Chapter 3
Limits and the
Derivative
Section 1
Introduction to Limits
(Part 1)

Introduction to Calculus!

How do Algebra and Calculus differ?
 Static vs. Dynamic
• Algebra: solve equations for a particular value of a variable (a static notion)
• Calculus: study how a change in one variable affects another variable (a dynamic notion)
 Calculus was independently developed by Isaac Newton
(1642-1727) of England and Gottfried von Leibniz (1646-
1716) of Germany to solve problems involving motion.

Today, Calculus is used when trying to understand dynamic phenomena with applications in science, business, economics, life science, and social science.
Barnett/Ziegler/Byleen Business Calculus 12e 2

Key Concepts

The two main concepts in Calculus are:
• The derivative
• The integral

Both concepts depend on the notion of limits .
Barnett/Ziegler/Byleen Business Calculus 12e 3

Learning Objectives for Section 3.1
Introduction to Limits
The student will learn about:
■ Limits: a graphical approach
■ Limits: a numerical approach
■ Limits: an algebraic approach
Barnett/Ziegler/Byleen Business Calculus 12e 4

Evaluating Limits

Graphical Approach
• Look at the graph to determine what value y is approaching as x approaches a particular value.
Barnett/Ziegler/Byleen Business Calculus 12e 5

Example 1

What is f(2)?
𝑓 2 = −1

What value is y approaching as x approaches 2?
y 𝑦 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ𝑒𝑠 𝑎 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 3 x
The limit of a function f(x) is the value it approaches when x gets really close to, but not equal to, a specific number.
lim 𝑥→2 𝑓(𝑥) = 3
Barnett/Ziegler/Byleen Business Calculus 12e 6

Example 2 𝑓 0
= 0 lim 𝑥→0 𝑓(𝑥)
= 0 𝑓 3
= undefined 𝑓 4
= 3 lim 𝑥→3 𝑓(𝑥) = 3 lim 𝑥→4 𝑓(𝑥) = 3
Barnett/Ziegler/Byleen Business Calculus 12e 7

Example 3
One-sided limits: 𝑓 1 = 2 lim 𝑥→1 − 𝑓(𝑥)
= 1 lim 𝑥→1 + 𝑓(𝑥)
= 2 lim 𝑥→1 𝑓(𝑥)
= Does not exist
(DNE)
A limit exists only if the left-hand limit equals the right-hand limit.
Barnett/Ziegler/Byleen Business Calculus 12e 8

lim 𝑥→0 −
1 𝑥
= −∞ lim 𝑥→0 +
1 𝑥
= +∞
Example 4
1 𝑓 𝑥 = 𝑥 lim 𝑥→0
1 𝑥
= 𝐷𝑁𝐸
Limit Video
Barnett/Ziegler/Byleen Business Calculus 12e 9

1 lim 𝑥→0 − 𝑥 2
= +∞
1 lim 𝑥→0 + 𝑥 2
= +∞
Example 5
1 𝑓 𝑥 = 𝑥 2
1 lim 𝑥→0 𝑥 2
= +∞
Barnett/Ziegler/Byleen Business Calculus 12e 10

Evaluating Limits

Numerical Approach
• Plug in values of x that are really close to the number it is approaching to determine the value that y is approaching.
Barnett/Ziegler/Byleen Business Calculus 12e 11

Example 6
0
 𝑓 𝑥 = 𝑥+1 𝑥+3
.1 −.0001 −.000001
.000001
.0001
.1
lim 𝑥→0 − 𝑓 𝑥 =?
lim 𝑥→0 + 𝑓 𝑥 =?
𝑓 .1 = 0.3548387
𝑓 −.1 = 0.310344
𝑓 −.0001 = 0.3333111
𝑓 .0001 = 0.3333555
𝑓 −.000001 = 0.33333311
1 lim 𝑥→0 − 𝑓 𝑥 =
3 𝑓 .000001 = 0.33333355
lim 𝑥→0 𝑓 𝑥 = lim 𝑥→0
1
3
+ 𝑓 𝑥 =
1
3
Barnett/Ziegler/Byleen Business Calculus 12e 12

Evaluating Limits
 Algebraic Approach
• “Plug and chug” method.
• Plug in the value of c . If you get zero in the denominator, then use an algebraic technique to rewrite f(x) into a form that works.
• Commonly used techniques: o Factor and cancel o Common denominator o Expand and simplify o Multiply by a conjugate
Barnett/Ziegler/Byleen Business Calculus 12e 13

“Plug and Chug”
 Example 7: Evaluate each limit. lim 𝑥→−1 𝑥 2 − 2𝑥 + 5 = (−1) 2 −2 −1 + 5 = 8 lim 𝑥→2
3𝑥 2 − 6 = 3(2) 2 −6 = 6 lim 𝑥→−2 𝑥 2 𝑥 2 + 1
=
(−2) 2
(−2) 2 +1
=
4
5
Barnett/Ziegler/Byleen Business Calculus 12e 14

“Plug and Chug”

Example 8: Evaluate each expression. 𝑓 𝑥 =
2𝑥 + 3 𝑖𝑓 𝑥 < 5
−𝑥 + 12 𝑥 > 5 lim 𝑥→2 𝑓 𝑥 = 2 2 + 3 = 7 lim 𝑥→7 𝑓 𝑥 = −7 + 12 = 5 lim 𝑥→5
− 𝑓 𝑥 = 2 5 + 3 = 13 lim 𝑥→5 + 𝑓 𝑥 = −(5) + 12 = 7 lim 𝑥→5 𝑓 𝑥 = 𝐷𝑁𝐸
Barnett/Ziegler/Byleen Business Calculus 12e 15

Factor and Cancel
Example 9: Evaluate the limit.
lim 𝑥→3 𝑥 2 − 9 𝑥 − 3
= lim 𝑥→3
= lim 𝑥→3
(𝑥 − 3)(𝑥 + 3)
= 3 + 3 𝑥 − 3 𝑥 + 3
= 6
Barnett/Ziegler/Byleen Business Calculus 12e 16

Common Denominator
Example 10: Evaluate the limit.
lim 𝑥→0
(3 + 𝑥) 𝑥
−1 −3 −1
= lim 𝑥→0
1
3 + 𝑥 − 𝑥
1
3
= lim 𝑥→0
3
3(3 + 𝑥)
−
3 − 3 − 𝑥 𝑥
3(3 + 𝑥)
= lim 𝑥→0 𝑥
−𝑥
3(3 + 𝑥)
= lim 𝑥→0 𝑥
1(3 + 𝑥)
3(3 + 𝑥)
= −
1
9
Barnett/Ziegler/Byleen Business Calculus 12e
−𝑥
= lim 𝑥→0
= lim 𝑥→0
3(3 + 𝑥)
−1
3(3 + 𝑥)
∙
1 𝑥
17

Expand and Simplify
Example 11: Evaluate the limit.
lim ℎ→0
(4 + ℎ) 2 ℎ
−16
= lim ℎ→0
(16 + 8ℎ + ℎ 2 ) − 16 ℎ
= lim ℎ→0
8ℎ + ℎ 2 ℎ
= lim ℎ→0
(8 + ℎ)
= 8
Barnett/Ziegler/Byleen Business Calculus 12e 18

Multiply by a Conjugate
Example 12: Evaluate the limit.
lim ℎ→7 ℎ + 2 − 3 ℎ − 7 ℎ + 2 − 3
∙ ℎ − 7 ℎ + 2 + 3
= lim ℎ→7 ℎ + 2 + 3
(ℎ + 2) − 9
= lim ℎ→7 (ℎ − 7)( ℎ + 2 + 3) ℎ − 7
= lim ℎ→7 (ℎ − 7)( ℎ + 2 + 3)
1
= lim ℎ→7 ( ℎ + 2 + 3)
=
1
6
Barnett/Ziegler/Byleen Business Calculus 12e 19

Homework
#3-1A: Pg 138 (6-10, 15, 17, 25,
29, 43, 46, 51, 53, 57)
Barnett/Ziegler/Byleen Business Calculus 12e 20

Chapter 3
Limits and the
Derivative
Section 1
Introduction to Limits
(Part 2)

Learning Objectives for Section 3.1
Introduction to Limits
The student will learn about:

Limits involving absolute value

Limits of difference quotients

Applications with limits
Barnett/Ziegler/Byleen Business Calculus 12e 22

Absolute Value
 All absolute value functions can be represented as a piecewise function:
• 𝑓 𝑥 = 𝑥 =
−𝑥 𝑥 < 0 𝑥 𝑥 ≥ 0
• 𝑓 𝑥 = 𝑥 − 4 =
−(𝑥 − 4) 𝑥 < 4 𝑥 − 4 𝑥 ≥ 4
• 𝑓 𝑥 = 2𝑥 + 4 =
− 2𝑥 + 4 𝑥 < −2
2𝑥 + 4 𝑥 ≥ −2
 Graph each function to verify the piecewise function.
Barnett/Ziegler/Byleen Business Calculus 12e 23

Absolute Value
 More Examples:
• 𝑓 𝑥 = 𝑥 𝑥 𝑥
= −𝑥 𝑥 𝑥 𝑥 < 0 𝑥 > 0
• 𝑓 𝑥 = 𝑥 𝑥+1 𝑥+1
−𝑥(𝑥+1)
= 𝑥+1 𝑥(𝑥+1) 𝑥+1 𝑥 <−1 𝑥 >−1
Barnett/Ziegler/Byleen Business Calculus 12e 24

Limits From a Graph
 Use the graph to evaluate each limit:
• 𝑓 𝑥 = 𝑥 − 3 y lim 𝑥→3
− 𝑓 𝑥 = 0 x lim 𝑥→3
+ 𝑓 𝑥 = 0 lim 𝑥→3 𝑓 𝑥 = 0
Barnett/Ziegler/Byleen Business Calculus 12e 25

Limits Algebraically
 𝑓 𝑥 = 𝑥 − 4 + 1
 Write the piecewise definition of f(x) .
• 𝑓 𝑥 = 𝑥 − 4 + 1 =
− 𝑥 − 4 + 1 𝑥 < 4 𝑥 − 4 + 1 𝑥 ≥ 4

Evaluate each limit. lim 𝑥→4
− 𝑓 𝑥 = − 4 − 4 + 1 = 1 lim 𝑥→4
+ 𝑓 𝑥 = 4 − 4 + 1 = 1 lim 𝑥→4 𝑓 𝑥 = 1
Barnett/Ziegler/Byleen Business Calculus 12e 26

Limits Algebraically
 𝑓 𝑥 = 𝑥 𝑥 𝑥
=
−𝑥 𝑥 𝑥

Evaluate each limit. 𝑥 < 0 𝑥 > 0 lim 𝑥→0
− 𝑓 𝑥 = 𝑥 lim 𝑥→0 − −𝑥
= lim 𝑥→0 −
−1 = −1 lim 𝑥→0
+ 𝑓 𝑥 = lim 𝑥→0 + 𝑥 𝑥
= lim 𝑥→0
+
1 = 1 lim 𝑥→0 𝑓 𝑥 = 𝐷𝑁𝐸
Barnett/Ziegler/Byleen Business Calculus 12e 27

Difference Quotients

One of the most important limits in Calculus is the limit of a difference quotient.
lim ℎ→0 𝑓 𝑎 + ℎ − 𝑓(𝑎) ℎ
Barnett/Ziegler/Byleen Business Calculus 12e 28

Difference Quotients
Let f ( x ) = 3 x
– 1 𝐸𝑣𝑎𝑙𝑢𝑎𝑡𝑒 lim ℎ→0 𝑓 𝑎 + ℎ − 𝑓(𝑎) ℎ lim ℎ→0
3 𝑎 + ℎ − 1 − [3𝑎 − 1] ℎ
3𝑎 + 3ℎ − 1 − [3𝑎 − 1]
= lim ℎ→0 ℎ
= lim ℎ→0
3𝑎 + 3ℎ − 1 − 3𝑎 + 1 ℎ
= lim ℎ→0
3ℎ ℎ
= lim ℎ→0
3 = 3
Barnett/Ziegler/Byleen Business Calculus 12e 29

Difference Quotients
Let 𝑓 𝑥 = 𝑥 .
𝐸𝑣𝑎𝑙𝑢𝑎𝑡𝑒: lim ℎ→0 𝑓 3 + ℎ − 𝑓(3) ℎ
3 + ℎ − 3
= lim ℎ→0 ℎ
=
3 + ℎ − 3
∙ ℎ
3 + ℎ + 3
= lim ℎ→0
(3 + ℎ) − 3
= lim ℎ→0 ℎ( 3 + ℎ + 3)
3 + ℎ + 3 ℎ
= lim ℎ→0 ℎ( 3 + ℎ + 3)
Barnett/Ziegler/Byleen Business Calculus 12e
1
= lim ℎ→0 ( 3 + ℎ + 3)
1
=
=
( 3 + 0 + 3)
1
2 3
=
6
3
30

Difference Quotients
Let 𝑓 𝑥 =
= lim ℎ→0
1 𝑥
1 𝑎 + ℎ
−
𝐸𝑣𝑎𝑙𝑢𝑎𝑡𝑒: lim ℎ→0
1 𝑎
= lim ℎ→0 𝑎 ℎ 𝑎(𝑎 + ℎ)
− ℎ
(𝑎 + ℎ) 𝑎(𝑎 + ℎ) 𝑓 𝑎 + ℎ − 𝑓(𝑎) ℎ
= lim ℎ→0
= lim ℎ→0
−ℎ 𝑎(𝑎 + ℎ)
−1 𝑎(𝑎 + ℎ) 𝑎 − 𝑎 − ℎ 𝑎(𝑎 + ℎ) =
−1
= lim ℎ→0 ℎ 𝑎(𝑎 + 0)
−1 −ℎ 𝑎(𝑎 + ℎ)
= lim ℎ→0 ℎ
Barnett/Ziegler/Byleen Business Calculus 12e
= 𝑎 2
∙
1 ℎ
31

Application

A long-distance telephone service charges $0.09 per minute for calls lasting 10 minutes or more and $0.18 per minute for calls lasting less than 10 minutes.
A) Write a piecewise function f(x) for a call lasting x minutes.
B) Graph f(x) over the interval (0, 40].
C) Evaluate: lim 𝑥→10 − 𝑓(𝑥) lim 𝑥→10 + 𝑓(𝑥) lim 𝑥→10 𝑓(𝑥)
Barnett/Ziegler/Byleen Business Calculus 12e 32

Solution 𝑓 𝑥 =
0.09𝑥 𝑥 ≥ 10
0.18𝑥 0 < 𝑥 < 10 r a s l l d o
Graph f(x) over the interval (0, 40] y
4
3
2
1 lim 𝑥→10
− lim 𝑥→10 + 𝑓 𝑥 = 𝑓 𝑥 =
1.8
0.9
10 20 minutes
30 40 x lim 𝑥→10 𝑓 𝑥 = 𝐷𝑁𝐸
Barnett/Ziegler/Byleen Business Calculus 12e 33

Homework
#3-1B: Pg 139 (44, 47, 49, 54,
58, 65, 67,
73, 74, 76, 83)
Barnett/Ziegler/Byleen Business Calculus 12e 34