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

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

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

8. 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 )
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34

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 )
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34

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

12. 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.
. . . . . .
V63.0121.041, 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:
1 cos x
lim+ = +∞ lim = −∞
x→0 x x→0 − 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:
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 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:
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 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
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.
. . . . . .
V63.0121.041, 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
f(x)
lim ?
x→a 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
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 10 / 34

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

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

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 )
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

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 )
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34

48. Another Example
Example
Find
x
lim
x→0 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
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.
. . . . . .
V63.0121.041, 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
5x2 + 3x − 1
Find lim if it exists.
x→∞ 3x2 + 7x + 27
. . . . . .
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
5x2 + 3x − 1
Find lim if it exists.
x→∞ 3x2 + 7x + 27
Solution
Using L’Hôpital:
5x2 + 3x − 1
lim
x→∞ 3x2 + 7x + 27
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34

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

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 19 / 34

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

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
. . . . . .
V63.0121.041, 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.
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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 21 / 34

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

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34

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?)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34

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

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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 23 / 34

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→∞
. . . . . .
V63.0121.041, 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
ex
lim = ∞.
x→∞ 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
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. . . . . . .
V63.0121.041, 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
ax
lim = ∞.
x→∞ 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
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.
. . . . . .
V63.0121.041, 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
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
. . . . . .
V63.0121.041, 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
ln x
lim = 0.
x→∞ xr
. . . . . .
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
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
. . . . . .
V63.0121.041, 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 ln x
x→0
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 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+
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34

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

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

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

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

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

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

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)
=∞
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 29 / 34

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

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+ .
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 30 / 34

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

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34

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 .
. . . . . .
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34

96. Another indeterminate power limit
Example
lim (3x)4x
x→0
. . . . . .
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 (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.
. . . . . .
V63.0121.041, 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 or ∞ .
0
∞
∞−∞ 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