25 Oct 2021
Problem 23
Page 111
Section 2.3: Calculating limits using limit laws
Chapter 2: Limits and Derivatives
Textbook ExpertVerified Tutor
25 Oct 2021
Given information
The given expression is:
Step-by-step explanation
Step 1.
Simplify the given expression.
Substitute 0 for 
in the given expression.
Therefore, it can be noticed that the value of the expression for 
is not defined.
The fifth law of limit is 
, if 
.
Evaluate the value of 
.
.
In the given expression it can be noticed that 
.
That is the denominator of the given expression becomes 0.
Therefore, the fifth law of limit cannot be applied to evaluate the limit.
Single Variable Calculus: Early Transcendentals
4th Edition, 2018
Stewart
ISBN: 9781337687805
Solutions
Chapter
Section
- Problem 1a
- Problem 1b
- Problem 1c
- Problem 1d
- Problem 1e
- Problem 1f
- Problem 2a
- Problem 2b
- Problem 2c
- Problem 2d
- Problem 2e
- Problem 2f
- Problem 3
- Problem 4
- Problem 5
- Problem 6
- Problem 7
- Problem 8a
- Problem 8b
- Problem 9
- Problem 10
- Problem 11
- Problem 12
- Problem 13
- Problem 14
- Problem 15
- Problem 16
- Problem 17
- Problem 18
- Problem 19
- Problem 20
- Problem 21
- Problem 22
- Problem 23
- Problem 24
- Problem 25a
- Problem 25b
- Problem 25c
- Problem 26a
- Problem 26b
- Problem 26c
- Problem 27
- Problem 28
- Problem 29
- Problem 30
- Problem 31
- Problem 32
- Problem 33
- Problem 34
- Problem 35
- Problem 36
- Problem 37a
- Problem 37b
- Problem 38a
- Problem 38b
- Problem 38c
- Problem 39a
- Problem 39b
- Problem 39c
- Problem 40a
- Problem 40b
- Problem 40c
- Problem 41
- Problem 42
- Problem 43
- Problem 44
- Problem 45
- Problem 46a
- Problem 46b
- Problem 47
- Problem 48
- Problem 49
- Problem 50