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# Lesson 17: Indeterminate forms and l'Hôpital's Rule (slides)

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

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### Lesson 17: Indeterminate forms and l'Hôpital's Rule (slides)

1. 1. Sec on 3.7 Indeterminate forms and lHôpital’s Rule V63.0121.011: Calculus I Professor Ma hew Leingang New York University March 30, 2011.
2. 2. Announcements Midterm has been returned. Please see FAQ on Blackboard (under ”Exams and Quizzes”) Quiz 3 this week in recita on on Sec on 2.6, 2.8, 3.1, 3.2
3. 3. Objectives Know when a limit is of indeterminate form: indeterminate quo ents: 0/0, ∞/∞ indeterminate products: 0×∞ indeterminate diﬀerences: ∞ − ∞ indeterminate powers: 00 , ∞0 , and 1∞ Resolve limits in indeterminate form
4. 4. Recall Recall the limit laws from Chapter 2. Limit of a sum is the sum of the limits
5. 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
6. 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
7. 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 quo ent is the quo ent of the limits ... whoops! This is true as long as you don’t try to divide by zero.
8. 8. More about dividing limits We know dividing by zero is bad. Most of the me, if an expression’s numerator approaches a ﬁnite nonzero number and denominator approaches zero, the quo ent has an inﬁnite. For example: 1 cos x lim+ = +∞ lim− = −∞ x→0 x x→0 x3
9. 9. Why 1/0 ̸= ∞ 1 Consider the func on 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 x→∞ not inﬁnite.
10. 10. Why 1/0 ̸= ∞ 1 Consider the func on 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 x→∞ not inﬁnite. Even less predictable: when numerator and denominator both go to zero.
11. 11. Experiments with funny limits sin2 x lim x→0 x .
12. 12. Experiments with funny limits sin2 x lim =0 x→0 x .
13. 13. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 x→0 sin x .
14. 14. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 does not exist x→0 sin x .
15. 15. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 does not exist x→0 sin x sin2 x . lim x→0 sin(x2 )
16. 16. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 does not exist x→0 sin x sin2 x . lim =1 x→0 sin(x2 )
17. 17. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 does not exist x→0 sin x sin2 x . lim =1 x→0 sin(x2 ) sin 3x lim x→0 sin x
18. 18. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 does not exist x→0 sin x sin2 x . lim =1 x→0 sin(x2 ) sin 3x lim =3 x→0 sin x
19. 19. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 does not exist x→0 sin x sin2 x . lim =1 x→0 sin(x2 ) sin 3x lim =3 x→0 sin x 0 All of these are of the form , and since we can get diﬀerent 0 answers in diﬀerent cases, we say this form is indeterminate.
20. 20. Language NoteIt 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 nonexistent 0 because this expression is undeﬁned. 0 The limit is of the form , which means 0 we cannot evaluate it with our limit laws.
21. 21. Indeterminate forms are like Tug Of War Which side wins depends on which side is stronger.
22. 22. Outline L’Hôpital’s Rule Rela ve Rates of Growth Other Indeterminate Limits Indeterminate Products Indeterminate Diﬀerences Indeterminate Powers
23. 23. The Linear Case Ques on f(x) If f and g are lines and f(a) = g(a) = 0, what is lim ? x→a g(x)
24. 24. The Linear Case Ques on f(x) If f and g are lines and f(a) = g(a) = 0, what is lim ? x→a g(x) Solu on The func ons f and g can be wri en in the form f(x) = m1 (x − a) g(x) = m2 (x − a) So f(x) m1 = g(x) m2
25. 25. The Linear Case, Illustrated y y = g(x) y = f(x) a f(x) g(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
26. 26. What then? But what if the func ons aren’t linear?
27. 27. What then? But what if the func ons aren’t linear? Can we approximate a func on near a point with a linear func on?
28. 28. What then? But what if the func ons aren’t linear? Can we approximate a func on near a point with a linear func on? What would be the slope of that linear func on?
29. 29. What then? But what if the func ons aren’t linear? Can we approximate a func on near a point with a linear func on? What would be the slope of that linear func on? The deriva ve!
30. 30. Theorem of the Day Theorem (L’Hopital’s Rule) Suppose f and g are diﬀeren able func ons 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 −∞.
31. 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 s pend to Johann Bernoulli, who proved this Guillaume François Antoine, theorem and named it a er him! Marquis de L’Hôpital (French, 1661–1704)
32. 32. Revisiting the previous examples Example sin2 x lim x→0 x
33. 33. Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim x→0 x x→0 1
34. 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
35. 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
36. 36. Revisiting the previous examples Example sin2 x lim x→0 sin x2
37. 37. Revisiting the previous examples Example numerator → 0 sin2 x. lim x→0 sin x2
38. 38. Revisiting the previous examples Example numerator → 0 sin2 x. lim . x→0 sin x2 denominator → 0
39. 39. Revisiting the previous examples Example numerator → 0 sin2 x. H sin x cos x 2 lim lim . = x→0 (cos x2 ) (2x) x→0 sin x2 denominator → 0
40. 40. Revisiting the previous examples Example numerator → 0 sin2 x H . sin x cos x 2 lim = lim x→0 sin x2 x→0 (cos x2 ) (2x)
41. 41. Revisiting the previous examples Example numerator → 0 sin2 x H . sin x cos x 2 lim = lim . x→0 sin x2 x→0 (cos x2 ) (2x ) denominator → 0
42. 42. Revisiting the previous examples Example numerator → 0 sin2 x H . cos2 x − sin2 x sin x cos x H 2 lim = lim lim . = x→0 cos x2 − 2x2 sin(x2 ) x→0 sin x2 x→0 (cos x2 ) (2x ) denominator → 0
43. 43. Revisiting the previous examples Example numerator → 1 sin2 x H sin x cos x H 2 cos2 x − sin2 x. lim = lim = lim x→0 sin x2 x→0 (cos x2 ) (2x) x→0 cos x2 − 2x2 sin(x2 )
44. 44. Revisiting the previous examples Example numerator → 1 sin2 x H sin x cos x H 2 cos2 x − sin2 x. lim = lim = lim . x→0 sin x2 x→0 (cos x2 ) (2x) x→0 cos x2 − 2x2 sin(x2 ) denominator → 1
45. 45. Revisiting the previous examples 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 ) (2x) x→0 cos x2 − 2x2 sin(x2 )
46. 46. Revisiting the previous examples Example sin 3x lim x→0 sin x
47. 47. Revisiting the previous examples Example sin 3x H 3 cos 3x lim = lim = 3. x→0 sin x x→0 cos x
48. 48. Another Example Example Find x lim x→0 cos x
49. 49. Beware of Red Herrings Example Find x lim x→0 cos x Solu on The limit of the denominator is 1, not 0, so L’Hôpital’s rule does not apply. The limit is 0.
50. 50. Outline L’Hôpital’s Rule Rela ve Rates of Growth Other Indeterminate Limits Indeterminate Products Indeterminate Diﬀerences Indeterminate Powers
51. 51. Limits of Rational Functionsrevisited Example 5x2 + 3x − 1 Find lim 2 if it exists. x→∞ 3x + 7x + 27
52. 52. Limits of Rational Functionsrevisited Example 5x2 + 3x − 1 Find lim 2 if it exists. x→∞ 3x + 7x + 27 Solu on Using L’Hôpital: 5x2 + 3x − 1 lim x→∞ 3x2 + 7x + 27
53. 53. Limits of Rational Functionsrevisited Example 5x2 + 3x − 1 Find lim 2 if it exists. x→∞ 3x + 7x + 27 Solu on Using L’Hôpital: 5x2 + 3x − 1 H 10x + 3 lim 2 = lim x→∞ 3x + 7x + 27 x→∞ 6x + 7
54. 54. Limits of Rational Functionsrevisited Example 5x2 + 3x − 1 Find lim 2 if it exists. x→∞ 3x + 7x + 27 Solu on Using L’Hôpital: 5x2 + 3x − 1 H 10x + 3 H 10 5 lim 2 = lim = lim = x→∞ 3x + 7x + 27 x→∞ 6x + 7 x→∞ 6 3
55. 55. Limits of Rational Functionsrevisited Example 5x2 + 3x − 1 Find lim 2 if it exists. x→∞ 3x + 7x + 27 Solu on Using L’Hôpital: 5x2 + 3x − 1 H 10x + 3 H 10 5 lim 2 = lim = lim = x→∞ 3x + 7x + 27 x→∞ 6x + 7 x→∞ 6 3
56. 56. Limits of Rational Functionsrevisited II Example 5x2 + 3x − 1 Find lim if it exists. x→∞ 7x + 27
57. 57. Limits of Rational Functionsrevisited II Example 5x2 + 3x − 1 Find lim if it exists. x→∞ 7x + 27 Solu on Using L’Hôpital: 5x2 + 3x − 1 lim x→∞ 7x + 27
58. 58. Limits of Rational Functionsrevisited II Example 5x2 + 3x − 1 Find lim if it exists. x→∞ 7x + 27 Solu on Using L’Hôpital: 5x2 + 3x − 1 H 10x + 3 lim = lim x→∞ 7x + 27 x→∞ 7
59. 59. Limits of Rational Functionsrevisited II Example 5x2 + 3x − 1 Find lim if it exists. x→∞ 7x + 27 Solu on Using L’Hôpital: 5x2 + 3x − 1 H 10x + 3 lim = lim =∞ x→∞ 7x + 27 x→∞ 7
60. 60. Limits of Rational Functionsrevisited III Example 4x + 7 Find lim if it exists. x→∞ 3x2 + 7x + 27
61. 61. Limits of Rational Functionsrevisited III Example 4x + 7 Find lim if it exists. x→∞ 3x2 + 7x + 27 Solu on Using L’Hôpital: 4x + 7 lim x→∞ 3x2 + 7x + 27
62. 62. Limits of Rational Functionsrevisited III Example 4x + 7 Find lim if it exists. x→∞ 3x2 + 7x + 27 Solu on Using L’Hôpital: 4x + 7 H 4 lim = lim x→∞ 3x2 + 7x + 27 x→∞ 6x + 7
63. 63. Limits of Rational Functionsrevisited III Example 4x + 7 Find lim if it exists. x→∞ 3x2 + 7x + 27 Solu on Using L’Hôpital: 4x + 7 H 4 lim = lim =0 x→∞ 3x2 + 7x + 27 x→∞ 6x + 7
64. 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 ra o of the leading coeﬃcients of x→∞ g(x) f and g.
65. 65. Exponential vs. geometric growth Example ex Find lim , if it exists. x→∞ x2
66. 66. Exponential vs. geometric growth Example ex Find lim , if it exists. x→∞ x2 Solu on We have ex H ex H ex lim = lim = lim = ∞. x→∞ x2 x→∞ 2x x→∞ 2
67. 67. Exponential vs. geometric growth Example ex What about lim ? x→∞ x3
68. 68. Exponential vs. geometric growth Example ex What about lim ? x→∞ x3 Answer S ll ∞. (Why?)
69. 69. Exponential vs. geometric growth Example ex What about lim ? x→∞ x3 Answer S ll ∞. (Why?) Solu on ex H ex H ex H ex lim = lim 2 = lim = lim = ∞. x→∞ x3 x→∞ 3x x→∞ 6x x→∞ 6
70. 70. Exponential vs. fractional powers Example ex Find lim √ , if it exists. x→∞ x
71. 71. Exponential vs. fractional powers Example ex Find lim √ , if it exists. x→∞ x Solu on (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 le -hand side must, too.
72. 72. Exponential vs. fractional powers Example ex Find lim √ , if it exists. x→∞ x Solu on (with L’Hôpital) ex ex √ lim √ = lim 1 −1/2 = lim 2 xex = ∞ x→∞ x x→∞ 2 x x→∞
73. 73. Exponential vs. any power Theorem ex Let r be any posi ve number. Then lim r = ∞. x→∞ x
74. 74. Exponential vs. any power Theorem ex Let r be any posi ve number. Then lim r = ∞. x→∞ x Proof. If r is a posi ve integer, then apply L’Hôpital’s rule r mes to the frac- on. You get ex H H ex lim = . . . = lim = ∞. x→∞ xr x→∞ r!
75. 75. Exponential vs. any power Theorem ex Let r be any posi ve number. Then lim r = ∞. x→∞ x Proof. If r is not an integer, let m be the smallest integer greater than r. Then ex ex if x 1, x x , so r m . The right-hand side tends to ∞ by the r m x x previous step.
76. 76. Any exponential vs. any power Theorem ax Let a 1 and r 0. Then lim r = ∞. x→∞ x
77. 77. Any exponential vs. any power Theorem ax Let a 1 and r 0. Then lim r = ∞. x→∞ x Proof. If r is a posi ve 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.
78. 78. Any exponential vs. any power Theorem ax Let a 1 and r 0. Then lim r = ∞. x→∞ x Proof. If r is a posi ve 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
79. 79. Logarithmic versus power growth Theorem ln x Let r be any posi ve number. Then lim = 0. x→∞ xr
80. 80. Logarithmic versus power growth Theorem ln x Let r be any posi ve number. Then lim = 0. x→∞ xr Proof. One applica on of L’Hôpital’s Rule here suﬃces: ln x H 1/x 1 lim = lim r−1 = lim r = 0. x→∞ xr x→∞ rx x→∞ rx
81. 81. Outline L’Hôpital’s Rule Rela ve Rates of Growth Other Indeterminate Limits Indeterminate Products Indeterminate Diﬀerences Indeterminate Powers
82. 82. Indeterminate products Example √ Find lim+ x ln x x→0 This limit is of the form 0 · (−∞).
83. 83. Indeterminate products Example √ Find lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solu on Jury-rig the expression to make an indeterminate quo ent. Then apply L’Hôpital’s Rule: √ lim x ln x x→0+
84. 84. Indeterminate products Example √ Find lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solu on Jury-rig the expression to make an indeterminate quo ent. Then apply L’Hôpital’s Rule: √ ln x lim x ln x = lim+ 1 √ x→0+ x→0 / x
85. 85. Indeterminate products Example √ Find lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solu on Jury-rig the expression to make an indeterminate quo ent. Then apply L’Hôpital’s Rule: √ ln x H x−1 lim x ln x = lim+ 1 √ = lim+ 1 −3/2 x→0+ x→0 / x x→0 − 2 x
86. 86. Indeterminate products Example √ Find lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solu on Jury-rig the expression to make an indeterminate quo ent. Then apply L’Hôpital’s Rule: √ ln x H x−1 √ lim x ln x = lim+ 1 √ = lim+ 1 −3/2 = lim+ −2 x x→0+ x→0 / x x→0 − 2 x x→0
87. 87. Indeterminate products Example √ Find lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solu on Jury-rig the expression to make an indeterminate quo ent. Then apply L’Hôpital’s Rule: √ ln x H x−1 √ lim x ln x = lim+ 1 √ = lim+ 1 −3/2 = lim+ −2 x = 0 x→0+ x→0 / x x→0 − 2 x x→0
88. 88. Indeterminate diﬀerences Example ( ) 1 lim − cot 2x x→0+ x This limit is of the form ∞ − ∞.
89. 89. Indeterminate Diﬀerences Solu on Again, rig it to make an indeterminate quo ent. ( ) 1 sin(2x) − x cos(2x) lim+ − cot 2x = lim+ x→0 x x→0 x sin(2x)
90. 90. Indeterminate Diﬀerences Solu on Again, rig it to make an indeterminate quo ent. ( ) 1 sin(2x) − x cos(2x) lim+ − cot 2x = lim+ x→0 x x→0 x sin(2x) H cos(2x) + 2x sin(2x) = lim+ x→0 2x cos(2x) + sin(2x)
91. 91. Indeterminate Diﬀerences Solu on Again, rig it to make an indeterminate quo ent. ( ) 1 sin(2x) − x cos(2x) lim+ − cot 2x = lim+ x→0 x x→0 x sin(2x) H cos(2x) + 2x sin(2x) = lim+ x→0 2x cos(2x) + sin(2x) =∞
92. 92. Indeterminate Diﬀerences Solu on Again, rig it to make an indeterminate quo ent. ( ) 1 sin(2x) − x cos(2x) lim+ − cot 2x = lim+ x→0 x x→0 x sin(2x) H cos(2x) + 2x sin(2x) = lim+ x→0 2x cos(2x) + sin(2x) =∞ The limit is +∞ because the numerator tends to 1 while the denominator tends to zero but remains posi ve.
93. 93. Checking your workThis all goes in the thought cloud tan 2x 1 lim = 1, so for small x, tan 2x ≈ 2x. So cot 2x ≈ and x→0 2x 2x 1 1 1 1 − cot 2x ≈ − = →∞ x x 2x 2x as x → 0+ .
94. 94. Indeterminate powers Example Find lim+ (1 − 2x)1/x x→0
95. 95. Indeterminate powers Example Find lim+ (1 − 2x)1/x x→0 Solu on 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
96. 96. Indeterminate powers Example Find lim+ (1 − 2x)1/x x→0 Solu on 0 This limit is of the form , so we can use L’Hôpital: 0 −2 ln(1 − 2x) H lim+ = lim+ 1−2x = −2 x→0 x x→0 1
97. 97. Indeterminate powers Example Find lim+ (1 − 2x)1/x x→0 Solu on 0 This limit is of the form , so we can use L’Hôpital: 0 −2 ln(1 − 2x) H lim+ = lim+ 1−2x = −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 .
98. 98. Another indeterminate power limit Example Find lim+ (3x)4x x→0
99. 99. Another indeterminate power limit Example Find lim+ (3x)4x x→0 Solu on ln(3x) ln lim+ (3x)4x = lim+ ln(3x)4x = lim+ 4x ln(3x) = lim+ 1/4x x→0 x→0 x→0 x→0 H 3/3x = lim+ −1 2 = lim+ (−4x) = 0 x→0 /4x x→0 So the answer is e0 = 1.
100. 100. 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 quo ent 00 take ln to make an indeterminate product ∞0 di o 1∞ di o
101. 101. Final Thoughts L’Hôpital’s Rule only works on indeterminate quo ents Luckily, most indeterminate limits can be transformed into indeterminate quo ents L’Hôpital’s Rule gives wrong answers for non-indeterminate limits!