James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

108 Chapter 2 Limits and Derivatives
use the cursor to estimate that the x-coordinate of the point of intersection of the line
y 5 2.2 and the curve y 5 x 3 2 5x 1 6 is about 0.911. Similarly, y 5 x 3 2 5x 1 6
intersects the line y 5 1.8 when x < 1.124. So, rounding toward 1 to be safe, we can
say that
if 0.92 , x , 1.12 then 1.8 , x 3 2 5x 1 6 , 2.2
This interval s0.92, 1.12d is not symmetric about x 5 1. The distance from x 5 1 to the
left endpoint is 1 2 0.92 5 0.08 and the distance to the right endpoint is 0.12. We can
choose to be the smaller of these numbers, that is, 5 0.08. Then we can rewrite our
inequalities in terms of distances as follows:
if | x 2 1 | , 0.08 then | sx 3 2 5x 1 6d 2 2 | , 0.2
This just says that by keeping x within 0.08 of 1, we are able to keep f sxd within 0.2
of 2.
Although we chose 5 0.08, any smaller positive value of would also have
worked.
■
TEC In Module 2.4/2.6 you can
explore the precise definition of a limit
both graphically and numerically.
The graphical procedure in Example 1 gives an illustration of the definition for
« 5 0.2, but it does not prove that the limit is equal to 2. A proof has to provide a for
every «.
In proving limit statements it may be helpful to think of the definition of limit as a
challenge. First it challenges you with a number «. Then you must be able to produce a
suitable . You have to be able to do this for every « . 0, not just a particular «.
Imagine a contest between two people, A and B, and imagine yourself to be B. Person
A stipulates that the fixed number L should be approximated by the values of f sxd to within
a degree of accuracy « (say, 0.01). Person B then responds by finding a number such
that if 0 , | x 2 a | , , then | f sxd 2 L | , «. Then A may become more exacting and
challenge B with a smaller value of « (say, 0.0001). Again B has to respond by finding a
corresponding . Usually the smaller the value of «, the smaller the corresponding value
of must be. If B always wins, no matter how small A makes «, then lim x l a f sxd 5 L.
ExamplE 2 Prove that lim
x l3
s4x 2 5d − 7.
SOLUtion
1. Preliminary analysis of the problem (guessing a value for ). Let « be a given
positive number. We want to find a number such that
if 0 , | x 2 3 | , then | s4x 2 5d 2 7 | , «
But | s4x 2 5d 2 7 | 5 | 4x 2 12 | 5 | 4sx 2 3d | 5 4 | x 2 3 | . Therefore we want
such that
if 0 , | x 2 3 | , then 4 | x 2 3 | , «
that is, if 0 , | x 2 3 | , then | x 2 3 | , « 4
This suggests that we should choose 5 «y4.
2. Proof (showing that this works). Given « . 0, choose 5 «y4. If
0 , | x 2 3 | , , then
| s4x 2 5d 2 7 | − | 4x 2 12 | − 4 | x 2 3 | , 4 − 4 S « 4D − «
Copyright 2016 Cengage Learning. All Rights Reserved. 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 eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.