This document outlines section 3.7 on indeterminate forms and L'Hôpital's Rule from a Calculus I course at NYU, dated November 4, 2010. It discusses concepts such as indeterminate quotients, products, differences, and powers, providing objectives and recall points related to limit laws and how to resolve limits that are indeterminate. The section includes theoretical explanations, examples of limits, and the related theorem of L'Hôpital, providing a foundation for understanding and applying these calculus principles.

1. Section 3.7
Indeterminate Forms and L’Hôpital’s
Rule
V63.0121.021, Calculus I
New York University
November 4, 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 2 / 35

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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 3 / 35

4. Recall
Recall the limit laws from Chapter 2.
Limit of a sum is the sum of the limits
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 4 / 35

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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 4 / 35

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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 4 / 35

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 quotient is the quotient of the limits ... whoops! This is
true as long as you don’t try to divide by zero.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 4 / 35

8. More about dividing limits
We know dividing by zero is bad.
Most of the time, if an expression’s numerator approaches a finite
nonzero number and denominator approaches zero, the quotient
has an infinite. For example:
1 cos x
lim+ = +∞ lim = −∞
x→0 x x→0 − x3
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 5 / 35

9. Why 1/0 ̸= ∞
1
Consider the function 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 not
x→∞
infinite.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 6 / 35

10. Why 1/0 ̸= ∞
1
Consider the function 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 not
x→∞
infinite.
Even less predictable: when numerator and denominator both go to
zero.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 6 / 35

11. Experiments with funny limits
sin2 x
lim
x→0 x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 7 / 35

12. Experiments with funny limits
sin2 x
lim =0
x→0 x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 7 / 35

13. Experiments with funny limits
sin2 x
lim =0
x→0 x
x
lim
x→0 sin2 x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 7 / 35

14. Experiments with funny limits
sin2 x
lim =0
x→0 x
x
lim does not exist
x→0 sin2 x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 7 / 35

15. 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 )
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 7 / 35

16. 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 )
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 7 / 35

17. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 7 / 35

18. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 7 / 35

19. 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 answers
0
in different cases, we say this form is indeterminate.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 7 / 35

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
0
nonexistent because this
expression is undefined.
0
The limit is of the form ,
0
which means we cannot
evaluate it with our limit
laws.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 8 / 35

21. Indeterminate forms are like Tug Of War
Which side wins depends on which side is stronger.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 9 / 35

22. Outline
L’Hôpital’s Rule
Relative Rates of Growth
Other Indeterminate Limits
Indeterminate Products
Indeterminate Differences
Indeterminate Powers
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 10 / 35

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)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 11 / 35

24. 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) m
= 1
g(x) m2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 11 / 35

25. 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) m
= = = 1
g(x) g(x) − g(a) (g(x) − g(a))/(x − a) m2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 12 / 35

26. What then?
But what if the functions aren’t linear?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 13 / 35

27. What then?
But what if the functions aren’t linear?
Can we approximate a function near a point with a linear function?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 13 / 35

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?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 13 / 35

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!
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 13 / 35

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 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 −∞.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 14 / 35

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 stipend to Johann
Bernoulli, who proved this
theorem and named it after
him! Guillaume François Antoine,
Marquis de L’Hôpital
(French, 1661–1704)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 15 / 35

32. Revisiting the previous examples
Example
sin2 x
lim
x→0 x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

33. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim
x→0 x x→0 1
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

34. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

35. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

36. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

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
lim
x→0 sin x2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

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
lim .
x→0 sin x2
. enominator → 0
d
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

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 2.
= lim ( )
x→0 sin x x→0 cos x2 (x)
2
. enominator → 0
d
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

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 sin x cos x
2
lim = lim ( )
x→0 sin x2 x→0 cos x2 (x)
2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

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 → 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

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 → 0
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 → 0
d
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

43. 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 )
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

44. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

45. 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 )
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

46. 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
lim
x→0 sin x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

47. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 16 / 35

48. Another Example
Example
Find
x
lim
x→0 cos x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 17 / 35

49. 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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 17 / 35

50. Outline
L’Hôpital’s Rule
Relative Rates of Growth
Other Indeterminate Limits
Indeterminate Products
Indeterminate Differences
Indeterminate Powers
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 18 / 35

51. Limits of Rational Functions revisited
Example
5x2 + 3x − 1
Find lim if it exists.
x→∞ 3x2 + 7x + 27
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 19 / 35

52. Limits of Rational Functions revisited
Example
5x2 + 3x − 1
Find lim if it exists.
x→∞ 3x2 + 7x + 27
Solution
Using L’Hôpital:
5x2 + 3x − 1
lim
x→∞ 3x2 + 7x + 27
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 19 / 35

53. Limits of Rational Functions revisited
Example
5x2 + 3x − 1
Find lim if it exists.
x→∞ 3x2 + 7x + 27
Solution
Using L’Hôpital:
5x2 + 3x − 1 H 10x + 3
lim = lim
x→∞ 3x2 + 7x + 27 x→∞ 6x + 7
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 19 / 35

54. Limits of Rational Functions revisited
Example
5x2 + 3x − 1
Find lim if it exists.
x→∞ 3x2 + 7x + 27
Solution
Using L’Hôpital:
5x2 + 3x − 1 H 10x + 3 H 10 5
lim = lim = lim =
x→∞ 3x2 + 7x + 27 x→∞ 6x + 7 x→∞ 6 3
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 19 / 35

55. Limits of Rational Functions revisited
Example
5x2 + 3x − 1
Find lim if it exists.
x→∞ 3x2 + 7x + 27
Solution
Using L’Hôpital:
5x2 + 3x − 1 H 10x + 3 H 10 5
lim = lim = lim =
x→∞ 3x2 + 7x + 27 x→∞ 6x + 7 x→∞ 6 3
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 19 / 35

56. Limits of Rational Functions revisited II
Example
5x2 + 3x − 1
Find lim if it exists.
x→∞ 7x + 27
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 20 / 35

57. Limits of Rational Functions revisited II
Example
5x2 + 3x − 1
Find lim if it exists.
x→∞ 7x + 27
Solution
Using L’Hôpital:
5x2 + 3x − 1
lim
x→∞ 7x + 27
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 20 / 35

58. Limits of Rational Functions revisited II
Example
5x2 + 3x − 1
Find lim if it exists.
x→∞ 7x + 27
Solution
Using L’Hôpital:
5x2 + 3x − 1 H 10x + 3
lim = lim
x→∞ 7x + 27 x→∞ 7
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 20 / 35

59. Limits of Rational Functions revisited II
Example
5x2 + 3x − 1
Find lim if it exists.
x→∞ 7x + 27
Solution
Using L’Hôpital:
5x2 + 3x − 1 H 10x + 3
lim = lim =∞
x→∞ 7x + 27 x→∞ 7
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 20 / 35

60. Limits of Rational Functions revisited III
Example
4x + 7
Find lim if it exists.
x→∞ 3x2 + 7x + 27
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 21 / 35

61. Limits of Rational Functions revisited III
Example
4x + 7
Find lim if it exists.
x→∞ 3x2 + 7x + 27
Solution
Using L’Hôpital:
4x + 7
lim
x→∞ 3x2 + 7x + 27
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 21 / 35

62. Limits of Rational Functions revisited III
Example
4x + 7
Find lim if it exists.
x→∞ 3x2 + 7x + 27
Solution
Using L’Hôpital:
4x + 7 H 4
lim = lim
x→∞ 3x2 + 7x + 27 x→∞ 6x + 7
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 21 / 35

63. Limits of Rational Functions revisited III
Example
4x + 7
Find lim if it exists.
x→∞ 3x2 + 7x + 27
Solution
Using L’Hôpital:
4x + 7 H 4
lim = lim =0
x→∞ 3x2 + 7x + 27 x→∞ 6x + 7
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 21 / 35

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 ratio of the leading coefficients of f
x→∞ g(x)
and g.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 22 / 35

65. Exponential versus geometric growth
Example
ex
Find lim , if it exists.
x→∞ x2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 23 / 35

66. Exponential versus geometric growth
Example
ex
Find lim , if it exists.
x→∞ x2
Solution
We have
ex H ex H ex
lim = lim = lim = ∞.
x→∞ x2 x→∞ 2x x→∞ 2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 23 / 35

67. Exponential versus geometric growth
Example
ex
Find lim , if it exists.
x→∞ x2
Solution
We have
ex H ex H ex
lim = lim = lim = ∞.
x→∞ x2 x→∞ 2x x→∞ 2
Example
ex
What about lim ?
x→∞ x3
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 23 / 35

68. Exponential versus geometric growth
Example
ex
Find lim , if it exists.
x→∞ x2
Solution
We have
ex H ex H ex
lim = lim = lim = ∞.
x→∞ x2 x→∞ 2x x→∞ 2
Example
ex
What about lim ?
x→∞ x3
Answer
Still ∞. (Why?)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 23 / 35

69. Exponential versus fractional powers
Example
ex
Find lim √ , if it exists.
x→∞ x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 24 / 35

70. Exponential versus fractional powers
Example
ex
Find lim √ , if it exists.
x→∞ x
Solution (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 left-hand side must, too.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 24 / 35

71. Exponential versus fractional powers
Example
ex
Find lim √ , if it exists.
x→∞ x
Solution (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 left-hand side must, too.
Solution (with L’Hôpital)
ex ex √
lim √ = lim 1 = lim 2 xex = ∞
x→∞ x x→∞ 2 x−1/2 x→∞
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 24 / 35

72. Exponential versus any power
Theorem
Let r be any positive number. Then
ex
lim = ∞.
x→∞ xr
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 25 / 35

73. Exponential versus any power
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!
If r is not an integer, let m be the smallest integer greater than r. Then
ex ex
if x 1, xr xm , so r m . The right-hand side tends to ∞ by the
x x
previous step. . . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 25 / 35

74. Any exponential versus any power
Theorem
Let a 1 and r 0. Then
ax
lim = ∞.
x→∞ xr
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 26 / 35

75. Any exponential versus any power
Theorem
Let a 1 and r 0. Then
ax
lim = ∞.
x→∞ xr
Proof.
If r is a positive 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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 26 / 35

76. Any exponential versus any power
Theorem
Let a 1 and r 0. Then
ax
lim = ∞.
x→∞ xr
Proof.
If r is a positive 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 26 / 35

77. Logarithmic versus power growth
Theorem
Let r be any positive number. Then
ln x
lim = 0.
x→∞ xr
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 27 / 35

78. Logarithmic versus power growth
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
limr
= lim r−1 = lim r = 0.
x→∞ x x→∞ rx x→∞ rx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 27 / 35

79. Outline
L’Hôpital’s Rule
Relative Rates of Growth
Other Indeterminate Limits
Indeterminate Products
Indeterminate Differences
Indeterminate Powers
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 28 / 35

80. Indeterminate products
Example
Find √
lim+ x ln x
x→0
This limit is of the form 0 · (−∞).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 29 / 35

81. 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+
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 29 / 35

82. 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+ 1 √
x→0+ x→0 / x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 29 / 35

83. 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+ 1 √ = lim+ 1
x→0+ x→0 / x x→0 − 2 x−3/2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 29 / 35

84. 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+ 1 √ = lim+ 1
x→0+ x→0 / x x→0 − 2 x−3/2
√
= lim+ −2 x
x→0
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 29 / 35

85. 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+ 1 √ = lim+ 1
x→0+ x→0 / x x→0 − 2 x−3/2
√
= lim+ −2 x = 0
x→0
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 29 / 35

86. Indeterminate differences
Example
( )
1
lim+ − cot 2x
x→0 x
This limit is of the form ∞ − ∞.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 30 / 35

87. 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)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 30 / 35

88. 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)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 30 / 35

89. 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)
=∞
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 30 / 35

90. 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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 30 / 35

91. 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+ .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 31 / 35

92. Indeterminate powers
Example
Find lim+ (1 − 2x)1/x
x→0
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 32 / 35

93. Indeterminate powers
Example
Find lim+ (1 − 2x)1/x
x→0
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 32 / 35

94. Indeterminate powers
Example
Find lim+ (1 − 2x)1/x
x→0
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
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 32 / 35

95. Indeterminate powers
Example
Find lim+ (1 − 2x)1/x
x→0
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
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 .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 32 / 35

96. Another indeterminate power limit
Example
lim (3x)4x
x→0
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 33 / 35

97. Another indeterminate power limit
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+ 1 = lim+ −1/4x2
x→0 /4x x→0
= lim+ (−4x) = 0
x→0
So the answer is e0 = 1.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 33 / 35

98. 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 quotient
00 take ln to make an indeterminate product
∞0 ditto
1∞ ditto
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 34 / 35

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!
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 35 / 35