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

1. Section 3.7
Indeterminate Forms and L’Hôpital’s
Rule
V63.0121.041, Calculus I
New York University
November 3, 2010
Announcements
Quiz 3 in recitation this week on Sections 2.6, 2.8, 3.1, and 3.2
. . . . . .

2. . . . . . .
Announcements
Quiz 3 in recitation this
week on Sections 2.6, 2.8,
3.1, and 3.2
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 2 / 34

3. . . . . . .
Objectives
Know when a limit is of
indeterminate form:
indeterminate quotients:
0/0, ∞/∞
indeterminate products:
0 × ∞
indeterminate
differences: ∞ − ∞
indeterminate powers:
00
, ∞0
, and 1∞
Resolve limits in
indeterminate form
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 3 / 34

4. . . . . . .
Experiments with funny limits
lim
x→0
sin2
x
x
.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34

5. . . . . . .
Experiments with funny limits
lim
x→0
sin2
x
x
= 0
.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34

6. . . . . . .
Experiments with funny limits
lim
x→0
sin2
x
x
= 0
lim
x→0
x
sin2
x
.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34

7. . . . . . .
Experiments with funny limits
lim
x→0
sin2
x
x
= 0
lim
x→0
x
sin2
x
does not exist
.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34

8. . . . . . .
Experiments with funny limits
lim
x→0
sin2
x
x
= 0
lim
x→0
x
sin2
x
does not exist
lim
x→0
sin2
x
sin(x2)
.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34

9. . . . . . .
Experiments with funny limits
lim
x→0
sin2
x
x
= 0
lim
x→0
x
sin2
x
does not exist
lim
x→0
sin2
x
sin(x2)
= 1
.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34

10. . . . . . .
Experiments with funny limits
lim
x→0
sin2
x
x
= 0
lim
x→0
x
sin2
x
does not exist
lim
x→0
sin2
x
sin(x2)
= 1
lim
x→0
sin 3x
sin x
.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34

11. . . . . . .
Experiments with funny limits
lim
x→0
sin2
x
x
= 0
lim
x→0
x
sin2
x
does not exist
lim
x→0
sin2
x
sin(x2)
= 1
lim
x→0
sin 3x
sin x
= 3
.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34

12. . . . . . .
Experiments with funny limits
lim
x→0
sin2
x
x
= 0
lim
x→0
x
sin2
x
does not exist
lim
x→0
sin2
x
sin(x2)
= 1
lim
x→0
sin 3x
sin x
= 3
.
All of these are of the form
0
0
, and since we can get different answers
in different cases, we say this form is indeterminate.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34

13. . . . . . .
Recall
Recall the limit laws from Chapter 2.
Limit of a sum is the sum of the limits
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 5 / 34

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
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 5 / 34

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
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 5 / 34

16. . . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 5 / 34

17. . . . . . .
More about dividing limits
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:
lim
x→0+
1
x
= +∞ lim
x→0−
cos x
x3
= −∞
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 6 / 34

18. . . . . . .
More about dividing limits
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:
lim
x→0+
1
x
= +∞ lim
x→0−
cos x
x3
= −∞
An exception would be something like
lim
x→∞
1
1
x sin x
= lim
x→∞
x csc x.
which does not exist and is not infinite.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 6 / 34

19. . . . . . .
More about dividing limits
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:
lim
x→0+
1
x
= +∞ lim
x→0−
cos x
x3
= −∞
An exception would be something like
lim
x→∞
1
1
x sin x
= lim
x→∞
x csc x.
which does not exist and is not infinite.
Even less predictable: numerator and denominator both go to
zero.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 6 / 34

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
case “is”
0
0
, and therefore
nonexistent because this
expression is undefined.
The limit is of the form
0
0
,
which means we cannot
evaluate it with our limit
laws.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 7 / 34

21. . . . . . .
Indeterminate forms are like Tug Of War
Which side wins depends on which side is stronger.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 8 / 34

22. . . . . . .
Outline
L’Hôpital’s Rule
Relative Rates of Growth
Other Indeterminate Limits
Indeterminate Products
Indeterminate Differences
Indeterminate Powers
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 9 / 34

23. . . . . . .
The Linear Case
Question
If f and g are lines and f(a) = g(a) = 0, what is
lim
x→a
f(x)
g(x)
?
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 10 / 34

24. . . . . . .
The Linear Case
Question
If f and g are lines and f(a) = g(a) = 0, what is
lim
x→a
f(x)
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)
g(x)
=
m1
m2
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 10 / 34

25. . . . . . .
The Linear Case, Illustrated
. .x
.y
.y = f(x)
.y = g(x)
.
.a
.
.x
.f(x)
.g(x)
f(x)
g(x)
=
f(x) − f(a)
g(x) − g(a)
=
(f(x) − f(a))/(x − a)
(g(x) − g(a))/(x − a)
=
m1
m2
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 11 / 34

26. . . . . . .
What then?
But what if the functions aren’t linear?
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 12 / 34

27. . . . . . .
What then?
But what if the functions aren’t linear?
Can we approximate a function near a point with a linear function?
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 12 / 34

28. . . . . . .
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?
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 12 / 34

29. . . . . . .
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? The derivative!
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 12 / 34

30. . . . . . .
Theorem of the Day
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
x→a
f(x) = 0 and lim
x→a
g(x) = 0
or
lim
x→a
f(x) = ±∞ and lim
x→a
g(x) = ±∞
Then
lim
x→a
f(x)
g(x)
= lim
x→a
f′
(x)
g′(x)
,
if the limit on the right-hand side is finite, ∞, or −∞.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 13 / 34

31. . . . . . .
Meet the Mathematician: L'H_pital
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
(French, 1661–1704)
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 14 / 34

32. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

33. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x cos x
1
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

34. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x
.
.sin x → 0
cos x
1
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

35. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x
.
.sin x → 0
cos x
1
= 0
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

36. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x
.
.sin x → 0
cos x
1
= 0
Example
lim
x→0
sin2
x
sin x2
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

37. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x cos x
1
= 0
Example
lim
x→0
sin2
x
.
.numerator → 0
sin x2
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

38. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x cos x
1
= 0
Example
lim
x→0
sin2
x
.
.numerator → 0
sin x2.
.denominator → 0
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

39. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x cos x
1
= 0
Example
lim
x→0
sin2
x
.
.numerator → 0
sin x2.
.denominator → 0
H
= lim
x→0
2 sin x cos x
(
cos x2
)
(2x)
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

40. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x cos x
1
= 0
Example
lim
x→0
sin2
x
sin x2
H
= lim
x→0
2 sin x cos x
.
.numerator → 0
(
cos x2
)
(2x)
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

41. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x cos x
1
= 0
Example
lim
x→0
sin2
x
sin x2
H
= lim
x→0
2 sin x cos x
.
.numerator → 0
(
cos x2
)
(2x
.
.denominator → 0
)
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

42. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x cos x
1
= 0
Example
lim
x→0
sin2
x
sin x2
H
= lim
x→0
2 sin x cos x
.
.numerator → 0
(
cos x2
)
(2x
.
.denominator → 0
)
H
= lim
x→0
cos2 x − sin2
x
cos x2 − 2x2 sin(x2)
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

43. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x cos x
1
= 0
Example
lim
x→0
sin2
x
sin x2
H
= lim
x→0
2 sin x cos x
(
cos x2
)
(2x)
H
= lim
x→0
cos2 x − sin2
x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

44. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x cos x
1
= 0
Example
lim
x→0
sin2
x
sin x2
H
= lim
x→0
2 sin x cos x
(
cos x2
)
(2x)
H
= lim
x→0
cos2 x − sin2
x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

45. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x cos x
1
= 0
Example
lim
x→0
sin2
x
sin x2
H
= lim
x→0
2 sin x cos x
(
cos x2
)
(2x)
H
= lim
x→0
cos2 x − sin2
x
cos x2 − 2x2 sin(x2)
= 1
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

46. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x cos x
1
= 0
Example
lim
x→0
sin2
x
sin x2
H
= lim
x→0
2 sin x cos x
(
cos x2
)
(2x)
H
= lim
x→0
cos2 x − sin2
x
cos x2 − 2x2 sin(x2)
= 1
Example
lim
x→0
sin 3x
sin x
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

47. . . . . . .
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x cos x
1
= 0
Example
lim
x→0
sin2
x
sin x2
H
= lim
x→0
2 sin x cos x
(
cos x2
)
(2x)
H
= lim
x→0
cos2 x − sin2
x
cos x2 − 2x2 sin(x2)
= 1
Example
lim
x→0
sin 3x
sin x
H
= lim
x→0
3 cos 3x
cos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

48. . . . . . .
Another Example
Example
Find
lim
x→0
x
cos x
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 16 / 34

49. . . . . . .
Beware of Red Herrings
Example
Find
lim
x→0
x
cos x
Solution
The limit of the denominator is 1, not 0, so L’Hôpital’s rule does not
apply. The limit is 0.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 16 / 34

50. . . . . . .
Outline
L’Hôpital’s Rule
Relative Rates of Growth
Other Indeterminate Limits
Indeterminate Products
Indeterminate Differences
Indeterminate Powers
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 17 / 34

51. . . . . . .
Limits of Rational Functions revisited
Example
Find lim
x→∞
5x2 + 3x − 1
3x2 + 7x + 27
if it exists.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34

52. . . . . . .
Limits of Rational Functions revisited
Example
Find lim
x→∞
5x2 + 3x − 1
3x2 + 7x + 27
if it exists.
Solution
Using L’Hôpital:
lim
x→∞
5x2 + 3x − 1
3x2 + 7x + 27
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34

53. . . . . . .
Limits of Rational Functions revisited
Example
Find lim
x→∞
5x2 + 3x − 1
3x2 + 7x + 27
if it exists.
Solution
Using L’Hôpital:
lim
x→∞
5x2 + 3x − 1
3x2 + 7x + 27
H
= lim
x→∞
10x + 3
6x + 7
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34

54. . . . . . .
Limits of Rational Functions revisited
Example
Find lim
x→∞
5x2 + 3x − 1
3x2 + 7x + 27
if it exists.
Solution
Using L’Hôpital:
lim
x→∞
5x2 + 3x − 1
3x2 + 7x + 27
H
= lim
x→∞
10x + 3
6x + 7
H
= lim
x→∞
10
6
=
5
3
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34

55. . . . . . .
Limits of Rational Functions revisited
Example
Find lim
x→∞
5x2 + 3x − 1
3x2 + 7x + 27
if it exists.
Solution
Using L’Hôpital:
lim
x→∞
5x2 + 3x − 1
3x2 + 7x + 27
H
= lim
x→∞
10x + 3
6x + 7
H
= lim
x→∞
10
6
=
5
3
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34

56. . . . . . .
Limits of Rational Functions revisited II
Example
Find lim
x→∞
5x2 + 3x − 1
7x + 27
if it exists.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 19 / 34

57. . . . . . .
Limits of Rational Functions revisited II
Example
Find lim
x→∞
5x2 + 3x − 1
7x + 27
if it exists.
Solution
Using L’Hôpital:
lim
x→∞
5x2 + 3x − 1
7x + 27
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 19 / 34

58. . . . . . .
Limits of Rational Functions revisited II
Example
Find lim
x→∞
5x2 + 3x − 1
7x + 27
if it exists.
Solution
Using L’Hôpital:
lim
x→∞
5x2 + 3x − 1
7x + 27
H
= lim
x→∞
10x + 3
7
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 19 / 34

59. . . . . . .
Limits of Rational Functions revisited II
Example
Find lim
x→∞
5x2 + 3x − 1
7x + 27
if it exists.
Solution
Using L’Hôpital:
lim
x→∞
5x2 + 3x − 1
7x + 27
H
= lim
x→∞
10x + 3
7
= ∞
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 19 / 34

60. . . . . . .
Limits of Rational Functions revisited III
Example
Find lim
x→∞
4x + 7
3x2 + 7x + 27
if it exists.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 20 / 34

61. . . . . . .
Limits of Rational Functions revisited III
Example
Find lim
x→∞
4x + 7
3x2 + 7x + 27
if it exists.
Solution
Using L’Hôpital:
lim
x→∞
4x + 7
3x2 + 7x + 27
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 20 / 34

62. . . . . . .
Limits of Rational Functions revisited III
Example
Find lim
x→∞
4x + 7
3x2 + 7x + 27
if it exists.
Solution
Using L’Hôpital:
lim
x→∞
4x + 7
3x2 + 7x + 27
H
= lim
x→∞
4
6x + 7
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 20 / 34

63. . . . . . .
Limits of Rational Functions revisited III
Example
Find lim
x→∞
4x + 7
3x2 + 7x + 27
if it exists.
Solution
Using L’Hôpital:
lim
x→∞
4x + 7
3x2 + 7x + 27
H
= lim
x→∞
4
6x + 7
= 0
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 20 / 34

64. . . . . . .
Limits of Rational Functions
Fact
Let f(x) and g(x) be polynomials of degree p and q.
If p q, then lim
x→∞
f(x)
g(x)
= ∞
If p q, then lim
x→∞
f(x)
g(x)
= 0
If p = q, then lim
x→∞
f(x)
g(x)
is the ratio of the leading coefficients of f
and g.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 21 / 34

65. . . . . . .
Exponential versus geometric growth
Example
Find lim
x→∞
ex
x2
, if it exists.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34

66. . . . . . .
Exponential versus geometric growth
Example
Find lim
x→∞
ex
x2
, if it exists.
Solution
We have
lim
x→∞
ex
x2
H
= lim
x→∞
ex
2x
H
= lim
x→∞
ex
2
= ∞.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34

67. . . . . . .
Exponential versus geometric growth
Example
Find lim
x→∞
ex
x2
, if it exists.
Solution
We have
lim
x→∞
ex
x2
H
= lim
x→∞
ex
2x
H
= lim
x→∞
ex
2
= ∞.
Example
What about lim
x→∞
ex
x3
?
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34

68. . . . . . .
Exponential versus geometric growth
Example
Find lim
x→∞
ex
x2
, if it exists.
Solution
We have
lim
x→∞
ex
x2
H
= lim
x→∞
ex
2x
H
= lim
x→∞
ex
2
= ∞.
Example
What about lim
x→∞
ex
x3
?
Answer
Still ∞. (Why?)
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34

69. . . . . . .
Exponential versus fractional powers
Example
Find lim
x→∞
ex
√
x
, if it exists.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 23 / 34

70. . . . . . .
Exponential versus fractional powers
Example
Find lim
x→∞
ex
√
x
, if it exists.
Solution (without L’Hôpital)
We have for all x 1, x1/2
x1
, so
ex
x1/2
ex
x
The right hand side tends to ∞, so the left-hand side must, too.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 23 / 34

71. . . . . . .
Exponential versus fractional powers
Example
Find lim
x→∞
ex
√
x
, if it exists.
Solution (without L’Hôpital)
We have for all x 1, x1/2
x1
, so
ex
x1/2
ex
x
The right hand side tends to ∞, so the left-hand side must, too.
Solution (with L’Hôpital)
lim
x→∞
ex
√
x
= lim
x→∞
ex
1
2 x−1/2
= lim
x→∞
2
√
xex
= ∞
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 23 / 34

72. . . . . . .
Exponential versus any power
Theorem
Let r be any positive number. Then
lim
x→∞
ex
xr
= ∞.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 24 / 34

73. . . . . . .
Exponential versus any power
Theorem
Let r be any positive number. Then
lim
x→∞
ex
xr
= ∞.
Proof.
If r is a positive integer, then apply L’Hôpital’s rule r times to the
fraction. You get
lim
x→∞
ex
xr
H
= . . .
H
= lim
x→∞
ex
r!
= ∞.
If r is not an integer, let m be the smallest integer greater than r. Then
if x 1, xr
xm
, so
ex
xr
ex
xm
. The right-hand side tends to ∞ by the
previous step.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 24 / 34

74. . . . . . .
Any exponential versus any power
Theorem
Let a 1 and r 0. Then
lim
x→∞
ax
xr
= ∞.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 25 / 34

75. . . . . . .
Any exponential versus any power
Theorem
Let a 1 and r 0. Then
lim
x→∞
ax
xr
= ∞.
Proof.
If r is a positive integer, we have
lim
x→∞
ax
xr
H
= . . .
H
= lim
x→∞
(ln a)rax
r!
= ∞.
If r isn’t an integer, we can compare it as before.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 25 / 34

76. . . . . . .
Any exponential versus any power
Theorem
Let a 1 and r 0. Then
lim
x→∞
ax
xr
= ∞.
Proof.
If r is a positive integer, we have
lim
x→∞
ax
xr
H
= . . .
H
= lim
x→∞
(ln a)rax
r!
= ∞.
If r isn’t an integer, we can compare it as before.
So even lim
x→∞
(1.00000001)x
x100000000
= ∞!
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 25 / 34

77. . . . . . .
Logarithmic versus power growth
Theorem
Let r be any positive number. Then
lim
x→∞
ln x
xr
= 0.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 26 / 34

78. . . . . . .
Logarithmic versus power growth
Theorem
Let r be any positive number. Then
lim
x→∞
ln x
xr
= 0.
Proof.
One application of L’Hôpital’s Rule here suffices:
lim
x→∞
ln x
xr
H
= lim
x→∞
1/x
rxr−1
= lim
x→∞
1
rxr
= 0.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 26 / 34

79. . . . . . .
Outline
L’Hôpital’s Rule
Relative Rates of Growth
Other Indeterminate Limits
Indeterminate Products
Indeterminate Differences
Indeterminate Powers
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 27 / 34

80. . . . . . .
Indeterminate products
Example
Find
lim
x→0+
√
x ln x
This limit is of the form 0 · (−∞).
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34

81. . . . . . .
Indeterminate products
Example
Find
lim
x→0+
√
x ln x
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→0+
√
x ln x
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34

82. . . . . . .
Indeterminate products
Example
Find
lim
x→0+
√
x ln x
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→0+
√
x ln x = lim
x→0+
ln x
1/
√
x
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34

83. . . . . . .
Indeterminate products
Example
Find
lim
x→0+
√
x ln x
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→0+
√
x ln x = lim
x→0+
ln x
1/
√
x
H
= lim
x→0+
x−1
−1
2 x−3/2
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34

84. . . . . . .
Indeterminate products
Example
Find
lim
x→0+
√
x ln x
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→0+
√
x ln x = lim
x→0+
ln x
1/
√
x
H
= lim
x→0+
x−1
−1
2 x−3/2
= lim
x→0+
−2
√
x
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34

85. . . . . . .
Indeterminate products
Example
Find
lim
x→0+
√
x ln x
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→0+
√
x ln x = lim
x→0+
ln x
1/
√
x
H
= lim
x→0+
x−1
−1
2 x−3/2
= lim
x→0+
−2
√
x = 0
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34

86. . . . . . .
Indeterminate differences
Example
lim
x→0+
(
1
x
− cot 2x
)
This limit is of the form ∞ − ∞.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 29 / 34

87. . . . . . .
Indeterminate differences
Example
lim
x→0+
(
1
x
− cot 2x
)
This limit is of the form ∞ − ∞.
Solution
Again, rig it to make an indeterminate quotient.
lim
x→0+
sin(2x) − x cos(2x)
x sin(2x)
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 29 / 34

88. . . . . . .
Indeterminate differences
Example
lim
x→0+
(
1
x
− cot 2x
)
This limit is of the form ∞ − ∞.
Solution
Again, rig it to make an indeterminate quotient.
lim
x→0+
sin(2x) − x cos(2x)
x sin(2x)
H
= lim
x→0+
cos(2x) + 2x sin(2x)
2x cos(2x) + sin(2x)
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 29 / 34

89. . . . . . .
Indeterminate differences
Example
lim
x→0+
(
1
x
− cot 2x
)
This limit is of the form ∞ − ∞.
Solution
Again, rig it to make an indeterminate quotient.
lim
x→0+
sin(2x) − x cos(2x)
x sin(2x)
H
= lim
x→0+
cos(2x) + 2x sin(2x)
2x cos(2x) + sin(2x)
= ∞
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 29 / 34

90. . . . . . .
Indeterminate differences
Example
lim
x→0+
(
1
x
− cot 2x
)
This limit is of the form ∞ − ∞.
Solution
Again, rig it to make an indeterminate quotient.
lim
x→0+
sin(2x) − x cos(2x)
x sin(2x)
H
= lim
x→0+
cos(2x) + 2x sin(2x)
2x cos(2x) + sin(2x)
= ∞
The limit is +∞ becuase the numerator tends to 1 while the
denominator tends to zero but remains positive.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 29 / 34

91. . . . . . .
Checking your work
.
.
lim
x→0
tan 2x
2x
= 1, so for small x,
tan 2x ≈ 2x. So cot 2x ≈
1
2x
and
1
x
− cot 2x ≈
1
x
−
1
2x
=
1
2x
→ ∞
as x → 0+
.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 30 / 34

92. . . . . . .
Indeterminate powers
Example
Find lim
x→0+
(1 − 2x)1/x
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34

93. . . . . . .
Indeterminate powers
Example
Find lim
x→0+
(1 − 2x)1/x
Take the logarithm:
ln
(
lim
x→0+
(1 − 2x)1/x
)
= lim
x→0+
ln
(
(1 − 2x)1/x
)
= lim
x→0+
ln(1 − 2x)
x
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34

94. . . . . . .
Indeterminate powers
Example
Find lim
x→0+
(1 − 2x)1/x
Take the logarithm:
ln
(
lim
x→0+
(1 − 2x)1/x
)
= lim
x→0+
ln
(
(1 − 2x)1/x
)
= lim
x→0+
ln(1 − 2x)
x
This limit is of the form
0
0
, so we can use L’Hôpital:
lim
x→0+
ln(1 − 2x)
x
H
= lim
x→0+
−2
1−2x
1
= −2
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34

95. . . . . . .
Indeterminate powers
Example
Find lim
x→0+
(1 − 2x)1/x
Take the logarithm:
ln
(
lim
x→0+
(1 − 2x)1/x
)
= lim
x→0+
ln
(
(1 − 2x)1/x
)
= lim
x→0+
ln(1 − 2x)
x
This limit is of the form
0
0
, so we can use L’Hôpital:
lim
x→0+
ln(1 − 2x)
x
H
= lim
x→0+
−2
1−2x
1
= −2
This is not the answer, it’s the log of the answer! So the answer we
want is e−2
.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34

96. . . . . . .
Another indeterminate power limit
Example
lim
x→0
(3x)4x
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 32 / 34

97. . . . . . .
Another indeterminate power limit
Example
lim
x→0
(3x)4x
Solution
ln lim
x→0+
(3x)4x
= lim
x→0+
ln(3x)4x
= lim
x→0+
4x ln(3x)
= lim
x→0+
ln(3x)
1/4x
H
= lim
x→0+
3/3x
−1/4x2
= lim
x→0+
(−4x) = 0
So the answer is e0
= 1.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 32 / 34

98. . . . . . .
Summary
Form Method
0
0 L’Hôpital’s rule directly
∞
∞ L’Hôpital’s rule directly
0 · ∞ jiggle to make 0
0 or ∞
∞.
∞ − ∞ combine to make an indeterminate product or quotient
00
take ln to make an indeterminate product
∞0 ditto
1∞
ditto
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 33 / 34

99. . . . . . .
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!
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 34 / 34