Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
How to do question 33
![EXAMPLE 4 Prove that lim x² = 9.
SOLUTION
1. Guessing a value for 8. Let ɛ > 0 be given. We have to find a number 8 >0
such that
if
0 < |x – 3|< 8
then
|x² – 9| < ɛ
To connect |x? – 9| with |x – 3| we write |x? – 9| = |(x + 3)(x – 3)|. Then
we want
0 < |x – 3|< 8
|x + 3||x – 3|< 8
if
then
Notice that if we can find a positive constant C such that |x + 3| < C, then
|x + 3||x – 3|< C[x – 3|
and we can make C|x – 3|< e by taking |x – 3|< 8/C, so we could choose
8 = ɛ/C.
We can find such a number C if we restrict x to lie in some interval centered at 3.
In fact, since we are interested only in values of x that are close to 3, it is reasonable
to assume that x is within a distance 1 from 3, that is, x – 3 < 1. Then 2 < x < 4,
so 5 <x + 3 <7. Thus we have |x + 3|< 7, and so C = 7 is a suitable choice for
the constant.
But now there are two restrictions on |x – 3|, namely
|x – 3|< 1
|x – 3|< =
and
To make sure that both of these inequalities are satisfied, we take ô to be the smaller of
the two numbers 1 and ɛ/7. The notation for this is 8 = min{1, ɛ/7}.
eserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the ellook and/or eChapter(s).
ent does not malerially affect the overall leaming experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it
SECTION 2.4 The Precise Definition of a Limit
111
2. Showing that this & works. Given ɛ > 0, let 8 = min{1, ɛ/7}. If
0< |x – 3|< 8, then |x – 3|<1 → 2<x<4 → |x + 3| < 7(as in part l).
We also have |x – 3|< 8/7, so
Ix-91-1x + 3||x -3| <7·늑3e
This shows that lim, 3 x? = 9.
As Example 4 shows, it is not always easy to prove that limit statements are true
using the ɛ, & definition. In fact, if we had been given a more complicated function such
as f(x) = (6x² – 8x + 9)/(2x? – 1), a proof would require a great deal of ingenuity.
Fortunately this is unnecessary because the Limit Laws stated in Section 2.3 can be
proved using Definition 2, and then the limits of complicated functions can be found
rigorously from the Limit Laws without resorting to the definition directly.
For instance, we prove the Sum Law: If lim,»af(x) = L and lim,»a g(x) = M both
exist, then
lim [f(x) + g(x)] = L + M](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fef63a737-eaa8-4b6a-be92-19a675838c28%2Ff0991354-6a50-4d10-b263-364b13c3354b%2F756afw_processed.png&w=3840&q=75)
Transcribed Image Text:EXAMPLE 4 Prove that lim x² = 9.
SOLUTION
1. Guessing a value for 8. Let ɛ > 0 be given. We have to find a number 8 >0
such that
if
0 < |x – 3|< 8
then
|x² – 9| < ɛ
To connect |x? – 9| with |x – 3| we write |x? – 9| = |(x + 3)(x – 3)|. Then
we want
0 < |x – 3|< 8
|x + 3||x – 3|< 8
if
then
Notice that if we can find a positive constant C such that |x + 3| < C, then
|x + 3||x – 3|< C[x – 3|
and we can make C|x – 3|< e by taking |x – 3|< 8/C, so we could choose
8 = ɛ/C.
We can find such a number C if we restrict x to lie in some interval centered at 3.
In fact, since we are interested only in values of x that are close to 3, it is reasonable
to assume that x is within a distance 1 from 3, that is, x – 3 < 1. Then 2 < x < 4,
so 5 <x + 3 <7. Thus we have |x + 3|< 7, and so C = 7 is a suitable choice for
the constant.
But now there are two restrictions on |x – 3|, namely
|x – 3|< 1
|x – 3|< =
and
To make sure that both of these inequalities are satisfied, we take ô to be the smaller of
the two numbers 1 and ɛ/7. The notation for this is 8 = min{1, ɛ/7}.
eserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the ellook and/or eChapter(s).
ent does not malerially affect the overall leaming experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it
SECTION 2.4 The Precise Definition of a Limit
111
2. Showing that this & works. Given ɛ > 0, let 8 = min{1, ɛ/7}. If
0< |x – 3|< 8, then |x – 3|<1 → 2<x<4 → |x + 3| < 7(as in part l).
We also have |x – 3|< 8/7, so
Ix-91-1x + 3||x -3| <7·늑3e
This shows that lim, 3 x? = 9.
As Example 4 shows, it is not always easy to prove that limit statements are true
using the ɛ, & definition. In fact, if we had been given a more complicated function such
as f(x) = (6x² – 8x + 9)/(2x? – 1), a proof would require a great deal of ingenuity.
Fortunately this is unnecessary because the Limit Laws stated in Section 2.3 can be
proved using Definition 2, and then the limits of complicated functions can be found
rigorously from the Limit Laws without resorting to the definition directly.
For instance, we prove the Sum Law: If lim,»af(x) = L and lim,»a g(x) = M both
exist, then
lim [f(x) + g(x)] = L + M
Transcribed Image Text:33. Verify that another possible choice of ô for showing that
limx-3 x? = 9 in Example 4 is ô
min{2, ɛ/8}.
=
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