L'Hôpital's Rule is not a magic bullet (or a sledgehammer) but it does allow us to find limits of indeterminate forms such as 0/0 and ∞/∞. With some algebra we can use it to resolve other indeterminate forms such as ∞-∞ and 0^0.

1. Section 3.7
Indeterminate Forms and L’Hôpital’s
Rule
V63.0121.034, Calculus I
November 2, 2009
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. . . . . .

2. Experiments with funny limits
sin2 x
lim
x→0+ x
.
. . . . . .

3. Experiments with funny limits
sin2 x
lim =0
x→0+ x
.
. . . . . .

4. Experiments with funny limits
sin2 x
lim =0
x→0+ x
x
lim
x→0 sin2 x
.
. . . . . .

5. Experiments with funny limits
sin2 x
lim =0
x→0+ x
x
lim does not exist
x→0 sin2 x
.
. . . . . .

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

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

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

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

10. Experiments with funny limits
sin2 x
lim =0
x→0+ x
x
lim does not exist
x→0 sin2 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.
. . . . . .

11. Outline
Indeterminate Forms
L’Hôpital’s Rule
Relative Rates of Growth
Application to Indeterminate Products
Application to Indeterminate Differences
Application to Indeterminate Powers
Summary
. . . . . .

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

13. 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
. . . . . .

14. 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
. . . . . .

15. 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 quotient is the quotient of the limits ... whoops!
This is true as long as you don’t try to divide by zero.
. . . . . .

16. We know dividing by zero is bad.
Most of the time, if an expression’s numerator approaches a
finite number and denominator approaches zero, the
quotient approaches some kind of infinity. For example:
1 cos x
lim = +∞ lim = −∞
x→0+ x x→0− x3
. . . . . .

17. We know dividing by zero is bad.
Most of the time, if an expression’s numerator approaches a
finite number and denominator approaches zero, the
quotient approaches some kind of infinity. For example:
1 cos x
lim = +∞ lim = −∞
x→0+ x x→0− x3
An exception would be something like
1
lim = lim x csc x.
x→∞ 1 sin x x→∞
x
which doesn’t exist.
. . . . . .

18. We know dividing by zero is bad.
Most of the time, if an expression’s numerator approaches a
finite number and denominator approaches zero, the
quotient approaches some kind of infinity. For example:
1 cos x
lim = +∞ lim = −∞
x→0+ x x→0− x3
An exception would be something like
1
lim = lim x csc x.
x→∞ 1 sin x x→∞
x
which doesn’t exist.
Even less predictable: numerator and denominator both go
to zero.
. . . . . .

19. 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
0
in each case “is” , and
0
therefore nonexistent
because this expression
is undefined.
The limit is of the form
0
, which means we
0
cannot evaluate it with
our limit laws.
. . . . . .

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

21. Outline
Indeterminate Forms
L’Hôpital’s Rule
Relative Rates of Growth
Application to Indeterminate Products
Application to Indeterminate Differences
Application to Indeterminate Powers
Summary
. . . . . .

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

23. The Linear Case
Question
If f and g are lines and f(a) = g(a) = 0, what is
f(x)
lim ?
x→a g(x)
Solution
The functions f and g can be written in the form
f(x) = m1 (x − a)
g(x) = m2 (x − a)
So
f (x ) m1
=
g (x ) m2
. . . . . .

24. The Linear Case, Illustrated
y
. y
. = g(x)
y
. = f(x)
g
. (x)
a
. f
.(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
. . . . . .

25. What then?
But what if the functions aren’t linear?
. . . . . .

26. What then?
But what if the functions aren’t linear?
Can we approximate a function near a point with a linear
function?
. . . . . .

27. What then?
But what if the functions aren’t linear?
Can we approximate a function near a point with a linear
function?
What would be the slope of that linear function?
. . . . . .

28. Theorem (L’Hopital’s Rule)
Suppose f and g are differentiable functions 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 −∞.
. . . . . .

29. Meet the Mathematician: L’Hôpital
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 stipend to Johann
Bernoulli, who proved
this theorem and named
it after him! Guillaume François Antoine,
Marquis de L’Hôpital
(1661–1704)
. . . . . .

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

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

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

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

34. Revisiting the previous examples
Example . in x → 0
s
sin2 x H 2 sin x.cos x
lim = lim =0
x→0 x x→0 1
Example
sin2 x
lim
x→0 sin x2
. . . . . .

35. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example
. umerator → 0
n
sin2 x.
lim
x→0 sin x2
. . . . . .

36. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example
. umerator → 0
n
sin2 x.
lim .
x→0 sin x2
. enominator → 0
d
. . . . . .

37. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example
. umerator → 0
n
sin2 x. H sin x cos x
2
lim 2.
= lim
x→0 sin x x→0 (cos x2 ) (x )
2
. enominator → 0
d
. . . . . .

38. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example
. umerator → 0
n
sin2 x H .
sin x cos x
2
lim = lim
x→0 sin x2 x→0 (cos x2 ) (x )
2
. . . . . .

39. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example
. umerator → 0
n
sin2 x H .
sin x cos x
2
lim = lim .)
x→0 sin x2 x→0 (cos x2 ) (x
2
. enominator → 0
d
. . . . . .

40. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example
. umerator → 0
n
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 ) (x
2
. enominator → 0
d
. . . . . .

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

42. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example
. umerator → 1
n
sin2 x H sin x cos x H
2 cos2 x − sin2 x.
lim = lim = lim .
x→0 sin x2 x→0 (cos x2 ) (x )
2 x→0 cos x2 − 2x2 sin(x2 )
. enominator → 1
d
. . . . . .

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

44. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
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 ) (x )
2 x→0 cos x2 − 2x2 sin(x2 )
Example
sin 3x H 3 cos 3x
lim = lim = 3.
x→0 sin x x→0 cos x
. . . . . .

45. Example
Find
x
lim
x→0 cos x
. . . . . .

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

47. Theorem
Let r be any positive number. Then
ex
lim = ∞.
x→∞ xr
. . . . . .

48. Theorem
Let r be any positive number. Then
ex
lim = ∞.
x→∞ xr
Proof.
If r is a positive integer, then apply L’Hôpital’s rule r times to the
fraction. You get
ex H H ex
lim = . . . = lim = ∞.
x→∞ xr x→∞ r!
For example, if r = 3, three invocations of L’Hôpital’s Rule give us
ex H ex H ex H ex
lim = lim = lim = lim =∞
x→∞ x3 x→∞ 3 · x2 x→∞ 2 · 3x x→∞ 1 · 2 · 3
. . . . . .

49. If r is not an integer, let m be the smallest integer greater than r.
Then if x 1, xr xm , so
ex ex
m.
xr x
The right-hand side tends to ∞ by the previous step.
. . . . . .

50. If r is not an integer, let m be the smallest integer greater than r.
Then if x 1, xr xm , so
ex ex
m.
xr x
The right-hand side tends to ∞ by the previous step. For
example, if r = 1/2, r 1 so for x 1
ex ex
x1/2 x
which gets arbitrarily large.
. . . . . .

51. Theorem
Let r be any positive number. Then
ln x
lim = 0.
x→∞ xr
. . . . . .

52. Theorem
Let r be any positive number. Then
ln x
lim = 0.
x→∞ xr
Proof.
One application 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
. . . . . .

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

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

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

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

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

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

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

60. Indeterminate differences
Example
( )
1
lim − cot 2x
x→0 x
This limit is of the form ∞ − ∞.
Solution
Again, rig it to make an indeterminate quotient.
sin(2x) − x cos(2x)
lim
x→0+ x sin(2x)
. . . . . .

61. Indeterminate differences
Example
( )
1
lim − cot 2x
x→0 x
This limit is of the form ∞ − ∞.
Solution
Again, rig it to make an indeterminate quotient.
sin(2x) − x cos(2x) H cos(2x) + 2x sin(2x)
lim = lim+
x→0+ x sin(2x) x→0 2x cos(2x) + sin(2x)
. . . . . .

62. Indeterminate differences
Example
( )
1
lim − cot 2x
x→0 x
This limit is of the form ∞ − ∞.
Solution
Again, rig it to make an indeterminate quotient.
sin(2x) − x cos(2x) H cos(2x) + 2x sin(2x)
lim = lim+
x→0+ x sin(2x) x→0 2x cos(2x) + sin(2x)
=∞
. . . . . .

63. Indeterminate differences
Example
( )
1
lim − cot 2x
x→0 x
This limit is of the form ∞ − ∞.
Solution
Again, rig it to make an indeterminate quotient.
sin(2x) − x cos(2x) H cos(2x) + 2x sin(2x)
lim = lim+
x→0+ x sin(2x) x→0 2x cos(2x) + sin(2x)
=∞
The limit is +∞ becuase the numerator tends to 1 while the
denominator tends to zero but remains positive.
. . . . . .

64. Checking your work
.
tan 2x
lim = 1, so for small x,
x→0 2x
1
tan 2x ≈ 2x. So cot 2x ≈ and
. 2x
1 1 1 1
− cot 2x ≈ − = →∞
x x 2x 2x
as x → 0+ .
. . . . . .

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

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

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

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

69. Example
lim (3x)4x
x→0
. . . . . .

70. Example
lim (3x)4x
x→0
Solution
ln lim (3x)4x = lim ln(3x)4x = lim 4x ln(3x)
x→0+ x→0+ x→0+
ln(3x) H 3/3x
= lim+ = lim+
x→0 1/4x x→0 −1/4x2
= lim (−4x) = 0
x→0+
So the answer is e0 = 1.
. . . . . .

71. Summary
Form Method
0
0 L’Hôpital’s rule directly
∞
∞ L’Hôpital’s rule directly
0 ∞
0·∞ jiggle to make 0 or ∞.
∞−∞ factor to make an indeterminate product
00 take ln to make an indeterminate product
∞0 ditto
1∞ ditto
. . . . . .

72. Final thoughts
L’Hôpital’s Rule only works on indeterminate quotients
Luckily, most indeterminate limits can be transformed into
indeterminate quotients
L’Hôpital’s Rule gives wrong answers for non-indeterminate
limits!
. . . . . .