- 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. 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. 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. Recall Recall the limit laws from Chapter 2. Limit of a sum is the sum of the limits
- 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. 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. 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. 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. 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. 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. Experiments with funny limits sin2 x lim x→0 x .
- 12. Experiments with funny limits sin2 x lim =0 x→0 x .
- 13. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 x→0 sin x .
- 14. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 does not exist x→0 sin x .
- 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. 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. 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. 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. 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. 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 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. Indeterminate forms are like Tug Of War Which side wins depends on which side is stronger.
- 22. Outline L’Hôpital’s Rule Rela ve Rates of Growth Other Indeterminate Limits Indeterminate Products Indeterminate Diﬀerences Indeterminate Powers
- 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. 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. 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. What then? But what if the func ons aren’t linear?
- 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. 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. 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. 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. 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. Revisiting the previous examples Example sin2 x lim x→0 x
- 33. Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim x→0 x x→0 1
- 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. 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. Revisiting the previous examples Example sin2 x lim x→0 sin x2
- 37. Revisiting the previous examples Example numerator → 0 sin2 x. lim x→0 sin x2
- 38. Revisiting the previous examples Example numerator → 0 sin2 x. lim . x→0 sin x2 denominator → 0
- 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. 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. 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. 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. 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. 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. 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. Revisiting the previous examples Example sin 3x lim x→0 sin x
- 47. Revisiting the previous examples Example sin 3x H 3 cos 3x lim = lim = 3. x→0 sin x x→0 cos x
- 48. Another Example Example Find x lim x→0 cos x
- 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. Outline L’Hôpital’s Rule Rela ve Rates of Growth Other Indeterminate Limits Indeterminate Products Indeterminate Diﬀerences Indeterminate Powers
- 51. Limits of Rational Functions revisited Example 5x2 + 3x − 1 Find lim 2 if it exists. x→∞ 3x + 7x + 27
- 52. Limits of Rational Functions revisited 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. Limits of Rational Functions revisited 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. Limits of Rational Functions revisited 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. Limits of Rational Functions revisited 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. Limits of Rational Functions revisited II Example 5x2 + 3x − 1 Find lim if it exists. x→∞ 7x + 27
- 57. Limits of Rational Functions revisited 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. Limits of Rational Functions revisited 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. Limits of Rational Functions revisited 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. Limits of Rational Functions revisited III Example 4x + 7 Find lim if it exists. x→∞ 3x2 + 7x + 27
- 61. Limits of Rational Functions revisited 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. Limits of Rational Functions revisited 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. Limits of Rational Functions revisited 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. 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. Exponential vs. geometric growth Example ex Find lim , if it exists. x→∞ x2
- 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. Exponential vs. geometric growth Example ex What about lim ? x→∞ x3
- 68. Exponential vs. geometric growth Example ex What about lim ? x→∞ x3 Answer S ll ∞. (Why?)
- 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. Exponential vs. fractional powers Example ex Find lim √ , if it exists. x→∞ x
- 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. 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. Exponential vs. any power Theorem ex Let r be any posi ve number. Then lim r = ∞. x→∞ x
- 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. 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. Any exponential vs. any power Theorem ax Let a 1 and r 0. Then lim r = ∞. x→∞ x
- 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. 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. Logarithmic versus power growth Theorem ln x Let r be any posi ve number. Then lim = 0. x→∞ xr
- 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. Outline L’Hôpital’s Rule Rela ve Rates of Growth Other Indeterminate Limits Indeterminate Products Indeterminate Diﬀerences Indeterminate Powers
- 82. Indeterminate products Example √ Find lim+ x ln x x→0 This limit is of the form 0 · (−∞).
- 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. 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. 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. 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. 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. Indeterminate diﬀerences Example ( ) 1 lim − cot 2x x→0+ x This limit is of the form ∞ − ∞.
- 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. 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. 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. 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. Checking your work This 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. Indeterminate powers Example Find lim+ (1 − 2x)1/x x→0
- 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. 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. 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. Another indeterminate power limit Example Find lim+ (3x)4x x→0
- 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. 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. 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!