chapter 1 limits and their properties
TRANSCRIPT
![Page 1: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/1.jpg)
Chapter 1 Limits and Their Properties
Lecture Note
1.1 A Preview of Calculus
1.2 Finding Limits Graphically and Numerically
1.3 Evaluating Limits Analytically
1.4 Continuity and One-Sided Limits
1.5 Infinite Limits
![Page 2: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/2.jpg)
1.1 A Preview of Calculus
1.2 Finding Limits Graphically and Numerically
1.3 Evaluating Limits Analytically
1.4 Continuity and One-Sided Limits
1.5 Infinite Limits
Chapter 1. Limits and Their PropertiesLecture Note
![Page 3: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/3.jpg)
Objectives.
Understand what calculus is and how it compares with precalculus.
Understand that the tangent line problem is basic to calculus.
Understand that the area problem is also basic to calculus
1.1 A Preview of Calculus
Lecture Note
![Page 4: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/4.jpg)
Calculus is the mathematics of change. For instance, calculus is the mathematics of velocities, accelerations, tangent lines, slopes, areas, volumes, arc lengths, centroids, curvatures, and a variety of other concepts that have enabled scientists, engineers, and economists to model real-life situations.
Although precalculus mathematics also deals with velocities, accelerations, tangent lines, slopes, and so on, there is a fundamental difference between precalculus mathematics and calculus.
Precalculus mathematics is more static, whereas calculus is more dynamic.
What is Calculus?
Lecture Note
![Page 5: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/5.jpg)
What is Calculus?: Examples
An object traveling at a constant velocity can be analyzed with precalculusmathematics. To analyze the velocity of an accelerating object, you need calculus.
The slope of a line can be analyzed with precalculus mathematics. To analyze the slope of a curve, you need calculus.
The curvature of a circle is constant and can be analyzed with precalculusmathematics. To analyze the variable curvature of a general curve, you need calculus.
The area of a rectangle can be analyzed with precalculus mathematics. To analyze the area under a general curve, you need calculus.
Lecture Note
![Page 6: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/6.jpg)
Each of these situations involves the same general strategyβ the reformulation of precalculus mathematics through the use of the limit process.
So, one way to answer the question βWhat is calculus?β is to say that calculus is a βlimit machineβ that involves three stages.
The first stage is precalculus mathematics, such as the slope of a line or the area of a rectangle.
The second stage is the limit process, and the third stage is a new calculus
formulation, such as a derivative or integral.
What is Calculus?
Lecture Note
PrecalculusMathematics Limit Process Calculus
![Page 7: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/7.jpg)
What is Calculus?
Lecture Note
![Page 8: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/8.jpg)
Lecture Note
What is Calculus?
![Page 9: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/9.jpg)
Lecture Note
What is Calculus?
![Page 10: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/10.jpg)
Lecture Note
What is Calculus?
![Page 11: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/11.jpg)
The Tangent Line / Area Problem
The notion of a limit is fundamental to the study of calculus.
Two classic problems in calculus using the limit notionβ the tangent line problem and the area problemβshould give you some idea of the way limits are used in calculus.
Lecture Note
![Page 12: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/12.jpg)
In the tangent line problem, you are given a functionf and a point P on its graph and are asked to find an equation of the tangent line to the graph at point P,as shown in the Figure.
The Tangent Line Problem
The notion of a limit is fundamental to the study of calculus.
Two classic problems in calculus using the limit notionβ the tangent line problem and the area problemβshould give you some idea of the way limits are used in calculus.
Lecture Note
![Page 13: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/13.jpg)
Except for cases involving a vertical tangent line, the problem of finding the tangent line at a point P is equivalent to finding the slope of the tangent line at P.
You can approximate this slope by using a line through the point of tangency and a second point on the curve, as shown in Figure. Such a line is called a secant line.
Lecture Note
The Tangent Line Problem
![Page 14: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/14.jpg)
If π π, π π is the point of tangency and
π π + Ξπ₯, π π + Ξπ₯
Is a second point on the graph of π, the slope of the secant line through these two points can be found using precalculus and is given by
ππππ =π π + π«π β π π
π + π«π β π=
π π + π«π β π π
π«π
Lecture Note
As point π approaches point π, the slopes of the secant lines approach the slope of the tangent line, as shown in the figure.
When such a βlimiting positionβ exists, the slope of the tangent line is said to be the limit of the slopes of the secant lines.
The Tangent Line Problem
![Page 15: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/15.jpg)
This process defines the derivative of the function π(π₯) at π₯ = π
π β²(π) = limΞπ₯β0
π π + Ξπ₯ β π π
π + Ξπ₯ β π= lim
Ξπ₯β0
π π + Ξπ₯ β π π
Ξπ₯
and the derivative of π(π₯) at π₯ = π measures the slope of the tangent line.
Lecture Note
The Tangent Line Problem
![Page 16: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/16.jpg)
A second classic problem in calculus is finding the area of a plane region that is bounded by the graphs of functions.
This problem can also be solved with a limit process.
In this case, the limit process is applied to the area of a rectangle to find the area of a general region.
Lecture Note
The Area Problem
![Page 17: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/17.jpg)
As a simple example, consider the region bounded by the graph of the function π¦ = π(π₯), the π₯-axis, and the vertical lines π₯ = π and π₯ = π, as shown in Figure 1.3.
Lecture Note
The Area Problem
![Page 18: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/18.jpg)
You can approximate the area of the region with several rectangular regions as shown in Figure 1.4.
Lecture Note
The Area Problem
As you increase the number of rectangles, the approximation tends to become better and better because the amount of area missed by the rectangles decreases.
Your goal is to determine the limit of the sum of the areas of the rectangles as the number of rectangles increases without bound.
![Page 19: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/19.jpg)
This process defines the definite integral of the function π(π₯) from π₯ = π to π₯ = π
π
π
π(π₯) ππ₯ = limπββ
(the sum of areas of rectangles) = limnββ
π=1
π
π π₯ππ β π
π
Lecture Note
The Area Problem
![Page 20: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/20.jpg)
1.1 A Preview of Calculus
1.2 Finding Limits Graphically and Numerically
1.3 Evaluating Limits Analytically
1.4 Continuity and One-Sided Limits
1.5 Infinite Limits
Chapter 1. Limits and Their PropertiesLecture Note
![Page 21: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/21.jpg)
Objectives
β’ Estimate a limit using a numerical or graphical approach
β’ Learn different ways that a limit can fail to exist
β’ Study and use a formal definition of limit
1.2 Finding Limits Graphically and Numerically
Lecture Note
![Page 22: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/22.jpg)
If f(x) becomes arbitrarily close to a single number L as x approaches c from either side, the limit of f(x), as x approaches c, is L. This limit is written as
limπ₯βπ
π π₯ = πΏ
An Introduction to Limits
Lecture Note
![Page 23: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/23.jpg)
limπ₯β1
π₯3 β 1
π₯ β 1
Example
To get an idea of the behavior of the graph of f near x = 1, you can use two sets of x-valuesβone set that approaches 1 from the left and one set that approaches 1 from the right, as shown in the table.
Lecture Note
![Page 24: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/24.jpg)
Example
The graph of f is a parabola that has a gap at the point (1, 3), as
Although x can not equal 1, you can move arbitrarily close to 1, and as a result f(x)moves arbitrarily close to 3.
limπ₯β1
π₯3 β 1
π₯ β 1= 3
limπ₯β1
π₯3 β 1
π₯ β 1
Lecture Note
![Page 25: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/25.jpg)
limπ₯β0
π₯
π₯ + 1 β 1
Example
Lecture Note
![Page 26: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/26.jpg)
limπ₯β0
π₯
π₯ + 1 β 1
Example
limπ₯β1
π₯
π₯ + 1 β 1= 2
Lecture Note
![Page 27: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/27.jpg)
Limits That Fail to Exist
Example
Discuss the existence of the limit limπ₯β0
π₯
π₯.
Because π₯ /π₯ approaches a different number from the right side of 0 than it approaches from the left side, the limit does not exist
Note that
π₯ = π₯, π₯ β₯ 0
βπ₯, π₯ < 0
Lecture Note
![Page 28: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/28.jpg)
Limits That Fail to Exist
Example
Discuss the existence of the limit limπ₯β0
1
π₯2.
Lecture Note
![Page 29: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/29.jpg)
Limits That Fail to Exist
Example
Discuss the existence of the limit limπ₯β0
sin1
π₯.
π₯2
π
2
3π
2
5π
2
7π
2
9π
2
11ππ₯ β 0
sin1
π₯1 β1 1 β1 1 β1 Limit does not exist.
As π₯ approaches 0, π(π₯) oscillates between β1 and 1. So the limit does not exist because no matter how small you choose πΏ, it is possible to choose π₯1and π₯2 within πΏ units of 0 such that sin 1/π₯1 = 1 and sin(1/π₯2) = β1.
Lecture Note
![Page 30: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/30.jpg)
Limits That Fail to Exist
Common Types of Behavior Associated with Nonexistence of a Limit
1. π(π₯) approaches a different number from the right side of π than it
approaches from the left side.
2. π(π₯) increases or decreases without bound as π₯ approaches π.
3. π(π₯) oscillates between two fixed values as π approaches π.
Lecture Note
![Page 31: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/31.jpg)
Formal Definition of Limit
Informal definition of limit: If f(x) becomes arbitrarily close to a single number L as x approaches c from either side, then the limit of f(x) as x approaches c is L, is written as
limπ₯βπ
π(π₯) = πΏ.
Formal definition of limit:Let π be a function defined on an open interval containing π (except possibly at π), and let πΏ be a real number. The statement
limπ₯βπ
π(π₯) = πΏ.
Means that for each π > 0 there exists a πΏ > 0 such that if 0 < π₯ β π < πΏ, then 0 < π π₯ β πΏ < π.
Lecture Note
![Page 32: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/32.jpg)
Example: Finding a πΉ for a given π
Example 6. Given the limit
limπ₯β3
(2π₯ β 5) = 1
find πΏ such that 2π₯ β 5 β 1 < 0.01 whenever 0 < π₯ β 3 < πΏ.
Lecture Note
Solution.In this problem, you are working with a given value of π = 0.01.To find an appropriate πΏ, try to establish a connection between the absolutevalues
2π₯ β 5 β 1 πππ π₯ β 3 .
Notice that2π₯ β 5 β 1 = 2π₯ β 6 = 2 π₯ β 3 .
Since 2π₯ β 5 β 1 < 0.01, you can choose πΏ = 1/2(0.01) = 0.005.This choice works because 0 < π₯ β 3 < 0.005 implies that
2π₯ β 5 β 1 = 2π₯ β 6 = 2 π₯ β 3 < 2 0.005 = 0.01.
![Page 33: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/33.jpg)
Example: Finding a πΉ for a given π
Given the limit
limπ₯β3
(2π₯ β 5) = 1
for πΏ such that 2π₯ β 5 β 1 < 0.01
whenever 0 < π₯ β 3 < πΏ.
![Page 34: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/34.jpg)
Example: Fusing the π β πΉ Definition of Limit
Example 7. Use π-πΏ definition of limit to prove thatlimπ₯β2
(3π₯ β 2) = 4
Solution.You must show that for each π > 0, there exists a πΏ > 0 such that
3π₯ β 2 β 4 < π π€βππππ£ππ π₯ β 2 < πΏ.Because your choice of πΏ depends on π, you need to establish a connectionbetween the absolute values
3π₯ β 2 β 4 πππ π₯ β 2Using
3π₯ β 2 β 4 = 3π₯ β 6 = 3 π₯ β 2 .we have, for a given π, you can choose πΏ = π/3. The choice works because0 < π₯ β 2 < πΏ = π/3 implies that
3π₯ β 2 β 4 = 3π₯ β 6 = 3 π₯ β 2 < 3 π/3) = π.
![Page 35: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/35.jpg)
Example: Fusing the π β πΉ Definition of Limit
Example 8. Use π-πΏ definition of limit to prove thatlimπ₯β2
π₯2 = 4
Solution.You must show that for each π > 0, there exists a πΏ > 0 such that
π₯2 β 4 < π π€βππππ£ππ π₯ β 2 < πΏ.To find an appropriate πΏ, begin by writing π₯2 β 4 = π₯ β 2 |π₯ + 2|. You areinterested in values of π₯ close to 2, so choose π₯ in the interval (1, 3). Tosatisfy this restriction, let πΏ < 1. Furthermore, for all π₯ in the interval (1,3),π₯ + 2 < 5 and thus π₯ + 2 < 5.
So, letting πΏ be the minimum of π/5 and 1, it follows that, whenever 0 <π₯ β 2 < πΏ, you have
π₯2 β 4 = π₯ β 2 π₯ + 2 >π
55 = π.
![Page 36: Chapter 1 Limits and Their Properties](https://reader030.vdocuments.us/reader030/viewer/2022012813/61c44eb396a52875b9689ab1/html5/thumbnails/36.jpg)
A few tips for using the π β πΉ Definition of Limit
β’ 95% of the time, you should start with your π term and algebraicallymanipulate it until you get something of the form π π₯ β πΏ = π π₯ β π₯ β π < π where π(π₯) is some function of π₯.
β’ Once you find he functionπ π₯ from the above hint, restrict πΏ (whichforces a restriction on π₯ so that you can get an upper bound for |π π₯ |.
β’ Use this upper bound for |π π₯ | to find πΏ. So if we algebraically get toπ π₯ β π₯ β π < π and π π₯ < π for some number π then if we can
make π π₯ β π < π then this will force π π₯ |π₯ β π| to be less than πbecause π π₯ π₯ β π < π π₯ β π < π.