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

136 Chapter 2 Limits and Derivatives
SOLUtion We rewrite the given inequality as
0.5 , 3x 2 2 x 2 2
5x 2 1 4x 1 1 , 0.7
1
y=0.7
y=0.5
0 15
FIGURE 17
y= 3≈-x-2
5≈+4x+1
We need to determine the values of x for which the given curve lies between the horizontal
lines y − 0.5 and y − 0.7. So we graph the curve and these lines in Figure 17.
Then we use the cursor to estimate that the curve crosses the line y − 0.5 when
x < 6.7. To the right of this number it seems that the curve stays between the lines
y − 0.5 and y − 0.7. Rounding up to be safe, we can say that
if x . 7 then Z
3x 2 2 x 2 2
5x 2 1 4x 1 1 2 0.6 Z , 0.1
In other words, for « − 0.1 we can choose N − 7 (or any larger number) in Defini ­
tion 7.
n
1
ExamplE 14 Use Definition 7 to prove that lim
x l ` x − 0.
SOLUtion Given « . 0, we want to find N such that
if x . N then Z
1
x 2 0 Z , «
In computing the limit we may assume that x . 0. Then 1yx , « &? x . 1y«. Let’s
choose N − 1y«. So
if x . N − 1 «
Therefore, by Definition 7,
then Z
1
x 2 0 Z − 1 x , «
1
lim
x l ` x − 0
Figure 18 illustrates the proof by showing some values of « and the corresponding
values of N.
y
y
y
∑=1
0 N=1
x
∑=0.2
0 N=5
x
∑=0.1
0 N=10
x
FIGURE 18
n
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