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

Section 2.6 Limits at Infinity; Horizontal Asymptotes 131
In computing the limit as x l 2`, we must remember that for x , 0, we have
sx 2 − | x | − 2x. So when we divide the numerator by x, for x , 0 we get
s2x 2 1 1
x
− s2x 2 1 1
2sx 2 − 2Î 2x 2 1 1
x 2 − 2Î2 1 1 x 2
Therefore
2Î2 1 1 s2x 2 1 1
x 2
lim
− lim
x l2` 3x 2 5 x l2`
3 2 5 x
−
2Î2 1 lim
x l2`
3 2 5 lim
x l2`
1
x
1
x 2
− 2 s2
3
y= œ„2
3
y=_ œ„2
3
y
x
Thus the line y − 2s2 y3 is also a horizontal asymptote.
A vertical asymptote is likely to occur when the denominator, 3x 2 5, is 0, that is,
when x − 5 3 . If x is close to 5 3 and x . 5 3 , then the denominator is close to 0 and 3x 2 5
is positive. The numerator s2x 2 1 1 is always positive, so f sxd is positive. Therefore
s2x 2 1 1
lim
x l s5y3d 1 3x 2 5
− `
(Notice that the numerator does not approach 0 as x l 5y3).
If x is close to 5 3 but x , 5 3 , then 3x 2 5 , 0 and so f sxd is large negative. Thus
x= 5 3
FIGURE 8
y − s2x2 1 1
3x 2 5
s2x
lim
2 1 1
x ls5y3d 2 3x 2 5
− 2`
The vertical asymptote is x − 5 3 . All three asymptotes are shown in Figure 8. n
ExamplE 5 Compute lim
x l` (sx 2 1 1 2 x).
We can think of the given function as
having a denominator of 1.
y
SOLUtion Because both sx 2 1 1 and x are large when x is large, it’s difficult to see
what happens to their difference, so we use algebra to rewrite the function. We first
multiply numerator and denominator by the conjugate radical:
lim (sx 2 1 1 2 x) − lim (sx 2 1 1 2
x l` x
x) ? sx 2 1 1 1 x
l` sx 2 1 1 1 x
sx 2 1 1d 2 x 2
− lim
x l` sx 2 1 1 1 x − lim
x l`
1
sx 2 1 1 1 x
1
y= œ„„„„„-x ≈+1
Notice that the denominator of this last expression (sx 2 1 1 1 x) becomes large as
x l ` (it’s bigger than x). So
0 1
x
lim (sx 1
2 1 1 2 x) − lim
x l` x l` sx 2 1 1 1 x − 0
FIGURE 9
Figure 9 illustrates this result.
n
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