O Read Hakata Tonko x = Ch. 1 Introduction - x O Homework Section x E 2.3 The Limit Laws b 4-6| bartleby Sin Blin Blin No X G square root symbo x + A https://openstax.org/books/calculus-volume-1/pages/2-3-the-limit-laws#fs-id1170571611196 < Calculus Volume 1 2.3 The Limit Laws E Table of contents Search this book 9 My highlights B Print wושיי >1 Functions and Graphs x – 3 lim 1-2 x2 – 2x = lim z-2 x (x – 2) * v2 Limits Introduction Step 2. Since z – 2 is the only part of the denominator that is zero when 2 is substituted, we then separate 1/(x – 2) from the rest of the function: 2.1 A Preview of Calculus 2.2 The Limit of a Function x - 3 lim 1 2.3 The Limit Laws * - 2 2.4 Continuity M Screenshot · now a Step 3. lim = - - and lim -1 = -00. Therefore, the product of (x – 3 2.5 The Precise Definition of a Limit 2. Screenshot taken Z2- 2-2 +oo: Show in folder Key Terms Key Equations x – 3 O t e = +oo. lim z-2 x2 - 27 23 The Lit La Key Concepts ing e fum KAKn the im Chapter Review Exercises •3 Derivatives 24 Cty as e t e •4 Applications of Derivatives CHECKPOINT 2.18 A •5 Integration t-e- - 9 MR .. »6 Applications of Integration Evaluate lim- -z+2 A| Table of Integrals 2+1 (r-1) COPY TO CLIPBOARD B|Table of Derivatives vI 5:18 O Read Hakata Tonko x = Ch. 1 Introduction O Homework Section x E 2.3 The Limit Laws b 4-6| bartleby Sin Blin Blin N G square root symbo x + A https://openstax.org/books/calculus-volume-1/pages/2-3-the-limit-laws#fs-id1170571611196 < Calculus Volume 1 2.3 The Limit Laws E Table of contents Search this book 9 My highlights B Print >1 Functions and Graphs Evaluating a Limit of the Form K/0, K +0 Using the Limit Laws v2 Limits Introduction Evaluate lim 2. * 2.1 A Preview of Calculus 2.2 The Limit of a Function [Hide Solution] 2.3 The Limit Laws Solution 2.4 Continuity Step 1. After substituting in x = 2, we see that this limit has the form –1/0. That is, as x approaches 2 from the 2.5 The Precise Definition of a Limit left, the numerator approaches -1; and the denominator approaches 0. Consequently, the magnitude of z(r-2) Key Terms becomes infinite. To get a better idea of what the limit is, we need to factor the denominator: x - 3 lim z+2 x2 – 2x Key Equations x - 3 = lim z2 x (x – 2) Key Concepts Chapter Review Exercises Step 2. Since x – 2 is the only part of the denominator that is zero when 2 is substituted, we then separate 1/(x – 2) from the rest of the function: •3 Derivatives •4 Applications of Derivatives 1 x - 3 = lim •5 Integration x - 2 »6 Applications of Integration Step 3. lim -3 and lim = -00. Therefore, the product of (x – 3) /x and 1/(x – 2) has a limit of A| Table of Integrals +oo: B|Table of Derivatives vi 5:18
O Read Hakata Tonko x = Ch. 1 Introduction - x O Homework Section x E 2.3 The Limit Laws b 4-6| bartleby Sin Blin Blin No X G square root symbo x + A https://openstax.org/books/calculus-volume-1/pages/2-3-the-limit-laws#fs-id1170571611196 < Calculus Volume 1 2.3 The Limit Laws E Table of contents Search this book 9 My highlights B Print wושיי >1 Functions and Graphs x – 3 lim 1-2 x2 – 2x = lim z-2 x (x – 2) * v2 Limits Introduction Step 2. Since z – 2 is the only part of the denominator that is zero when 2 is substituted, we then separate 1/(x – 2) from the rest of the function: 2.1 A Preview of Calculus 2.2 The Limit of a Function x - 3 lim 1 2.3 The Limit Laws * - 2 2.4 Continuity M Screenshot · now a Step 3. lim = - - and lim -1 = -00. Therefore, the product of (x – 3 2.5 The Precise Definition of a Limit 2. Screenshot taken Z2- 2-2 +oo: Show in folder Key Terms Key Equations x – 3 O t e = +oo. lim z-2 x2 - 27 23 The Lit La Key Concepts ing e fum KAKn the im Chapter Review Exercises •3 Derivatives 24 Cty as e t e •4 Applications of Derivatives CHECKPOINT 2.18 A •5 Integration t-e- - 9 MR .. »6 Applications of Integration Evaluate lim- -z+2 A| Table of Integrals 2+1 (r-1) COPY TO CLIPBOARD B|Table of Derivatives vI 5:18 O Read Hakata Tonko x = Ch. 1 Introduction O Homework Section x E 2.3 The Limit Laws b 4-6| bartleby Sin Blin Blin N G square root symbo x + A https://openstax.org/books/calculus-volume-1/pages/2-3-the-limit-laws#fs-id1170571611196 < Calculus Volume 1 2.3 The Limit Laws E Table of contents Search this book 9 My highlights B Print >1 Functions and Graphs Evaluating a Limit of the Form K/0, K +0 Using the Limit Laws v2 Limits Introduction Evaluate lim 2. * 2.1 A Preview of Calculus 2.2 The Limit of a Function [Hide Solution] 2.3 The Limit Laws Solution 2.4 Continuity Step 1. After substituting in x = 2, we see that this limit has the form –1/0. That is, as x approaches 2 from the 2.5 The Precise Definition of a Limit left, the numerator approaches -1; and the denominator approaches 0. Consequently, the magnitude of z(r-2) Key Terms becomes infinite. To get a better idea of what the limit is, we need to factor the denominator: x - 3 lim z+2 x2 – 2x Key Equations x - 3 = lim z2 x (x – 2) Key Concepts Chapter Review Exercises Step 2. Since x – 2 is the only part of the denominator that is zero when 2 is substituted, we then separate 1/(x – 2) from the rest of the function: •3 Derivatives •4 Applications of Derivatives 1 x - 3 = lim •5 Integration x - 2 »6 Applications of Integration Step 3. lim -3 and lim = -00. Therefore, the product of (x – 3) /x and 1/(x – 2) has a limit of A| Table of Integrals +oo: B|Table of Derivatives vi 5:18
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.6: Inequalities
Problem 77E
Related questions
Question
In the following exercises, use direct substitution to obtain an undefined expression. Then, use the method of Example 2.23 to simplify the function to help determine the limit.
105. lim 2x^2 + 7x − 4 / x^2 + x − 2
x→1−
106. lim 2x^2 + 7x −4 / x^2 + x − 2
x→1+
Transcribed Image Text:O Read Hakata Tonko x
= Ch. 1 Introduction - x
O Homework Section x
E 2.3 The Limit Laws
b 4-6| bartleby
Sin Blin Blin No X
G square root symbo x +
A https://openstax.org/books/calculus-volume-1/pages/2-3-the-limit-laws#fs-id1170571611196
< Calculus Volume 1
2.3 The Limit Laws
E Table of contents
Search this book
9 My highlights B Print
wושיי
>1 Functions and Graphs
x – 3
lim
1-2 x2 – 2x
= lim
z-2 x (x – 2) *
v2 Limits
Introduction
Step 2. Since z – 2 is the only part of the denominator that is zero when 2 is substituted, we then separate
1/(x – 2) from the rest of the function:
2.1 A Preview of Calculus
2.2 The Limit of a Function
x - 3
lim
1
2.3 The Limit Laws
* - 2
2.4 Continuity
M Screenshot · now a
Step 3. lim = -
- and lim -1
= -00. Therefore, the product of (x – 3
2.5 The Precise Definition of a Limit
2.
Screenshot taken
Z2- 2-2
+oo:
Show in folder
Key Terms
Key Equations
x – 3
O t
e
= +oo.
lim
z-2 x2 - 27
23 The Lit La
Key Concepts
ing e fum KAKn the im
Chapter Review Exercises
•3 Derivatives
24 Cty
as e
t
e
•4 Applications of Derivatives
CHECKPOINT 2.18
A
•5 Integration
t-e-
-
9 MR ..
»6 Applications of Integration
Evaluate lim-
-z+2
A| Table of Integrals
2+1 (r-1)
COPY TO CLIPBOARD
B|Table of Derivatives
vI 5:18
![O Read Hakata Tonko x
= Ch. 1 Introduction
O Homework Section x
E 2.3 The Limit Laws
b 4-6| bartleby
Sin Blin Blin N
G square root symbo x
+
A https://openstax.org/books/calculus-volume-1/pages/2-3-the-limit-laws#fs-id1170571611196
< Calculus Volume 1
2.3 The Limit Laws
E Table of contents
Search this book
9 My highlights
B Print
>1 Functions and Graphs
Evaluating a Limit of the Form K/0, K +0 Using the Limit Laws
v2 Limits
Introduction
Evaluate lim 2. *
2.1 A Preview of Calculus
2.2 The Limit of a Function
[Hide Solution]
2.3 The Limit Laws
Solution
2.4 Continuity
Step 1. After substituting in x = 2, we see that this limit has the form –1/0. That is, as x approaches 2 from the
2.5 The Precise Definition of a Limit
left, the numerator approaches -1; and the denominator approaches 0. Consequently, the magnitude of
z(r-2)
Key Terms
becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:
x - 3
lim
z+2 x2 – 2x
Key Equations
x - 3
= lim
z2 x (x – 2)
Key Concepts
Chapter Review Exercises
Step 2. Since x – 2 is the only part of the denominator that is zero when 2 is substituted, we then separate
1/(x – 2) from the rest of the function:
•3 Derivatives
•4 Applications of Derivatives
1
x - 3
= lim
•5 Integration
x - 2
»6 Applications of Integration
Step 3. lim -3
and lim
= -00. Therefore, the product of (x – 3) /x and 1/(x – 2) has a limit of
A| Table of Integrals
+oo:
B|Table of Derivatives
vi 5:18](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F54c876d5-59b0-44a0-83c7-bb99bd887b31%2Fdb4088d4-f287-48c2-be63-016a424123f1%2Fouang5t_processed.png&w=3840&q=75)
Transcribed Image Text:O Read Hakata Tonko x
= Ch. 1 Introduction
O Homework Section x
E 2.3 The Limit Laws
b 4-6| bartleby
Sin Blin Blin N
G square root symbo x
+
A https://openstax.org/books/calculus-volume-1/pages/2-3-the-limit-laws#fs-id1170571611196
< Calculus Volume 1
2.3 The Limit Laws
E Table of contents
Search this book
9 My highlights
B Print
>1 Functions and Graphs
Evaluating a Limit of the Form K/0, K +0 Using the Limit Laws
v2 Limits
Introduction
Evaluate lim 2. *
2.1 A Preview of Calculus
2.2 The Limit of a Function
[Hide Solution]
2.3 The Limit Laws
Solution
2.4 Continuity
Step 1. After substituting in x = 2, we see that this limit has the form –1/0. That is, as x approaches 2 from the
2.5 The Precise Definition of a Limit
left, the numerator approaches -1; and the denominator approaches 0. Consequently, the magnitude of
z(r-2)
Key Terms
becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:
x - 3
lim
z+2 x2 – 2x
Key Equations
x - 3
= lim
z2 x (x – 2)
Key Concepts
Chapter Review Exercises
Step 2. Since x – 2 is the only part of the denominator that is zero when 2 is substituted, we then separate
1/(x – 2) from the rest of the function:
•3 Derivatives
•4 Applications of Derivatives
1
x - 3
= lim
•5 Integration
x - 2
»6 Applications of Integration
Step 3. lim -3
and lim
= -00. Therefore, the product of (x – 3) /x and 1/(x – 2) has a limit of
A| Table of Integrals
+oo:
B|Table of Derivatives
vi 5:18
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