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

Section 4.4 Indeterminate Forms and l’Hospital’s Rule 309
Indeterminate Differences
If lim x l a f sxd − ` and lim x l a tsxd − `, then the limit
lim f f sxd 2 tsxdg
x l a
is called an indeterminate form of type ` 2 `. Again there is a contest between f and
t. Will the answer be ` ( f wins) or will it be 2` (t wins) or will they compromise on a
finite number? To find out, we try to convert the difference into a quotient (for instance,
by using a common denominator, or rationalization, or factoring out a common factor)
so that we have an indeterminate form of type 0 0 or `y`.
ExamplE 7 Compute lim
x l1 1S 1
ln x 2 1
x 2 1D.
SOLUtion First notice that 1ysln xd l ` and 1ysx 2 1d l ` as x l 1 1 , so the limit
is indeterminate of type ` 2 `. Here we can start with a common denominator:
1S lim
1
x l1 ln x 2 1
x 2 1 2 ln x
− lim
x 2 1D
x l1 1 sx 2 1d ln x
Both numerator and denominator have a limit of 0, so l’Hospital’s Rule applies, giving
x 2 1 2 ln x
lim
x l1 1 sx 2 1d ln x
− lim
x l1 1
1 2 1 x
sx 2 1d ? 1 x 1 ln x
− lim
x l1 1 x 2 1
x 2 1 1 x ln x
Again we have an indeterminate limit of type 0 0 , so we apply l’Hospital’s Rule a second
time:
x 2 1
lim
x l1 1 x 2 1 1 x ln x − lim 1
x l1 1 1 1 x ? 1 x 1 ln x
− lim
x l1 1 1
2 1 ln x − 1 2
n
ExamplE 8 Calculate lim
xl`
se x 2 xd.
SOLUtion This is an indeterminate difference because both e x and x approach infinity.
We would expect the limit to be infinity because e x l ` much faster than x. But we can
verify this by factoring out x:
e x 2 x − xS ex
x 2 1 D
The term e x yx l ` as x l ` by l’Hospital’s Rule and so we now have a product in
which both factors grow large:
lim
xl` sex 2 xd −
xl`FxS lim
ex
x
DG 2 1 − `
n
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