MATH 206 JAMES STEWART
CALCULUS
METRIC VERSION, 8 E
Early Transcendentals
SECTION 4.4
INDETERMINATE FORMS AND
l’HOSPITAL RULE
Examples: 1-7, 9, 10
Exercises: 10, 25, 27, 31, 32, 44, 47, 52,
57, 58, 63
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P3
P4
ex
Example 2 Find lim 2
x x
Solution Since lim e x and lim x 2
x x
the limit is an indeterminate form of type ,so we can apply l'Hospital Rule:
e x
d
ex
lim 2 lim dx
x x x d
dx
x 2
ex
lim again apply l'Hospital Rule
x 2x
d
ex
lim dx
x d
2x
dx
ex
lim
x 2
P5
ln x
Example 3 Calculate lim
x x
Solution Since lim ln x and lim x
x x
the limit is an indeterminate form of type ,so we can apply l'Hospital Rule:
d
ln x ln x
lim lim dx
x x x d 12
x
dx
1
lim x
x 1
1 2
x
2
2
lim 1
x 1
2
x
2
lim
x x
P6
0
tan x x
Example 4 Find lim
x 0 x3
Solution Since lim tan x x 0 and lim x 3 0, the limit is
x 0 x 0
0
an indeterminate form of type , we can apply l'Hospital Rule
0
d
tan x x tan x x
lim lim dx
x3 d
x 3
x 0 x 0
dx
sec 2 x 1 0
lim 2
again apply l'Hospital Rule:
x 0 3x 0
2sec 2 x tan x
lim
x 0 6x
sec 2 x tan x 0
lim again apply l'Hospital Rule:
x 0 3x 0
lim
2sec x .sec x tan x tan x sec 2 x .sec 2 x
x 0 3
2sec 2 x tan 2 x sec 4 x
lim
x 0 3
0 1 1
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3 3
sin x
Example 5 Find lim
x 1 cos x
Solution Since lim sin x 0 and lim 1 cos x 2
x x
sin x 0
lim 0
x 1 cos x 2
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Indeterminate Products
Indeterminate form of type 0
If lim f x 0 and lim g x or
x a x a
then it is not clear what the value of lim f x g x if any, will be.
x a
We can deal with it by writing
f x g x
lim f x g x lim OR lim
x a x a 1 x a 1
g x f x
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Example 6 Evaluate lim x ln x
x 0
Solution Since lim x 0 and lim ln x
x 0 x 0
the limit is an indeterminate form of type 0 ,
ln x
lim x ln x lim
x 0 x 0 1
x
d
ln x
lim dx
x 0
d
1
dx x
1
lim x
x 0 1
2
x
lim x 0
x 0
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Indeterminate Differences
If lim f x and lim g x , then the limit
x a x a
lim f x g x is called an indeterminate form of type -
x a
Product Rule:
d dv du
uv u v
dx dx dx
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1 1
Example 7 Evaluate lim
x 1 ln x x 1
1 1
Solution Since lim and lim , the limit is an indeterminate form of type ,
x 1 ln x x 1 x 1
1 1 x 1 ln x 0
lim lim we can apply l'Hospital Rule:
x 1 ln x x 1 x 1
ln x x 1 0
d 1
dx x 1 ln x 1
x
= lim xlim
1 x 1
x 1
d
dx
ln x x 1
x ln x
x 1
x
lim
x 1 x 1 x ln x
x
x 1 0
lim agian apply l'Hospital Rule:
x 1 x 1 x ln x 0
1
lim
x 1 1
1 x x ln x
1 1
lim
x 1 1 1 ln x 2
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Indeterminate Powers
Several indeterminate forms arise from the limit lim f x
g x
x a
1.lim f x 0 and lim g x 0 type 0 0
x a x a
2.lim f x and lim g x 0 type 0
x a x a
3.lim f x 1 and lim g x type 1
x a x a
Each of these three cases can be treated either by taking the natural logarithm
g x
Let y f x , then ln y g x ln f x
or by writing the function as exponential
g x
f x
g x ln f x
e
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Evaluate lim 1 sin 4x
cot x
Example 9
x 0
Solution Since lim 1 sin 4x 1 and cot x , thus the limit
x 0
is an indeterminate form of type 1 .
Let y 1 sin 4x
cot x
ln y ln 1 sin 4x
cot x
lim ln y lim ln 1 sin 4x
cot x
x 0 x 0
lim cot x ln 1 sin 4x 0
x 0
ln 1 sin 4x 0
lim we can apply l'Hospital Rule
x 0 tan x 0
d
dx ln 1 sin 4x
1
cos 4 x 4 4
= lim lim 1 sin 4x 2
d
tan x
x 0 x 0 sec x
1
dx
lim ln y 4
x 0
lim ln y
x 0
Note that y e ln y
, thus lim y e e 4
x 0
lim 1 sin 4x e 4
cot x
x 0 P14
x
Example 10 Find lim x
x 0
Solution Note that the limit is an indeterminate form of type 0 0 ,
Let y x x
ln y ln x
x
lim ln y lim ln x
x
x 0 x 0
lim x ln x 0
x 0
ln x
lim we can apply l'Hospital Rule:
x 0 1
x
1
1 x2
lim x
1 xlim
0
xlim
x
x 0
2 x 1 0
x
lim ln y
x 0
lim ln y 0. Thus lim y e e 0 1
x 0 x 0
lim x
x
1
x 0 P15
x 3 8
Exercise 10 Find lim
x 2 x 2
0
Solution Note that the limit is an indeterminate form of type ,
0
so we can apply l'Hospital Rule
x 8
3
d
x 3 8
3x 2
lim dx lim 3 2 12
2
lim
x 2 x 2 x 2 d
x 2
x 2 1
dx
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1 2x 1 4x
Exercise 25 Find lim
x 0 x
0
Solution Note that the limit is an indeterminate form of type ,
0
so we can apply l'Hospital Rule:
d
1 1
d
1 2x 1 4x 2
1 1 2
lim
1 2x 2
1 4x 2
lim
dx dx
x 0 x 0 d
x
x
dx
1 1
1 1
lim 1 2x 2 1 4x 2 4
2
x 0 2 2
1 1
lim 1 2x 2
1 4x 2
2
x 0
1 2
lim 1 2 3
x 0 1 2x 1 4x
1 2x 1 4x
lim 3
x 0 x
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e x 1 x
Exercise 27 Find lim
x 0 x2
0
Solution Note that the limit is an indeterminate form of type ,
0
so we can apply l'Hospital Rule
e 1 x
x
d
dx
e x 1 x
lim lim
x2 d
x 2
x 0 x 0
dx
e x 1 0
lim , agian use I'Hospital Rule
x 0 2x 0
d
d
x
e x 1
e x
1
lim lim
x 0 d x 0 2
2x 2
dx
e x 1 x 1
lim 2
.
x 0 x 2
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sin 1 x
Exercise 31 Find lim
x 0 x
0
Solution Note that the limit is an indeterminate form of type ,
0
so we can apply l'Hospital Rule
1
sin x
d
sin 1 x
lim lim dx
x 0 x 0 d
x
x
dx
1
lim 1 x
2 1
lim 1
x 0 1 x 0
1 x 2
sin 1 x
lim 1
x 0 x
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ln x
2
Exercise 32 Calculate lim
x x
Solution Note that the limit is an indeterminate form of type ,
so we can apply l'Hospital Rule
d
2
ln x ln x
2
lim lim dx
x x d
x
x
dx
1
lim 2 ln x
x
x
2 ln x
lim , use l'Hospital rule again
x x
1
2
2 lim x lim 0
x 1 x x
ln x
2
lim 0 P20
x x
Exercise 44 Evaluate lim sin x ln x
x 0
Solution Note that lim sin x 0 and lim ln x ,
x 0 x 0
the limit is an indeterminate product form,
ln x
lim sin x ln x lim
1
, so we can apply l'Hospital Rule
x 0 x 0
sin x
d
ln x
lim dx
d
csc x
x 0
dx
1 1
x x
lim 1
1
lim
x 0
cot x csc x
x 0
sin x tan x
sin x tan x 0
lim , so we can apply l'Hospital Rule
x 0 x 0
d
dx sin x tan x cos x tan x sin x sec 2 x
lim lim
d x 0
x
x 0 1
dx
0 P21
3 x 2
Exercise 47 Find lim x e
x
Solution
x x3
lim x 2
2
3
lim x e , so we can apply l'Hospital Rule
x x
e
d
dx
x 3
3x 2
lim lim x 2
x d
dx
ex
2 x
e 2x
3x
lim x 2 , again apply l'Hospital Rule
x
e 2
d
3x 3x
lim dx
2 lim
x x d
x
2e 2e x
2
dx
3 3
lim x2
lim x2
0
x x
2 e 2x 4x e
x
lim x e 0.
2
3
x
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Exercise 52 Evaluate lim csc x cot x
x 0
Solution Note that the limit is an indeterminate form of type ,
1 cos x
lim csc x cot x lim
x 0 x 0
sin x sin x
1 cos x 0
lim , so we can apply l'Hospital Rule
x 0
sin x 0
sin x
lim 0
x 0
cos x
lim csc x cot x 0
x 0
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x
Exercise 57 Find lim x
x 0
Solution Note that the limit is an indeterminate form of type 0 0 ,
ln y ln x ln y x ln x
x
Let y x x
lim ln y lim x ln x 0
x 0 x 0
ln x
lim , so we can apply l'Hospital Rule
x 0 1
x
d ln x
1
= lim dx lim x
x 0 d x 0
1 1
1 x 2 1
dx
x 2
2
lim
x 0
2
x 2
1
lim 2 x
x 0 lim 2
x 0
x 0
lim ln y
x 0
Thus, lim y e e 0 1
x 0
lim x
x
1 P24
x 0
Exercise 58 Find lim tan 2x
x
x 0
Solution Note that the limit is an indeterminate form of type 0 0 ,
y tan 2x ln y ln tan 2x
x x
Let
lim ln y lim ln tan 2x lim x ln tan 2x
x
x 0 x 0 x 0
ln tan 2x
lim , so we can apply l'Hospital Rule
x 0 1
x
1 d
tan 2x dx tan 2x
1
tan 2x 2sec 2x
2
lim lim
x 0 1 x 0 1
x2 x 2
1
2x 2 2x 2
2
2x sec 2x 2
cos 2
2 x cos 2x
lim lim lim 2
x 0 tan 2x x 0 sin 2x x 0
cos 2x sin 2x
cos 2x
2x 2 1 2x 2
lim xlim
0
sec 2x xlim
0 sin 2 x
x 0
cos 2 x sin 2 x
4x
1 lim 0
x 0 2cos 2x
lim ln y
Note that lim y e x 0
e0 1 lim tan 2x 1
x
x 0 x 0 P25
1
x
Exercise 63 Find lim x
x
Solution Note that the limit is an indeterminate form of type 0 ,
1 1
1
Let y x x
ln y ln x x
ln y ln x
x
ln x
lim ln y lim , so we can apply l'Hospital Rule
x x x
d
dx ln x 1
= lim x 0
lim
x
d
x x
dx
lim ln y
Thus lim y e x
e 0 1
x
1
lim x x 1
x
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