L'Hôpital's Rule allows us to resolve limits of indeterminate form: 0/0, infinity/infinity, infinity-infinity, 0^0, 1^infinity, and infinity^0

1. Sec on 3.7
Indeterminate forms and lHôpital’s
Rule
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University
March 30, 2011
.

2. Announcements
Midterm has been
returned. Please see FAQ
on Blackboard (under
”Exams and Quizzes”)
Quiz 3 this week in
recita on on Sec on 2.6,
2.8, 3.1, 3.2

3. Objectives
Know when a limit is of
indeterminate form:
indeterminate quo ents:
0/0, ∞/∞
indeterminate products:
0×∞
indeterminate
differences: ∞ − ∞
indeterminate powers:
00 , ∞0 , and 1∞
Resolve limits in
indeterminate form

4. Recall
Recall the limit laws from Chapter 2.
Limit of a sum is the sum of the limits

5. Recall
Recall the limit laws from Chapter 2.
Limit of a sum is the sum of the limits
Limit of a difference is the difference of the limits

6. Recall
Recall the limit laws from Chapter 2.
Limit of a sum is the sum of the limits
Limit of a difference is the difference of the limits
Limit of a product is the product of the limits

7. Recall
Recall the limit laws from Chapter 2.
Limit of a sum is the sum of the limits
Limit of a difference is the difference of the limits
Limit of a product is the product of the limits
Limit of a quo ent is the quo ent of the limits ... whoops! This
is true as long as you don’t try to divide by zero.

8. More about dividing limits
We know dividing by zero is bad.
Most of the me, if an expression’s numerator approaches a
finite nonzero number and denominator approaches zero, the
quo ent has an infinite. For example:
1 cos x
lim+ = +∞ lim− = −∞
x→0 x x→0 x3

9. Why 1/0 ̸= ∞
1
Consider the func on f(x) = 1 .
x sin x
y
.
x
Then lim f(x) is of the form 1/0, but the limit does not exist and is
x→∞
not infinite.

10. Why 1/0 ̸= ∞
1
Consider the func on f(x) = 1 .
x sin x
y
.
x
Then lim f(x) is of the form 1/0, but the limit does not exist and is
x→∞
not infinite.
Even less predictable: when numerator and denominator both go to
zero.

11. Experiments with funny limits
sin2 x
lim
x→0 x
.

12. Experiments with funny limits
sin2 x
lim =0
x→0 x
.

13. Experiments with funny limits
sin2 x
lim =0
x→0 x
x
lim 2
x→0 sin x
.

14. Experiments with funny limits
sin2 x
lim =0
x→0 x
x
lim 2 does not exist
x→0 sin x
.

15. Experiments with funny limits
sin2 x
lim =0
x→0 x
x
lim 2 does not exist
x→0 sin x
sin2 x .
lim
x→0 sin(x2 )

16. Experiments with funny limits
sin2 x
lim =0
x→0 x
x
lim 2 does not exist
x→0 sin x
sin2 x .
lim =1
x→0 sin(x2 )

17. Experiments with funny limits
sin2 x
lim =0
x→0 x
x
lim 2 does not exist
x→0 sin x
sin2 x .
lim =1
x→0 sin(x2 )
sin 3x
lim
x→0 sin x

18. Experiments with funny limits
sin2 x
lim =0
x→0 x
x
lim 2 does not exist
x→0 sin x
sin2 x .
lim =1
x→0 sin(x2 )
sin 3x
lim =3
x→0 sin x

19. Experiments with funny limits
sin2 x
lim =0
x→0 x
x
lim 2 does not exist
x→0 sin x
sin2 x .
lim =1
x→0 sin(x2 )
sin 3x
lim =3
x→0 sin x
0
All of these are of the form , and since we can get different
0
answers in different cases, we say this form is indeterminate.

20. Language Note
It depends on what the meaning of the word “is” is
Be careful with the language here. We
are not saying that the limit in each
0
case “is” , and therefore nonexistent
0
because this expression is undefined.
0
The limit is of the form , which means
0
we cannot evaluate it with our limit
laws.

21. Indeterminate forms are like Tug Of War
Which side wins depends on which side is stronger.

22. Outline
L’Hôpital’s Rule
Rela ve Rates of Growth
Other Indeterminate Limits
Indeterminate Products
Indeterminate Differences
Indeterminate Powers

23. The Linear Case
Ques on
f(x)
If f and g are lines and f(a) = g(a) = 0, what is lim ?
x→a g(x)

24. The Linear Case
Ques on
f(x)
If f and g are lines and f(a) = g(a) = 0, what is lim ?
x→a g(x)
Solu on
The func ons f and g can be wri en in the form
f(x) = m1 (x − a) g(x) = m2 (x − a)
So
f(x) m1
=
g(x) m2

25. The Linear Case, Illustrated
y y = g(x)
y = f(x)
a f(x) g(x)
. x
x
f(x) f(x) − f(a) (f(x) − f(a))/(x − a) m1
= = =
g(x) g(x) − g(a) (g(x) − g(a))/(x − a) m2

26. What then?
But what if the func ons aren’t linear?

27. What then?
But what if the func ons aren’t linear?
Can we approximate a func on near a point with a linear
func on?

28. What then?
But what if the func ons aren’t linear?
Can we approximate a func on near a point with a linear
func on?
What would be the slope of that linear func on?

29. What then?
But what if the func ons aren’t linear?
Can we approximate a func on near a point with a linear
func on?
What would be the slope of that linear func on? The
deriva ve!

30. Theorem of the Day
Theorem (L’Hopital’s Rule)
Suppose f and g are differen able func ons and g′ (x) ̸= 0 near a
(except possibly at a). Suppose that
lim f(x) = 0 and lim g(x) = 0
x→a x→a
or lim f(x) = ±∞ and lim g(x) = ±∞
x→a x→a
Then
f(x) f′ (x)
lim = lim ′ ,
x→a g(x) x→a g (x)
if the limit on the right-hand side is finite, ∞, or −∞.

31. Meet the Mathematician
wanted to be a military man, but
poor eyesight forced him into
math
did some math on his own
(solved the “brachistocrone
problem”)
paid a s pend to Johann
Bernoulli, who proved this Guillaume François Antoine,
theorem and named it a er him! Marquis de L’Hôpital
(French, 1661–1704)

32. Revisiting the previous examples
Example
sin2 x
lim
x→0 x

33. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim
x→0 x x→0 1

34. Revisiting the previous examples
Example sin x → 0
sin2 x H 2 sin x.cos x
lim = lim
x→0 x x→0 1

35. Revisiting the previous examples
Example sin x → 0
sin2 x H 2 sin x.cos x
lim = lim =0
x→0 x x→0 1

36. Revisiting the previous examples
Example
sin2 x
lim
x→0 sin x2

37. Revisiting the previous examples
Example
numerator → 0
sin2 x.
lim
x→0 sin x2

38. Revisiting the previous examples
Example
numerator → 0
sin2 x.
lim .
x→0 sin x2
denominator → 0

39. Revisiting the previous examples
Example
numerator → 0
sin2 x. H sin x cos x
2
lim lim
. = x→0 (cos x2 ) (2x)
x→0 sin x2
denominator → 0

40. Revisiting the previous examples
Example
numerator → 0
sin2 x H .
sin x cos x
2
lim = lim
x→0 sin x2 x→0 (cos x2 ) (2x)

41. Revisiting the previous examples
Example
numerator → 0
sin2 x H .
sin x cos x
2
lim = lim .
x→0 sin x2 x→0 (cos x2 ) (2x )
denominator → 0

42. Revisiting the previous examples
Example
numerator → 0
sin2 x H . cos2 x − sin2 x
sin x cos x H
2
lim = lim lim
. = x→0 cos x2 − 2x2 sin(x2 )
x→0 sin x2 x→0 (cos x2 ) (2x )
denominator → 0

43. Revisiting the previous examples
Example
numerator → 1
sin2 x H sin x cos x H
2 cos2 x − sin2 x.
lim = lim = lim
x→0 sin x2 x→0 (cos x2 ) (2x)
x→0 cos x2 − 2x2 sin(x2 )

44. Revisiting the previous examples
Example
numerator → 1
sin2 x H sin x cos x H
2 cos2 x − sin2 x.
lim = lim = lim .
x→0 sin x2 x→0 (cos x2 ) (2x)
x→0 cos x2 − 2x2 sin(x2 )
denominator → 1

45. Revisiting the previous examples
Example
sin2 x H sin x cos x H
2 cos2 x − sin2 x
lim = lim = lim =1
x→0 sin x2 x→0 (cos x2 ) (2x)
x→0 cos x2 − 2x2 sin(x2 )

46. Revisiting the previous examples
Example
sin 3x
lim
x→0 sin x

47. Revisiting the previous examples
Example
sin 3x H 3 cos 3x
lim = lim = 3.
x→0 sin x x→0 cos x

48. Another Example
Example
Find
x
lim
x→0 cos x

49. Beware of Red Herrings
Example
Find
x
lim
x→0 cos x
Solu on
The limit of the denominator is 1, not 0, so L’Hôpital’s rule does not
apply. The limit is 0.

50. Outline
L’Hôpital’s Rule
Rela ve Rates of Growth
Other Indeterminate Limits
Indeterminate Products
Indeterminate Differences
Indeterminate Powers

51. Limits of Rational Functions
revisited
Example
5x2 + 3x − 1
Find lim 2 if it exists.
x→∞ 3x + 7x + 27

52. Limits of Rational Functions
revisited
Example
5x2 + 3x − 1
Find lim 2 if it exists.
x→∞ 3x + 7x + 27
Solu on
Using L’Hôpital:
5x2 + 3x − 1
lim
x→∞ 3x2 + 7x + 27

53. Limits of Rational Functions
revisited
Example
5x2 + 3x − 1
Find lim 2 if it exists.
x→∞ 3x + 7x + 27
Solu on
Using L’Hôpital:
5x2 + 3x − 1 H 10x + 3
lim 2 = lim
x→∞ 3x + 7x + 27 x→∞ 6x + 7

54. Limits of Rational Functions
revisited
Example
5x2 + 3x − 1
Find lim 2 if it exists.
x→∞ 3x + 7x + 27
Solu on
Using L’Hôpital:
5x2 + 3x − 1 H 10x + 3 H 10 5
lim 2 = lim = lim =
x→∞ 3x + 7x + 27 x→∞ 6x + 7 x→∞ 6 3

55. Limits of Rational Functions
revisited
Example
5x2 + 3x − 1
Find lim 2 if it exists.
x→∞ 3x + 7x + 27
Solu on
Using L’Hôpital:
5x2 + 3x − 1 H 10x + 3 H 10 5
lim 2 = lim = lim =
x→∞ 3x + 7x + 27 x→∞ 6x + 7 x→∞ 6 3

56. Limits of Rational Functions
revisited II
Example
5x2 + 3x − 1
Find lim if it exists.
x→∞ 7x + 27

57. Limits of Rational Functions
revisited II
Example
5x2 + 3x − 1
Find lim if it exists.
x→∞ 7x + 27
Solu on
Using L’Hôpital:
5x2 + 3x − 1
lim
x→∞ 7x + 27

58. Limits of Rational Functions
revisited II
Example
5x2 + 3x − 1
Find lim if it exists.
x→∞ 7x + 27
Solu on
Using L’Hôpital:
5x2 + 3x − 1 H 10x + 3
lim = lim
x→∞ 7x + 27 x→∞ 7

59. Limits of Rational Functions
revisited II
Example
5x2 + 3x − 1
Find lim if it exists.
x→∞ 7x + 27
Solu on
Using L’Hôpital:
5x2 + 3x − 1 H 10x + 3
lim = lim =∞
x→∞ 7x + 27 x→∞ 7

60. Limits of Rational Functions
revisited III
Example
4x + 7
Find lim if it exists.
x→∞ 3x2 + 7x + 27

61. Limits of Rational Functions
revisited III
Example
4x + 7
Find lim if it exists.
x→∞ 3x2 + 7x + 27
Solu on
Using L’Hôpital:
4x + 7
lim
x→∞ 3x2 + 7x + 27

62. Limits of Rational Functions
revisited III
Example
4x + 7
Find lim if it exists.
x→∞ 3x2 + 7x + 27
Solu on
Using L’Hôpital:
4x + 7 H 4
lim = lim
x→∞ 3x2 + 7x + 27 x→∞ 6x + 7

63. Limits of Rational Functions
revisited III
Example
4x + 7
Find lim if it exists.
x→∞ 3x2 + 7x + 27
Solu on
Using L’Hôpital:
4x + 7 H 4
lim = lim =0
x→∞ 3x2 + 7x + 27 x→∞ 6x + 7

64. Limits of Rational Functions
Fact
Let f(x) and g(x) be polynomials of degree p and q.
f(x)
If p q, then lim =∞
x→∞ g(x)
f(x)
If p q, then lim =0
x→∞ g(x)
f(x)
If p = q, then lim is the ra o of the leading coefficients of
x→∞ g(x)
f and g.

65. Exponential vs. geometric growth
Example
ex
Find lim , if it exists.
x→∞ x2

66. Exponential vs. geometric growth
Example
ex
Find lim , if it exists.
x→∞ x2
Solu on
We have
ex H ex H ex
lim = lim = lim = ∞.
x→∞ x2 x→∞ 2x x→∞ 2

67. Exponential vs. geometric growth
Example
ex
What about lim ?
x→∞ x3

68. Exponential vs. geometric growth
Example
ex
What about lim ?
x→∞ x3
Answer
S ll ∞. (Why?)

69. Exponential vs. geometric growth
Example
ex
What about lim ?
x→∞ x3
Answer
S ll ∞. (Why?)
Solu on
ex H ex H ex H ex
lim = lim 2 = lim = lim = ∞.
x→∞ x3 x→∞ 3x x→∞ 6x x→∞ 6

70. Exponential vs. fractional powers
Example
ex
Find lim √ , if it exists.
x→∞ x

71. Exponential vs. fractional powers
Example
ex
Find lim √ , if it exists.
x→∞ x
Solu on (without L’Hôpital)
We have for all x 1, x1/2 x1 , so
ex ex
x1/2 x
The right hand side tends to ∞, so the le -hand side must, too.

72. Exponential vs. fractional powers
Example
ex
Find lim √ , if it exists.
x→∞ x
Solu on (with L’Hôpital)
ex ex √
lim √ = lim 1 −1/2 = lim 2 xex = ∞
x→∞ x x→∞ 2 x x→∞

73. Exponential vs. any power
Theorem
ex
Let r be any posi ve number. Then lim r = ∞.
x→∞ x

74. Exponential vs. any power
Theorem
ex
Let r be any posi ve number. Then lim r = ∞.
x→∞ x
Proof.
If r is a posi ve integer, then apply L’Hôpital’s rule r mes to the frac-
on. You get
ex H H ex
lim = . . . = lim = ∞.
x→∞ xr x→∞ r!

75. Exponential vs. any power
Theorem
ex
Let r be any posi ve number. Then lim r = ∞.
x→∞ x
Proof.
If r is not an integer, let m be the smallest integer greater than r. Then
ex ex
if x 1, x x , so r m . The right-hand side tends to ∞ by the
r m
x x
previous step.

76. Any exponential vs. any power
Theorem
ax
Let a 1 and r 0. Then lim r = ∞.
x→∞ x

77. Any exponential vs. any power
Theorem
ax
Let a 1 and r 0. Then lim r = ∞.
x→∞ x
Proof.
If r is a posi ve integer, we have
ax H H (ln a)r ax
lim = . . . = lim = ∞.
x→∞ xr x→∞ r!
If r isn’t an integer, we can compare it as before.

78. Any exponential vs. any power
Theorem
ax
Let a 1 and r 0. Then lim r = ∞.
x→∞ x
Proof.
If r is a posi ve integer, we have
ax H H (ln a)r ax
lim = . . . = lim = ∞.
x→∞ xr x→∞ r!
If r isn’t an integer, we can compare it as before.
(1.00000001)x
So even lim = ∞!
x→∞ x100000000

79. Logarithmic versus power growth
Theorem
ln x
Let r be any posi ve number. Then lim = 0.
x→∞ xr

80. Logarithmic versus power growth
Theorem
ln x
Let r be any posi ve number. Then lim = 0.
x→∞ xr
Proof.
One applica on of L’Hôpital’s Rule here suffices:
ln x H 1/x 1
lim = lim r−1 = lim r = 0.
x→∞ xr x→∞ rx x→∞ rx

81. Outline
L’Hôpital’s Rule
Rela ve Rates of Growth
Other Indeterminate Limits
Indeterminate Products
Indeterminate Differences
Indeterminate Powers

82. Indeterminate products
Example
√
Find lim+ x ln x
x→0
This limit is of the form 0 · (−∞).

83. Indeterminate products
Example
√
Find lim+ x ln x
x→0
This limit is of the form 0 · (−∞).
Solu on
Jury-rig the expression to make an indeterminate quo ent. Then
apply L’Hôpital’s Rule:
√
lim x ln x
x→0+

84. Indeterminate products
Example
√
Find lim+ x ln x
x→0
This limit is of the form 0 · (−∞).
Solu on
Jury-rig the expression to make an indeterminate quo ent. Then
apply L’Hôpital’s Rule:
√ ln x
lim x ln x = lim+ 1 √
x→0+ x→0 / x

85. Indeterminate products
Example
√
Find lim+ x ln x
x→0
This limit is of the form 0 · (−∞).
Solu on
Jury-rig the expression to make an indeterminate quo ent. Then
apply L’Hôpital’s Rule:
√ ln x H x−1
lim x ln x = lim+ 1 √ = lim+ 1 −3/2
x→0+ x→0 / x x→0 − 2 x

86. Indeterminate products
Example
√
Find lim+ x ln x
x→0
This limit is of the form 0 · (−∞).
Solu on
Jury-rig the expression to make an indeterminate quo ent. Then
apply L’Hôpital’s Rule:
√ ln x H x−1 √
lim x ln x = lim+ 1 √ = lim+ 1 −3/2 = lim+ −2 x
x→0+ x→0 / x x→0 − 2 x x→0

87. Indeterminate products
Example
√
Find lim+ x ln x
x→0
This limit is of the form 0 · (−∞).
Solu on
Jury-rig the expression to make an indeterminate quo ent. Then
apply L’Hôpital’s Rule:
√ ln x H x−1 √
lim x ln x = lim+ 1 √ = lim+ 1 −3/2 = lim+ −2 x = 0
x→0+ x→0 / x x→0 − 2 x x→0

88. Indeterminate differences
Example
( )
1
lim − cot 2x
x→0+ x
This limit is of the form ∞ − ∞.

89. Indeterminate Differences
Solu on
Again, rig it to make an indeterminate quo ent.
( )
1 sin(2x) − x cos(2x)
lim+ − cot 2x = lim+
x→0 x x→0 x sin(2x)

90. Indeterminate Differences
Solu on
Again, rig it to make an indeterminate quo ent.
( )
1 sin(2x) − x cos(2x)
lim+ − cot 2x = lim+
x→0 x x→0 x sin(2x)
H cos(2x) + 2x sin(2x)
= lim+
x→0 2x cos(2x) + sin(2x)

91. Indeterminate Differences
Solu on
Again, rig it to make an indeterminate quo ent.
( )
1 sin(2x) − x cos(2x)
lim+ − cot 2x = lim+
x→0 x x→0 x sin(2x)
H cos(2x) + 2x sin(2x)
= lim+
x→0 2x cos(2x) + sin(2x)
=∞

92. Indeterminate Differences
Solu on
Again, rig it to make an indeterminate quo ent.
( )
1 sin(2x) − x cos(2x)
lim+ − cot 2x = lim+
x→0 x x→0 x sin(2x)
H cos(2x) + 2x sin(2x)
= lim+
x→0 2x cos(2x) + sin(2x)
=∞
The limit is +∞ because the numerator tends to 1 while the
denominator tends to zero but remains posi ve.

93. Checking your work
This all goes in the thought cloud
tan 2x 1
lim = 1, so for small x, tan 2x ≈ 2x. So cot 2x ≈ and
x→0 2x 2x
1 1 1 1
− cot 2x ≈ − = →∞
x x 2x 2x
as x → 0+ .

94. Indeterminate powers
Example
Find lim+ (1 − 2x)1/x
x→0

95. Indeterminate powers
Example
Find lim+ (1 − 2x)1/x
x→0
Solu on
Take the logarithm:
( ) ( ) ln(1 − 2x)
ln lim+ (1 − 2x)1/x
= lim+ ln (1 − 2x)1/x
= lim+
x→0 x→0 x→0 x

96. Indeterminate powers
Example
Find lim+ (1 − 2x)1/x
x→0
Solu on
0
This limit is of the form , so we can use L’Hôpital:
0
−2
ln(1 − 2x) H
lim+ = lim+ 1−2x = −2
x→0 x x→0 1

97. Indeterminate powers
Example
Find lim+ (1 − 2x)1/x
x→0
Solu on
0
This limit is of the form , so we can use L’Hôpital:
0
−2
ln(1 − 2x) H
lim+ = lim+ 1−2x = −2
x→0 x x→0 1
This is not the answer, it’s the log of the answer! So the answer we
want is e−2 .

98. Another indeterminate power limit
Example
Find lim+ (3x)4x
x→0

99. Another indeterminate power limit
Example
Find lim+ (3x)4x
x→0
Solu on
ln(3x)
ln lim+ (3x)4x = lim+ ln(3x)4x = lim+ 4x ln(3x) = lim+ 1/4x
x→0 x→0 x→0 x→0
H 3/3x
= lim+ −1 2 = lim+ (−4x) = 0
x→0 /4x x→0
So the answer is e0 = 1.

100. Summary
Form Method
0
0 L’Hôpital’s rule directly
∞
∞ L’Hôpital’s rule directly
∞
0·∞ jiggle to make 0 or
0 ∞.
∞ − ∞ combine to make an indeterminate product or quo ent
00 take ln to make an indeterminate product
∞0 di o
1∞ di o

101. Final Thoughts
L’Hôpital’s Rule only works on indeterminate quo ents
Luckily, most indeterminate limits can be transformed into
indeterminate quo ents
L’Hôpital’s Rule gives wrong answers for non-indeterminate
limits!