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UAE_Math_Grade_12_Vol_1_SE_718383_ch3
GO01962-Smith-v1.cls
July 4, 2016
Rather than apply l’Ĥopital’s Rule to this last expression, we first simplify the
expression, by multiplying top and bottom by (x + 1). We now have
1 − 1
[ ] x + 1 ( )
lim 1 − 1 x + 1
x→0 ln (x + 1) x = lim ( 1 ) x + 1
x→0
x + 1 x + ln (x + 1)(1)
(x + 1) − 1 ( )
0
= lim
x→0 x + (x + 1) ln (x + 1) 0
d (x)
dx
= lim By l’Ĥopital’s Rule.
x→0 d [x + (x + 1) ln (x + 1)]
dx
1 1
= lim = ,
x→0 1 2
1 + (1) ln (x + 1) + (x + 1)
(x + 1)
which is consistent with Figure 4.19.

y EXAMPLE 2.7 The Indeterminate Form 0 ⋅ ∞
( 1 )
0.4 Evaluate lim ln x .
x→∞ x
0.3
Solution This limit has the indeterminate form (0 ⋅ ∞). From the graph in
Figure 4.20, it appears that the function is decreasing very slowly toward 0 as
0.2 x → ∞. It’s easy to rewrite this in the form ∞ and then apply l’Ĥopital’s Rule.
Note that ∞
0.1 ( 1 ) ln x ( ∞ )
lim ln x = lim
x→∞ x x→∞ x ∞
x
20 40 60 80 100 d ln x
dx
FIGURE 4.20 = lim d By l’Ĥopital’s Rule.
x→∞
y = 1 ln x dx x
x
1
x 0
= lim = = 0.
x→∞ 1 1
∞
0
0
Note: If lim[f(x)] g(x) has one of the indeterminate forms 0 , ∞ or 1 , then, letting
x→c
y = [f(x)] g(x) ,wehavefor f(x) > 0 that
ln y = ln[f(x)] g(x) = g(x) ln [f(x)],
y
so that lim ln y = lim{g(x) ln [f(x)]} will have the indeterminate form 0 ⋅ ∞, which we can
x→c x→c
20 deal with as in example 2.7.
15
EXAMPLE 2.8 The Indeterminate Form 1 ∞
10 1
Evaluate lim x x−1 .
x→1 + ∞
Copyright © McGraw-Hill Education 0.5 FIGURE 4.21 2.0 x graph in Figure 4.21, it appears that the limit is somewhere around 3. We define
5
Solution First, note that this limit has the indeterminate form (1 ). From the
1
1.0
1.5
y = x x−1 , so that
1
1
ln y = ln x x−1 =
ln x.
1
x − 1
y = x x−1
245