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∞
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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4. Experiments with funny limits
sin2 x
lim
x→0 x
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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5. Experiments with funny limits
sin2 x
lim =0
x→0 x
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
o June 7, 2010 4 / 26
6. Experiments with funny limits
sin2 x
lim =0
x→0 x
x
lim
x→0 sin2 x
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
o June 7, 2010 4 / 26
7. Experiments with funny limits
sin2 x
lim =0
x→0 x
x
lim does not exist
x→0 sin2 x
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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(x 2 )
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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(x 2 )
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
o June 7, 2010 4 / 26
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(x 2 )
sin 3x
lim
x→0 sin x
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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(x 2 )
sin 3x
lim =3
x→0 sin x
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
o June 7, 2010 4 / 26
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(x 2 )
sin 3x
lim =3
x→0 sin x
0
All of these are of the form , and since we can get different answers in
0
different cases, we say this form is indeterminate.
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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13. Recall
Recall the limit laws from Chapter 2.
Limit of a sum is the sum of the limits
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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.
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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 doesn’t exist.
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
o June 7, 2010 6 / 26
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 doesn’t exist.
Even less predictable: numerator and denominator both go to zero.
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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
, and therefore nonexistent
0
because this expression is
undefined.
0
The limit is of the form ,
0
which means we cannot
evaluate it with our limit
laws.
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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21. Indeterminate forms are like Tug Of War
Which side wins depends on which side is stronger.
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22. Outline
L’Hˆpital’s Rule
o
Other Indeterminate Limits
Indeterminate Products
Indeterminate Differences
Indeterminate Powers
Summary
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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)
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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) m1
=
g (x) m2
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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) m1
= = =
g (x) g (x) − g (a) (g (x) − g (a))/(x − a) m2
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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26. What then?
But what if the functions aren’t linear?
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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27. What then?
But what if the functions aren’t linear?
Can we approximate a function near a point with a linear function?
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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?
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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!
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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 −∞.
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31. Meet the Mathematician: L’Hˆpital
o
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 Fran¸ois Antoine,
c
Marquis de L’Hˆpital
o
(French, 1661–1704)
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32. Revisiting the previous examples
Example
sin2 x
lim
x→0 x
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33. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim
x→0 x x→0 1
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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34. Revisiting the previous examples
Example sin x → 0
sin2 x H 2 sin x cos x
lim = lim
x→0 x x→0 1
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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35. Revisiting the previous examples
Example sin x → 0
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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36. Revisiting the previous examples
Example sin x → 0
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example
sin2 x
lim
x→0 sin x 2
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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37. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example numerator → 0
sin2 x
lim
x→0 sin x 2
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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38. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example numerator → 0
sin2 x
lim
x→0 sin x 2
denominator → 0
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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39. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example numerator → 0
sin2 x H 2 sin x cos x
¡
lim 2
= lim
x→0 sin x x→0 (cos x 2 ) (2x )
¡
denominator → 0
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
o June 7, 2010 15 / 26
40. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example numerator → 0
sin2 x H 2 sin x cos x
¡
lim 2
= lim
x→0 sin x x→0 (cos x 2 ) (2x )
¡
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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41. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example numerator → 0
sin2 x H 2 sin x cos x
¡
lim 2
= lim
x→0 sin x x→0 (cos x 2 ) (2x )
¡
denominator → 0
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
o June 7, 2010 15 / 26
42. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example numerator → 0
sin2 x H 2 sin x cos x H
¡ cos2 x − sin2 x
lim = lim = lim
x→0 sin x 2 x→0 (cos x 2 ) (2x )
¡ x→0 cos x 2 − 2x 2 sin(x 2 )
denominator → 0
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
o June 7, 2010 15 / 26
43. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example numerator → 1
sin2 x H 2 sin x cos x H
¡ cos2 x − sin2 x
lim = lim = lim
x→0 sin x 2 x→0 (cos x 2 ) (2x )
¡ x→0 cos x 2 − 2x 2 sin(x 2 )
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
o June 7, 2010 15 / 26
44. Revisiting the previous examples
Example
sin2 x H 2 sin x cos x
lim = lim =0
x→0 x x→0 1
Example numerator → 1
sin2 x H 2 sin x cos x H
¡ cos2 x − sin2 x
lim = lim = lim
x→0 sin x 2 x→0 (cos x 2 ) (2x )
¡ x→0 cos x 2 − 2x 2 sin(x 2 )
denominator → 1
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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 2 sin x cos x H
¡ cos2 x − sin2 x
lim = lim = lim =1
x→0 sin x 2 x→0 (cos x 2 ) (2x )
¡ x→0 cos x 2 − 2x 2 sin(x 2 )
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
o June 7, 2010 15 / 26
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 2 sin x cos x H
¡ cos2 x − sin2 x
lim = lim = lim =1
x→0 sin x 2 x→0 (cos x 2 ) (2x )
¡ x→0 cos x 2 − 2x 2 sin(x 2 )
Example
sin 3x H 3 cos 3x
lim = lim = 3.
x→0 sin x x→0 cos x
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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47. Another Example
Example
Find
x
lim
x→0 cos x
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48. Beware of Red Herrings
Example
Find
x
lim
x→0 cos x
Solution
The limit of the denominator is 1, not 0, so L’Hˆpital’s rule does not
o
apply. The limit is 0.
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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49. Outline
L’Hˆpital’s Rule
o
Other Indeterminate Limits
Indeterminate Products
Indeterminate Differences
Indeterminate Powers
Summary
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50. Indeterminate products
Example
Find √
lim+ x ln x
x→0
This limit is of the form 0 · (−∞).
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51. 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’Hˆpital’s Rule:
o
√
lim x ln x
x→0+
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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52. 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’Hˆpital’s Rule:
o
√ ln x
lim x ln x = lim+ 1/√x
x→0+ x→0
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
o June 7, 2010 18 / 26
53. 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’Hˆpital’s Rule:
o
√ ln x H x −1
lim x ln x = lim+ √ = lim+
x→0+ x→0 1/ x x→0 − 1 x −3/2
2
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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’Hˆpital’s Rule:
o
√ 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
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
o June 7, 2010 18 / 26
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’Hˆpital’s Rule:
o
√ 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
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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56. Indeterminate differences
Example
1
lim+ − cot 2x
x→0 x
This limit is of the form ∞ − ∞.
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57. 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)
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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58. 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)
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
o June 7, 2010 19 / 26
59. 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)
=∞
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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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) 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.
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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61. 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+ .
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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62. Indeterminate powers
Example
Find lim+ (1 − 2x)1/x
x→0
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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63. 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
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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64. 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’Hˆpital:
o
0
−2
ln(1 − 2x) H 1−2x
lim+ = lim+ = −2
x→0 x x→0 1
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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65. 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’Hˆpital:
o
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 .
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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66. Another indeterminate power limit
Example
lim (3x)4x
x→0
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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67. 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+ 1/4x
= lim+ −1/4x 2
x→0 x→0
= lim+ (−4x) = 0
x→0
So the answer is e 0 = 1.
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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68. Summary
Form Method
0
0 L’Hˆpital’s rule directly
o
∞
∞ L’Hˆpital’s rule directly
o
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
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
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69. Final Thoughts
L’Hˆpital’s Rule only works on indeterminate quotients
o
Luckily, most indeterminate limits can be transformed into
indeterminate quotients
L’Hˆpital’s Rule gives wrong answers for non-indeterminate limits!
o
V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule
o June 7, 2010 24 / 26
