the limit equal the function value?

Practice 2.3Graph Interpretation
  1. A graph shows f(4) = 3 but an open circle at (4, 3) with the curve approaching y = 8 from both sides. State the limit and the function value. What type of discontinuity is present?
  2. A graph shows a function where the left trace approaches y = −2 and the right trace approaches y = −2 as x → 0, but there is no circle at x = 0 at all (function undefined). Does the limit exist? Is the function continuous at x = 0?
  3. A graph shows a function increasing steeply toward +∞ on both sides of x = 2. Describe the limit behavior at x = 2.
Section 2.4

One-Sided Limits

One-Sided Limit Notation

lim(x→a⁻) f(x) = L — the left-hand limit: x approaches a from values less than a.

lim(x→a⁺) f(x) = L — the right-hand limit: x approaches a from values greater than a.

The superscript minus (⁻) indicates approach from the left; plus (⁺) from the right.

Example 2.4.1
One-Sided Limits from a Graph
  • GraphA piecewise function: f(x) = x² for x < 2 (open circle at (2, 4)); f(x) = 5 for x ≥ 2 (filled circle at (2, 5)).
  • AnalyzeLeft: following the x² branch toward x = 2, output approaches 4. Right: constant branch at y = 5 approaches 5. Function value: f(2) = 5.
  • Interpret lim(x→2⁻) f(x) = 4 lim(x→2⁺) f(x) = 5 f(2) = 5 Because the one-sided limits differ, the two-sided limit does not exist at x = 2.
Left limit = 4  ·  Right limit = 5  ·  Two-sided limit does not exist.
Example 2.4.2
One-Sided Limits for the Absolute Value Function
  • ProblemFind lim(x→0⁻) |x|/x and lim(x→0⁺) |x|/x.
  • ThinkWhen x < 0: |x| = −x, so |x|/x = −x/x = −1. When x > 0: |x| = x, so |x|/x = x/x = 1.
  • Analyze lim(x→0⁻) |x|/x = −1 lim(x→0⁺) |x|/x = 1
  • InterpretLeft and right limits differ, so lim(x→0) |x|/x does not exist. The function jumps from −1 to +1 at x = 0.
Left limit = −1  ·  Right limit = 1  ·  Two-sided limit does not exist.
Practice 2.4One-Sided Limits
  1. f(x) = x² for x < 1; f(x) = 2x for x ≥ 1. Find lim(x→1⁻) and lim(x→1⁺).
  2. g(x) = 3 for x ≤ 0; g(x) = x + 1 for x > 0. Find both one-sided limits at x = 0.
  3. Find lim(x→2⁻) and lim(x→2⁺) of f(x) = |x − 2|/(x − 2).
  4. h(x) = x + 5 for x < −1; h(x) = x² + 3 for x ≥ −1. Find both one-sided limits at x = −1.
Section 2.5

Two-Sided Limits

The Two-Sided Limit Theorem
Existence Condition

lim(x→a) f(x) = L if and only if lim(x→a⁻) f(x) = L AND lim(x→a⁺) f(x) = L.

The two-sided limit exists exactly when both one-sided limits exist and are equal. If they differ, the two-sided limit does not exist.

Example 2.5.1
Confirming a Two-Sided Limit from a Piecewise Function
  • ProblemLet f(x) = x + 1 for x < 3; f(x) = 4 for x = 3; f(x) = 2x − 2 for x > 3. Find lim(x→3) f(x).
  • ThinkCompute both one-sided limits and check whether they agree. The function value at x = 3 is provided but does not determine the limit.
  • Analyze Left: lim(x→3⁻) (x + 1) = 3 + 1 = 4 Right: lim(x→3⁺) (2x − 2) = 2(3) − 2 = 4 Both one-sided limits equal 4.
  • InterpretThe two-sided limit is 4. Note: f(3) = 4 as well, which happens to match, but this is not required for the limit to exist.
lim(x→3) f(x) = 4
Example 2.5.2
A Two-Sided Limit That Does Not Exist
  • ProblemLet g(x) = x² for x < 1; g(x) = x + 3 for x ≥ 1. Find lim(x→1) g(x) if it exists.
  • Analyze Left: lim(x→1⁻) x² = 1 Right: lim(x→1⁺) (x + 3) = 4
  • InterpretThe one-sided limits differ (1 ≠ 4), so lim(x→1) g(x) does not exist. The function has a jump discontinuity at x = 1.
lim(x→1) g(x) does not exist — jump at x = 1.
Practice 2.5Two-Sided Limits
  1. For f(x) = 3x − 1 (x < 2) and f(x) = x² − 1 (x ≥ 2): find both one-sided limits at x = 2 and determine whether the two-sided limit exists.
  2. A function has lim(x→0⁻) = −3 and lim(x→0⁺) = −3. Does the two-sided limit exist? What is it?
  3. For p(x) = |x − 4| / (x − 4): find the one-sided limits at x = 4. Does lim(x→4) p(x) exist?
Sections 2.6 – 2.8
Evaluating Limits Algebraically
Direct Substitution · Indeterminate Forms · Factoring · Rationalization · Rational Functions
Section 2.6

Limits by Direct Substitution

In many cases, a limit can be found simply by substituting the value of a into the function. This is valid when the function is continuous at a — a concept defined precisely in Section 2.11. For now, direct substitution works reliably for polynomial, rational (when the denominator is nonzero), trigonometric, and exponential functions at most points.

Limit Laws

If lim(x→a) f(x) and lim(x→a) g(x) both exist:

Example 2.6.1
Direct Substitution — Polynomial
  • ProblemFind lim(x→4) (3x² − 5x + 2).
  • ThinkPolynomials are continuous everywhere. Direct substitution works.
  • Calculate3(4)² − 5(4) + 2 = 3(16) − 20 + 2 = 48 − 20 + 2 = 30
Answer: 30
Example 2.6.2
Direct Substitution — Rational Function (Nonzero Denominator)
  • ProblemFind lim(x→−1) (2x² + 3)/(x + 5).
  • ThinkAt x = −1, denominator = −1 + 5 = 4 ≠ 0. Direct substitution is valid.
  • Calculate(2(−1)² + 3) / (−1 + 5) = (2 + 3) / 4 = 5/4
Answer: 5/4
Example 2.6.3
Direct Substitution — Trigonometric
  • ProblemFind lim(x→π/2) sin x.
  • ThinkThe sine function is continuous everywhere. Substitute directly.
  • Calculatesin(π/2) = 1
Answer: 1
Practice 2.6Direct Substitution
  1. lim(x→2) (x³ − 4x + 1)
  2. lim(x→−3) (x² + x − 6)/(x + 2)
  3. lim(x→π) cos(x)
  4. lim(x→0) e^(2x)
  5. lim(x→4) √(3x + 4)
  6. lim(x→1) (x⁴ − 2x + 3)
Section 2.7

Limits Requiring Simplification

When direct substitution produces 0/0, the result is called an indeterminate form. This does not mean the limit is zero or undefined — it means more work is needed. The most common strategy is to factor and cancel the troublesome factor, then substitute. When square roots are involved, rationalizing the numerator or denominator is often the key step.

Indeterminate Form 0/0

When substitution gives 0/0, do not stop. This form does not equal zero or any other number on its own. It is a signal that algebraic simplification is required — factoring, rationalization, or another technique — before the limit can be evaluated.

Example 2.7.1
Factoring to Remove the Indeterminate Form
  • ProblemFind lim(x→3) (x² − 9)/(x − 3).
  • ThinkAt x = 3: numerator = 0 and denominator = 0. Indeterminate form. Factor x² − 9 = (x − 3)(x + 3).
  • Calculate lim(x→3) (x−3)(x+3)/(x−3) = lim(x→3) (x+3) [x ≠ 3 in limit process] = 3 + 3 = 6
  • CheckThe simplified function x + 3 is continuous at x = 3, so direct substitution is valid after cancellation.
Answer: 6. This confirms the table estimate from Section 2.2.
Example 2.7.2
Sum of Cubes
  • ProblemFind lim(x→−2) (x³ + 8)/(x + 2).
  • ThinkAt x = −2: numerator = −8 + 8 = 0, denominator = 0. Factor as sum of cubes: x³ + 8 = (x + 2)(x² − 2x + 4).
  • Calculate lim(x→−2) (x+2)(x²−2x+4)/(x+2) = lim(x→−2) (x²−2x+4) = 4 + 4 + 4 = 12
Answer: 12
Example 2.7.3
Rationalizing the Numerator
  • ProblemFind lim(x→0) (√(x + 9) − 3)/x.
  • ThinkAt x = 0: numerator = √9 − 3 = 0, denominator = 0. Rationalize by multiplying numerator and denominator by the conjugate √(x+9) + 3.
  • Calculate · [√(x+9) + 3] / [√(x+9) + 3] = (x+9 − 9) / [x(√(x+9) + 3)] = x / [x(√(x+9) + 3)] = 1 / (√(x+9) + 3) Now substitute x = 0: 1 / (√9 + 3) = 1 / (3 + 3) = 1/6
Answer: 1/6
Conjugate Technique

The conjugate of (√u − c) is (√u + c). Their product is u − c², which eliminates the square root. Use this whenever a square root in a numerator or denominator creates a 0/0 form.

Practice 2.7Simplification and Factoring
  1. lim(x→2) (x² − 4)/(x − 2)
  2. lim(x→−1) (x² + 3x + 2)/(x + 1)
  3. lim(x→3) (x² − x − 6)/(x − 3)
  4. lim(x→0) (√(x+4) − 2)/x
  5. lim(x→9) (√x − 3)/(x − 9)
  6. lim(h→0) [(x+h)² − x²]/h (simplify in x)
Section 2.8

Limits Involving Rational Functions

Rational functions — quotients of polynomials — require careful attention near points where the denominator equals zero. The behavior of the numerator at those points determines what happens. Two distinct cases arise.

CaseNumerator at x = aDenominator at x = aResult
ReducibleAlso equals zeroEquals zeroIndeterminate form 0/0 — factor and cancel. Limit may be finite.
IrreducibleNonzero (equals some c ≠ 0)Equals zeroInfinite limit — the function grows without bound near x = a. See Section 2.9.
Example 2.8.1
Reducible Rational Function
  • ProblemFind lim(x→5) (x² − 25)/(x² − x − 20).
  • ThinkAt x = 5: numerator = 0, denominator = 25 − 5 − 20 = 0. Factor both. x² − 25 = (x−5)(x+5). x² − x − 20 = (x−5)(x+4).
  • Calculate = lim(x→5) (x−5)(x+5) / (x−5)(x+4) = lim(x→5) (x+5)/(x+4) = 10/9
Answer: 10/9
Example 2.8.2
Irreducible — Infinite Limit
  • ProblemFind lim(x→2) (x + 3)/(x − 2).
  • ThinkAt x = 2: numerator = 5 ≠ 0, denominator = 0. This is not an indeterminate form — the function grows without bound. Check the sign from each side.
  • Analyze x → 2⁻: (x − 2) is small and negative → fraction → −∞ x → 2⁺: (x − 2) is small and positive → fraction → +∞
  • InterpretBecause the one-sided limits differ in sign, lim(x→2) (x+3)/(x−2) does not exist. There is a vertical asymptote at x = 2.
lim(x→2⁻) = −∞  ·  lim(x→2⁺) = +∞  ·  Two-sided limit does not exist  ·  Vertical asymptote at x = 2
Practice 2.8Rational Function Limits
  1. lim(x→5) (x² − 25)/(x − 5)
  2. lim(x→4) (x − 4)/(x² − 16)
  3. lim(x→3) (x + 1)/(x − 3) — find one-sided limits
  4. lim(x→−2) (x² − 4)/(x + 2)
Sections 2.9 – 2.10
Limits and Infinite Behavior
Infinite Limits · Vertical Asymptotes · Limits at Infinity · Horizontal Asymptotes
Section 2.9

Infinite Limits and Vertical Asymptotes

An infinite limit occurs when a function's values grow without bound — positively or negatively — as the input approaches some finite value. Writing lim(x→a) f(x) = +∞ is a precise statement: it means that f(x) can be made arbitrarily large by taking x close enough to a. Infinity is not a number, but this notation communicates a specific behavior.

A vertical asymptote at x = a exists when at least one of the one-sided limits at a is ±∞. On a graph, the function's curve approaches this vertical line but never crosses it.

Example 2.9.1
Vertical Asymptote of 1/x
  • ProblemFind the one-sided limits of f(x) = 1/x as x → 0.
  • Analyze x → 0⁺: denominator small and positive → 1/x → +∞ x → 0⁻: denominator small and negative → 1/x → −∞
  • InterpretThere is a vertical asymptote at x = 0. The two-sided limit does not exist because the one-sided limits have opposite signs.
lim(x→0⁺) 1/x = +∞  ·  lim(x→0⁻) 1/x = −∞  ·  Vertical asymptote: x = 0
Graph Instruction — Section 2.9

Draw the graph of y = 1/x. Show the right branch rising steeply toward +∞ as x approaches 0 from the right, and the left branch dropping steeply toward −∞ as x approaches 0 from the left. Draw the vertical asymptote x = 0 and the horizontal asymptote y = 0 as dashed lines. Label both asymptotes. Ask students: from which direction does the function approach +∞? From which does it approach −∞?

Example 2.9.2
Finding Vertical Asymptotes of a Rational Function
  • ProblemFind all vertical asymptotes of f(x) = 3/[(x−1)(x+4)].
  • ThinkSet the denominator equal to zero and check that the numerator is nonzero there.
  • Analyze(x−1)(x+4) = 0 → x = 1 and x = −4. The numerator, 3, is never zero. Both points produce infinite limits.
  • InterpretVertical asymptotes at x = 1 and x = −4.
Vertical asymptotes at x = 1 and x = −4.
Asymptote vs. Hole

If the numerator is also zero where the denominator is zero — if both share a common factor — there may be a removable discontinuity (a hole) rather than a vertical asymptote. Always factor and check before identifying an asymptote.

Practice 2.9Infinite Limits
  1. Find vertical asymptotes of f(x) = 1/(x² − 9).
  2. Find lim(x→3⁺) and lim(x→3⁻) of 2/(x − 3).
  3. Does f(x) = (x² − 4)/(x − 2) have a vertical asymptote at x = 2? Explain.
  4. Find vertical asymptotes of g(x) = (x + 1)/[(x − 2)(x + 5)].
Section 2.10

Limits at Infinity and Horizontal Asymptotes

A limit at infinity describes the behavior of a function as x itself grows without bound — the output values as you move far to the right (x → +∞) or far to the left (x → −∞). When a limit at infinity equals a finite number L, the line y = L is a horizontal asymptote. This represents the end behavior of the function.

Key Building Blocks

For any n > 0:   lim(x→+∞) 1/xⁿ = 0   and   lim(x→−∞) 1/xⁿ = 0

These follow from the fact that as x grows without bound, any fixed power in the denominator dominates. All other limits at infinity for rational functions can be reduced to these building blocks.

Example 2.10.1
Limit of a Rational Function at Infinity
  • ProblemFind lim(x→+∞) (3x² − 2x + 1)/(5x² + 7).
  • ThinkAs x → +∞, the highest-degree terms dominate. Divide numerator and denominator by x², the highest power present.
  • Calculate = lim(x→+∞) [3 − 2/x + 1/x²] / [5 + 7/x²] As x → +∞: 2/x → 0, 1/x² → 0, 7/x² → 0 = (3 − 0 + 0) / (5 + 0) = 3/5
  • InterpretThe horizontal asymptote is y = 3/5. As x grows very large or very negative, the function approaches this value.
Answer: 3/5  ·  Horizontal asymptote: y = 3/5
Example 2.10.2
Numerator Degree Greater Than Denominator
  • ProblemFind lim(x→+∞) (x³ + 1)/(x² + 1).
  • ThinkDivide by x²: the numerator becomes x + 1/x², which grows without bound as x → +∞.
  • InterpretThe limit is +∞. There is no horizontal asymptote. The function grows without bound in the positive direction.
Limit = +∞  ·  No horizontal asymptote.

Degree Comparison Summary

Degree comparisonLimit at ±∞Horizontal Asymptote
Numerator degree < Denominator degree0y = 0
Numerator degree = Denominator degreeRatio of leading coefficientsy = aₙ/bₘ
Numerator degree > Denominator degree±∞None
Practice 2.10Limits at Infinity
  1. lim(x→+∞) (2x² − 3)/(x² + 5)
  2. lim(x→−∞) (x³ − 2x)/(x³ + 1)
  3. lim(x→+∞) (5x − 1)/(x² + 3)
  4. lim(x→+∞) (7x³ + 2)/(3x² − x)
  5. lim(x→+∞) (4x + 9)/(2x − 1)
  6. lim(x→−∞) eˣ
Sections 2.11 – 2.13
Continuity and the Intermediate Value Theorem
Continuity at a Point · Continuity on an Interval · IVT
Section 2.11

Continuity at a Point

The concept of a limit sets the stage for defining continuity precisely. Informally, a function is continuous at a point if its graph has no break, hole, or jump there. But the formal definition requires all three of the following conditions to hold simultaneously.

Definition
Continuity at a Point

A function f is continuous at x = a if all three conditions hold:

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 fails, the function has a discontinuity at x = a.

Types of Discontinuities

TypeWhat FailsGraph Appearance
RemovableEither f(a) is undefined, or f(a) ≠ limit (but limit exists)A hole in the curve — a single missing or misplaced point
JumpBoth one-sided limits exist but are unequalA vertical gap — the curve jumps from one height to another
InfiniteOne or both one-sided limits are ±∞A vertical asymptote — the curve rises or drops without bound
Graph Instruction — Section 2.11

Draw three separate labeled graphs, one for each type of discontinuity. Removable: a smooth curve with a single open circle at a point, with a filled circle at a different height (or no filled circle at all). Jump: a piecewise function with two horizontal branches at different heights meeting at x = a, with open and filled circles to indicate which endpoints are included. Infinite: a rational function with a vertical asymptote, showing the curve approaching +∞ from one side and −∞ from the other. Label each graph clearly with the type of discontinuity shown.

Example 2.11.1
Testing Continuity at a Point
  • ProblemIs f(x) = (x² − 4)/(x − 2) continuous at x = 2?
  • Test 1Is f(2) defined? At x = 2, denominator = 0. So f(2) is not defined. The first condition fails immediately.
  • InterpretThe function is not continuous at x = 2. This is a removable discontinuity: the limit exists (equals 4, as shown earlier by factoring), but the function is undefined at that point.
Not continuous at x = 2 — removable discontinuity (hole at (2, 4)).
Example 2.11.2
Repairing a Removable Discontinuity
  • ProblemDefine g(x) = (x² − 4)/(x − 2) for x ≠ 2, and g(2) = c. Find c so that g is continuous at x = 2.
  • ThinkFor continuity, we need g(2) = lim(x→2) g(x). From Section 2.7, this limit equals 4.
  • InterpretSetting c = 4 fills the hole and makes g continuous at x = 2. This is why removable discontinuities are called removable — they can be repaired by redefining the function at one point.
c = 4. Setting g(2) = 4 makes g continuous at x = 2.
Example 2.11.3
Piecewise Function — All Three Conditions
  • ProblemIs h(x) = 2x + 1 for x < 1; h(x) = x² + 2 for x ≥ 1, continuous at x = 1?
  • Test 1h(1) = 1² + 2 = 3. Defined. ✓
  • Test 2 lim(x→1⁻) (2x+1) = 3 lim(x→1⁺) (x²+2) = 3 Two-sided limit = 3. Exists. ✓
  • Test 3lim(x→1) h(x) = 3 = h(1). ✓
  • InterpretAll three conditions hold. The function is continuous at x = 1.
h is continuous at x = 1.
Practice 2.11Continuity at a Point
  1. Is f(x) = (x − 3)/(x² − 9) continuous at x = 3? If not, identify the type of discontinuity.
  2. For g(x) = x² − 1 (x < 2) and g(x) = 2x + 1 (x ≥ 2): test all three continuity conditions at x = 2.
  3. Find the value of k that makes h(x) = kx + 3 for x < 1; h(x) = x² + 2 for x ≥ 1, continuous at x = 1.
  4. A function has f(5) = 8 but lim(x→5) f(x) = 3. Identify the type of discontinuity and explain which condition fails.
Section 2.12

Continuity on an Interval

Continuity on an Interval

A function is continuous on an open interval (a, b) if it is continuous at every point inside the interval.

A function is continuous on a closed interval [a, b] if it is continuous on (a, b), and also satisfies: lim(x→a⁺) f(x) = f(a) and lim(x→b⁻) f(x) = f(b). (One-sided continuity at the endpoints.)

Standard Continuous Function Families

Function FamilyContinuous Where
Polynomial functionsAll real numbers (−∞, +∞)
Rational functionsEverywhere the denominator is nonzero
sin x and cos xAll real numbers
eˣ and aˣ (a > 0)All real numbers
ln x(0, +∞)
√x[0, +∞)
Example 2.12.1
Finding the Interval of Continuity
  • ProblemOn what intervals is f(x) = (x + 1)/(x² − 5x + 6) continuous?
  • ThinkFactor the denominator: x² − 5x + 6 = (x − 2)(x − 3). The function is undefined at x = 2 and x = 3.
  • InterpretContinuous on (−∞, 2), (2, 3), and (3, +∞). The discontinuities at x = 2 and x = 3 are infinite (numerator is nonzero at both points).
Continuous on (−∞, 2) ∪ (2, 3) ∪ (3, +∞).
Practice 2.12Intervals of Continuity
  1. Find intervals of continuity for f(x) = x/(x² − x − 6).
  2. Is g(x) = √(4 − x²) continuous on [−2, 2]? Explain.
  3. Find intervals where h(x) = 1/(x² + 1) is continuous.
  4. Where is f(x) = ln(x − 3) continuous?
Section 2.13

The Intermediate Value Theorem as an Introduction

One of the most elegant consequences of continuity on a closed interval is the Intermediate Value Theorem (IVT). The theorem formalizes a geometric observation: if a continuous function starts at one height and ends at another, it must pass through every height in between. It cannot skip over values without lifting the pen.

Theorem
Intermediate Value Theorem (IVT)

Let f be continuous on the closed interval [a, b], and let N be any number between f(a) and f(b), with f(a) ≠ f(b). Then there exists at least one number c in the open interval (a, b) such that f(c) = N.

What the IVT Does and Does Not Say

Does say: If f is continuous on [a, b] and N is between f(a) and f(b), then at least one c exists with f(c) = N.

Does not say: There is exactly one such c. There may be many. The IVT also does not provide a method for finding c — only that c exists. And it requires continuity. A function with a jump can skip over values without being caught.

Graph Instruction — Section 2.13

Draw a continuous curve from (a, f(a)) to (b, f(b)) where f(a) is below a dashed horizontal line at height N and f(b) is above it. Show that the curve must cross the dashed line at least once, labeling the crossing point c. Then draw a second graph with the same endpoints but including a jump discontinuity that allows the curve to skip over the value N entirely — illustrating why the continuity hypothesis is essential.

Example 2.13.1
Guaranteeing a Zero
  • ProblemShow that f(x) = x³ − x − 2 has at least one zero on the interval [1, 2].
  • ThinkIf f(1) < 0 and f(2) > 0, and f is continuous on [1, 2], then by the IVT there must exist some c ∈ (1, 2) with f(c) = 0.
  • Check Continuityf(x) = x³ − x − 2 is a polynomial, so it is continuous everywhere, including on [1, 2].
  • Check Values f(1) = 1 − 1 − 2 = −2 (negative) f(2) = 8 − 2 − 2 = 4 (positive) Since 0 is between f(1) = −2 and f(2) = 4, the IVT applies.
  • InterpretBy the IVT, there exists at least one c ∈ (1, 2) with f(c) = 0. The equation x³ − x − 2 = 0 has at least one real solution between 1 and 2. (Technology gives c ≈ 1.521.)
By the IVT: at least one zero exists in (1, 2). The IVT proves existence; it does not find the exact value.
Example 2.13.2
Guaranteeing an Intermediate Value
  • ProblemShow that g(x) = cos x takes the value 0.5 on the interval [0, π].
  • Check g(0) = cos(0) = 1 g(π) = cos(π) = −1 The value 0.5 lies between 1 and −1.
  • Continuitycos x is continuous everywhere, so on [0, π]. The IVT applies.
  • InterpretThere exists some c ∈ (0, π) with cos c = 0.5. In this case c = π/3 ≈ 1.047.
By the IVT: cos(π/3) = 0.5. The IVT guaranteed the value existed before we found it explicitly.
Practice 2.13Intermediate Value Theorem
  1. Show that f(x) = x³ + 2x − 5 has at least one zero on [1, 2]. State all required hypotheses and verify them.
  2. A continuous function satisfies f(0) = −4 and f(6) = 11. Does f take the value 0 on [0, 6]? Explain using the IVT.
  3. Explain why the IVT cannot be applied to the function f(x) = 1/(x − 2) on the interval [1, 3] to conclude the function takes every value between f(1) and f(3).
  4. A population function P(t) is continuous, with P(0) = 2500 and P(12) = 8700. Explain why the population must have equaled exactly 5000 at some point during the 12-month study.
Section 2.14

Common Limit and Continuity Mistakes

Students learning limits and continuity encounter specific conceptual and algebraic traps. The following are real, recurring errors — not generic warnings. Each one identifies the mistake precisely and explains the correct reasoning.

Mistake 1 — Confusing the Limit with the Function Value
"lim(x→2) f(x) is the same as f(2)."

This is true only when f is continuous at x = 2. In general, the limit and the function value are separate. A function can have a limit at a point where it is undefined, or have a function value that differs from its limit. Always ask: does the three-condition continuity definition hold?

Mistake 2 — Concluding a Limit Does Not Exist Because the Function Is Undefined
"f(2) is undefined, so lim(x→2) f(x) does not exist."

Not true. A limit exists when the function approaches a value from both sides — regardless of whether the function is defined at that point. f(x) = (x² − 4)/(x − 2) is undefined at x = 2 but has limit 4 there.

Mistake 3 — Treating 0/0 as Zero or as Undefined
"lim(x→3) (x² − 9)/(x − 3) = 0/0, so the limit is 0 (or does not exist)."

The expression 0/0 is an indeterminate form. It does not equal zero. It is a signal that more algebraic work is needed — factoring, rationalization, or another technique. After simplification, the limit in this case is 6.

Mistake 4 — Assuming All Rational Functions Have Vertical Asymptotes Where the Denominator Is Zero
"The denominator of (x² − 9)/(x − 3) is zero at x = 3, so there is a vertical asymptote there."

If both numerator and denominator are zero at that point, factor first. In this case the factor (x − 3) cancels, producing a removable discontinuity (a hole), not an asymptote. Vertical asymptotes only arise when the numerator is nonzero and the denominator is zero.

Mistake 5 — Confusing Infinite Limits with Limits at Infinity
Using "vertical asymptote" and "horizontal asymptote" interchangeably with "infinite limit" and "limit at infinity."

An infinite limit occurs as x approaches a finite number and the function grows without bound — this produces a vertical asymptote. A limit at infinity occurs as x itself grows without bound — this produces a horizontal asymptote. The direction of approach is completely different.

Mistake 6 — Computing Only One-Sided Limit and Stopping
"lim(x→1⁻) f(x) = 4, so lim(x→1) f(x) = 4."

For the two-sided limit to exist, both one-sided limits must exist and be equal. Always check both sides. Finding the left-hand limit alone is not enough to conclude the two-sided limit exists or what its value is.

Mistake 7 — Applying IVT Without Verifying Continuity
Applying the IVT to a piecewise function or rational function on an interval containing a discontinuity.

The IVT requires continuity on the entire closed interval [a, b]. A function with a jump discontinuity can skip over an intermediate value entirely. Before invoking the IVT, always verify that f is continuous on the full interval being used.

Mistake 8 — Thinking ±∞ Means the Limit Exists
"lim(x→0⁺) 1/x = +∞, so the limit exists and equals +∞."

When a limit "equals" +∞ or −∞, this notation describes behavior (the function grows without bound) — it does not mean the limit equals a real number. The limit does not exist as a finite value. The notation is a precise way of describing divergence, not a declaration that infinity is the answer.

Mistake 9 — Forgetting to Check Whether the Three-Condition Test is Complete
"The limit exists at x = a, so the function is continuous there."

Continuity requires three things: the function is defined at a, the limit exists at a, and the two agree. A function can have a perfectly good limit at a point where it is still not continuous — for example, if f(a) is defined but f(a) ≠ lim f(x).

Mistake 10 — Forgetting That Cancellation Is Valid Only Because x ≠ a in a Limit
Canceling (x − a) without noting why this step is justified.

When evaluating lim(x→a), x approaches a but never equals a. This is precisely why the factor (x − a) can be canceled — the cancellation happens for all x near a, but not at a. Understanding this is what distinguishes the limit process from simply evaluating the function at a point.

Chapter 2 — Back Matter

Chapter Summary

Key Terms Review

limit
The value L that f(x) approaches as x approaches a, written lim(x→a) f(x) = L. Defined by approach behavior, not function value at a.
one-sided limit
A limit that considers approach from only one direction — left (x→a⁻) or right (x→a⁺).
two-sided limit
lim(x→a) f(x); exists if and only if both one-sided limits exist and are equal.
approach behavior
The pattern of output values that f(x) exhibits as x gets closer and closer to a target value, from either side.
indeterminate form
An expression such as 0/0 that requires further algebraic analysis before the limit can be determined.
direct substitution
Evaluating a limit by substituting the target value directly; valid when the function is continuous at that point.
infinite limit
A limit in which f(x) grows or drops without bound as x approaches a finite value a.
vertical asymptote
A vertical line x = a near which the function grows without bound. Occurs when at least one one-sided limit at a is ±∞.
limit at infinity
lim(x→+∞) f(x) or lim(x→−∞) f(x); describes the function's behavior as x grows without bound.
horizontal asymptote
A line y = L that the graph approaches as x → ±∞, occurring when the limit at infinity equals L.
continuity at a point
The property that f(a) is defined, lim(x→a) f(x) exists, and the two are equal. All three conditions are required.
removable discontinuity
A discontinuity where the limit exists but either the function is undefined or the function value differs from the limit. Appears as a hole on the graph.
jump discontinuity
A discontinuity where both one-sided limits exist but are not equal. The graph jumps from one value to another.
infinite discontinuity
A discontinuity where one or both one-sided limits are ±∞. The graph has a vertical asymptote at that point.
continuity on an interval
A function is continuous on (a, b) if continuous at every interior point; on [a, b] also requires one-sided continuity at the endpoints.
Intermediate Value Theorem
If f is continuous on [a, b] and N is between f(a) and f(b), then there exists at least one c ∈ (a, b) with f(c) = N.

Limit Estimation Practice

Part AConcept Check
  1. Explain the difference between f(2) and lim(x→2) f(x). Give an example where they are equal and an example where they differ.
  2. Is it possible for a limit to exist at a point where the function is undefined? Explain and give an example.
  3. What does it mean when lim(x→a⁻) f(x) = L₁ and lim(x→a⁺) f(x) = L₂ with L₁ ≠ L₂? What can you conclude about the two-sided limit?
  4. A student claims that 0/0 = 0. Explain why this is wrong in the context of limits.
  5. What three conditions must all hold for a function to be continuous at x = a?
Part BAlgebraic Limits — Direct Substitution
  1. lim(x→2) (x³ − 4x + 1)
  2. lim(x→−3) (x² + x − 6)/(x + 2)
  3. lim(x→π) cos x
  4. lim(x→0) e^(2x)
  5. lim(x→4) √(3x + 4)
  6. lim(x→1) (x⁴ − 2x + 5)
Part CFactoring and Simplification
  1. lim(x→2) (x² − 4)/(x − 2)
  2. lim(x→−1) (x² + 3x + 2)/(x + 1)
  3. lim(x→3) (x² − x − 6)/(x − 3)
  4. lim(x→5) (x² − 25)/(x² − 4x − 5)
  5. lim(x→−2) (x³ + 8)/(x + 2)
  6. lim(h→0) [(x+h)² − x²]/h
Part DRationalization
  1. lim(x→0) (√(x+4) − 2)/x
  2. lim(x→9) (√x − 3)/(x − 9)
  3. lim(x→0) (√(1+x) − 1)/x
  4. lim(x→16) (√x − 4)/(x − 16)
Part EOne-Sided Limits
  1. Let f(x) = x² for x < 2 and f(x) = 3x − 2 for x ≥ 2. Find lim(x→2⁻) f(x), lim(x→2⁺) f(x), and determine whether the two-sided limit exists.
  2. Let g(x) = 5 for x ≤ 0 and g(x) = x + 2 for x > 0. Find one-sided limits at x = 0 and determine whether g is continuous there.
  3. Find lim(x→1⁻) |x−1|/(x−1) and lim(x→1⁺) |x−1|/(x−1).
Part FInfinite Limits and Asymptotes
  1. Find vertical asymptotes of f(x) = 1/(x² − 9).
  2. Find lim(x→3⁺) and lim(x→3⁻) of 2/(x − 3).
  3. Does f(x) = (x² − 4)/(x − 2) have a vertical asymptote at x = 2? Explain fully.
  4. Find vertical asymptotes of g(x) = 3/[(x−1)(x+4)].
Part GLimits at Infinity
  1. lim(x→+∞) (2x² − 3)/(x² + 5)
  2. lim(x→−∞) (x³ − 2x)/(x³ + 1)
  3. lim(x→+∞) (5x − 1)/(x² + 3)
  4. lim(x→+∞) (7x³ + 2)/(3x² − x)
  5. lim(x→+∞) (4x + 9)/(2x − 1)
  6. lim(x→−∞) e^x
Part HContinuity
  1. Determine whether f(x) = (x − 3)/(x² − 9) is continuous at x = 3. If not, identify the type of discontinuity.
  2. Determine whether g(x) = x² − 1 for x < 2 and g(x) = 2x + 1 for x ≥ 2, is continuous at x = 2.
  3. Find k so that h(x) = kx + 3 for x < 1 and h(x) = x² + 2 for x ≥ 1, is continuous at x = 1.
  4. On what intervals is f(x) = x/(x² − x − 6) continuous?
  5. Sketch a possible graph of a function with a removable discontinuity at x = 1, a jump discontinuity at x = 3, and a vertical asymptote at x = 5.

Applications

  1. IVT and Population. A wildlife researcher models a population that is P(0) = 2500 at the start of a study and P(12) = 8700 twelve months later, where P is a continuous function. Explain why the population must have equaled exactly 5000 at some point during the study. Identify the theorem and verify the hypotheses.
  2. Temperature. A cooling liquid's temperature is measured continuously. At t = 0, the temperature is 95°C; at t = 30, it is 22°C. Using the IVT, explain why the temperature must have passed through 60°C at some point during the 30 minutes.
  3. Economics. A company's monthly revenue function R(x), where x is units sold, is continuous for x ≥ 0. If R(0) = 0 and R(1000) = 45000, what does the IVT say about the behavior of R between 0 and 1000 units?
  4. Limits in Marginal Analysis. Marginal cost is defined as MC(q) = lim(h→0) [C(q+h) − C(q)]/h, where C(q) is total cost. Interpret this limit in context: what does it measure? Why must it be expressed as a limit rather than as a simple difference?
  5. Physics — Instantaneous Velocity. A ball's position in meters at time t seconds is given by s(t) = −4.9t² + 12t. The average velocity over the interval [1, 1+h] is [s(1+h) − s(1)]/h. Write and simplify this expression, then take the limit as h → 0 to find the instantaneous velocity at t = 1. What are the units?

Error Analysis

Each problem contains an error. Identify the error and explain the correct approach.

Problem 1
lim(x→3) (x² − 9)/(x − 3): "The denominator is zero, so the limit does not exist."

What error is being made? What is the correct limit and what technique resolves it?

Problem 2
"f(2) = 7, therefore lim(x→2) f(x) = 7."

What information is missing? What additional evidence is needed to conclude the limit equals 7?

Problem 3
Simplify (x − 5)/(x − 5) to get 1, then declare f(x) = (x − 5)/(x − 5) is continuous at x = 5.

Identify the error. The simplified function and the original are not the same. Explain the distinction and what type of discontinuity exists.

Problem 4
"The horizontal asymptote of (x² + 1)/(x + 1) is y = x because leading terms are x² and x."

What does the degree comparison actually say here? What is the correct end behavior?

Problem 5
"lim(x→0) 1/x² = +∞, so the limit exists and equals infinity."

Explain why this phrasing is flawed even though the behavior described (the function grows without bound) is correct. What does it mean precisely to say a limit "equals +∞"?

Chapter 2 Checkpoint

Chapter 2 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 2
Part A — Notation and Meaning
  1. Write the limit statement that means: "As x approaches 5, the function f(x) approaches 12."
  2. A function g satisfies g(3) = 8 and lim(x→3) g(x) = 8. Is g continuous at x = 3? Name the three conditions and verify each.
  3. A function h satisfies h(4) = 1 and lim(x→4) h(x) = 6. Identify the type of discontinuity at x = 4.
Part B — Estimating Limits
  1. A table shows f(x) values approaching 6 from both sides as x → 2. What is lim(x→2) f(x)?
  2. A graph shows lim(x→1⁻) f(x) = 4 and lim(x→1⁺) f(x) = 4, but f(1) is undefined. Does the two-sided limit exist? Is the function continuous at x = 1?
Part C — Algebraic Limits
  1. Evaluate: lim(x→5) (2x² − 3x + 4)
  2. Evaluate: lim(x→4) (x² − 16)/(x − 4)
  3. Evaluate: lim(x→0) (√(x+1) − 1)/x
  4. Evaluate: lim(x→−3) (x² + 5x + 6)/(x + 3)
Part D — One-Sided Limits
  1. Let f(x) = 3x − 1 for x < 2 and f(x) = x² + 1 for x ≥ 2. Find lim(x→2⁻), lim(x→2⁺), and lim(x→2) if it exists. Is f continuous at x = 2?
Part E — Infinite Limits and Asymptotes
  1. Find all vertical asymptotes of f(x) = (x + 2)/(x² − x − 6).
  2. Find lim(x→+∞) (4x² − 7)/(2x² + 1) and identify any horizontal asymptote.
Part F — Continuity
  1. Find the value(s) of k so that f(x) = k²x + 1 for x < 1; f(x) = kx + 4 for x ≥ 1 is continuous at x = 1.
  2. On what intervals is g(x) = (x + 3)/(x² − 4) continuous?
Part G — Intermediate Value Theorem
  1. Show that f(x) = x³ + 2x − 5 has at least one zero on [1, 2]. Identify the theorem, verify the continuity hypothesis, compute f(1) and f(2), and explain the conclusion.
  2. A continuous function p satisfies p(0) = −4 and p(6) = 11. Does p take the value 0 on [0, 6]? Explain fully.
Looking Ahead
Chapter 3: The Derivative

Chapter 2 built the foundation: limits describe approach behavior, and continuity tells us when functions behave without breaks. These ideas will be used immediately in Chapter 3, which defines the derivative as a limit. The derivative of a function at a point is:

f′(a) = lim(h→0) [f(a+h) − f(a)] / h

Every concept from Chapter 2 — limit notation, approach behavior, indeterminate forms, factoring, rationalization — will appear directly in this definition. Chapter 3 uses limits to define instantaneous rate of change, interpret the derivative as a slope, study the derivative as a function, and develop the graphical relationship between f and f′.

MGU Library Connection
  • Limit Laws Reference — Appendix D of this textbook
  • Continuity Reference — Appendix E of this textbook
  • Algebra II and Trigonometry and Precalculus: rational functions, function behavior, and asymptotes
  • MGU Physics module: instantaneous velocity as a limit of average velocity
  • MGU Data Science Foundations and Scientific Computing: convergence of sequences and approximation algorithms use the same controlled approach behavior introduced here
  • MGU Dictionary entries: limit, continuity, asymptote, Intermediate Value Theorem