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

Section 2.6 Limits at Infinity; Horizontal Asymptotes 133
ings are attached to the following symbols:
lim f sxd − `
x l2`
lim f sxd − 2` lim
x l`
f sxd − 2`
x l2`
y
y=˛
ExamplE 9 Find lim
x l `
x 3 and lim
x l2` x 3 .
SOLUtion When x becomes large, x 3 also becomes large. For instance,
0
FIGURE 11
lim
x l ` x 3 − `, lim
x l2` x 3 − 2`
x
10 3 − 1000 100 3 − 1,000,000 1000 3 − 1,000,000,000
In fact, we can make x 3 as big as we like by requiring x to be large enough. Therefore
we can write
lim
x l ` x 3 − `
Similarly, when x is large negative, so is x 3 . Thus
y
y=´
lim x 3 − 2`
x l2`
These limit statements can also be seen from the graph of y − x 3 in Figure 11.
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Looking at Figure 10 we see that
100
0
1
FIGURE 12
e x is much larger than x 3
when x is large.
y=˛
x
lim e x − `
x l`
but, as Figure 12 demonstrates, y − e x becomes large as x l ` at a much faster rate
than y − x 3 .
ExamplE 10 Find lim
x l` sx 2 2 xd.
SOLUtion It would be wrong to write
lim sx 2 2 xd − lim x 2 2 lim x − ` 2 `
x l` x l` x l`
The Limit Laws can’t be applied to infinite limits because ` is not a number
(` 2 ` can’t be defined). However, we can write
lim sx 2 2 xd − lim xsx 2 1d − `
x l` x l`
because both x and x 2 1 become arbitrarily large and so their product does too.
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ExamplE 11 Find lim
x l `
x 2 1 x
3 2 x .
SOLUtion As in Example 3, we divide the numerator and denominator by the highest
power of x in the denominator, which is just x:
lim
x l`
x 2 1 x
3 2 x − lim
x l`
x 1 1
3
x 2 1 − 2`
because x 1 1 l ` and 3yx 2 1 l 0 2 1 − 21 as x l `.
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