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# Lesson 17: Indeterminate Forms and L'Hopital's Rule (Section 041 slides)

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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

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### Lesson 17: Indeterminate Forms and L'Hopital's Rule (Section 041 slides)

1. 1. Section 3.7 Indeterminate Forms and L’Hôpital’s Rule V63.0121.041, Calculus I New York University November 3, 2010Announcements Quiz 3 in recitation this week on Sections 2.6, 2.8, 3.1, and 3.2 . . . . . .
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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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 0All of these are of the form , and since we can get different answers 0in 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. 13. RecallRecall 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. 14. RecallRecall 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. 15. RecallRecall 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. 16. RecallRecall 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. 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. 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. 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. 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 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. 21. Indeterminate forms are like Tug Of WarWhich 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. 22. OutlineL’Hôpital’s RuleRelative Rates of GrowthOther 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. 23. The Linear CaseQuestionIf 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. 24. The Linear CaseQuestionIf f and g are lines and f(a) = g(a) = 0, what is f(x) lim ? x→a g(x)SolutionThe 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. 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. 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. 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. 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. 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. 30. Theorem of the DayTheorem (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→aor lim f(x) = ±∞ and lim g(x) = ±∞ x→a x→aThen 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. 31. Meet the Mathematician: LH_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. 32. Revisiting the previous examplesExample 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. 33. Revisiting the previous examplesExample 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. 34. Revisiting the previous examplesExample . 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. 35. Revisiting the previous examplesExample . 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. 36. Revisiting the previous examplesExample . in x → 0 s . sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1Example 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. 37. Revisiting the previous examplesExample sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1Example . 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. 38. Revisiting the previous examplesExample sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1Example . 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. 39. Revisiting the previous examplesExample sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1Example . 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. 40. Revisiting the previous examplesExample sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1Example . 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. 41. Revisiting the previous examplesExample sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1Example . 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. 42. Revisiting the previous examplesExample sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1Example . 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. 43. Revisiting the previous examplesExample sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1Example . 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. 44. Revisiting the previous examplesExample sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1Example . 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. 45. Revisiting the previous examplesExample sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1Example 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. 46. Revisiting the previous examplesExample sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1Example 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. 47. Revisiting the previous examplesExample sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1Example 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. 48. Another ExampleExampleFind 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. 49. Beware of Red HerringsExampleFind x lim x→0 cos xSolutionThe limit of the denominator is 1, not 0, so L’Hôpital’s rule does notapply. The limit is 0. . . . . . . V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 16 / 34
50. 50. OutlineL’Hôpital’s RuleRelative Rates of GrowthOther 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. 51. Limits of Rational Functions revisitedExample 5x2 + 3x − 1Find 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. 52. Limits of Rational Functions revisitedExample 5x2 + 3x − 1Find lim if it exists. x→∞ 3x2 + 7x + 27SolutionUsing 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. 53. Limits of Rational Functions revisitedExample 5x2 + 3x − 1Find lim if it exists. x→∞ 3x2 + 7x + 27SolutionUsing 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. 54. Limits of Rational Functions revisitedExample 5x2 + 3x − 1Find lim if it exists. x→∞ 3x2 + 7x + 27SolutionUsing 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. 55. Limits of Rational Functions revisitedExample 5x2 + 3x − 1Find lim if it exists. x→∞ 3x2 + 7x + 27SolutionUsing 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. 56. Limits of Rational Functions revisited IIExample 5x2 + 3x − 1Find 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. 57. Limits of Rational Functions revisited IIExample 5x2 + 3x − 1Find lim if it exists. x→∞ 7x + 27SolutionUsing 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. 58. Limits of Rational Functions revisited IIExample 5x2 + 3x − 1Find lim if it exists. x→∞ 7x + 27SolutionUsing 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. 59. Limits of Rational Functions revisited IIExample 5x2 + 3x − 1Find lim if it exists. x→∞ 7x + 27SolutionUsing 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. 60. Limits of Rational Functions revisited IIIExample 4x + 7Find 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. 61. Limits of Rational Functions revisited IIIExample 4x + 7Find lim if it exists. x→∞ 3x2 + 7x + 27SolutionUsing 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. 62. Limits of Rational Functions revisited IIIExample 4x + 7Find lim if it exists. x→∞ 3x2 + 7x + 27SolutionUsing 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. 63. Limits of Rational Functions revisited IIIExample 4x + 7Find lim if it exists. x→∞ 3x2 + 7x + 27SolutionUsing 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. 64. Limits of Rational FunctionsFactLet 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. 65. Exponential versus geometric growthExample exFind 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. 66. Exponential versus geometric growthExample exFind lim , if it exists. x→∞ x2SolutionWe 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. 67. Exponential versus geometric growthExample exFind lim , if it exists. x→∞ x2SolutionWe have ex H ex H ex lim = lim = lim = ∞. x→∞ x2 x→∞ 2x x→∞ 2Example exWhat about lim ? x→∞ x3 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34
68. 68. Exponential versus geometric growthExample exFind lim , if it exists. x→∞ x2SolutionWe have ex H ex H ex lim = lim = lim = ∞. x→∞ x2 x→∞ 2x x→∞ 2Example exWhat about lim ? x→∞ x3AnswerStill ∞. (Why?) . . . . . . V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34
69. 69. Exponential versus fractional powersExample exFind 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. 70. Exponential versus fractional powersExample exFind lim √ , if it exists. x→∞ xSolution (without L’Hôpital)We have for all x 1, x1/2 x1 , so ex ex x1/2 xThe 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. 71. Exponential versus fractional powersExample exFind lim √ , if it exists. x→∞ xSolution (without L’Hôpital)We have for all x 1, x1/2 x1 , so ex ex x1/2 xThe 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. 72. Exponential versus any powerTheoremLet 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. 73. Exponential versus any powerTheoremLet r be any positive number. Then ex lim = ∞. x→∞ xrProof.If r is a positive integer, then apply L’Hôpital’s rule r times to thefraction. 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 exif x 1, xr xm , so r m . The right-hand side tends to ∞ by the x xprevious step. . . . . . . V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 24 / 34
74. 74. Any exponential versus any powerTheoremLet 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. 75. Any exponential versus any powerTheoremLet a 1 and r 0. Then ax lim = ∞. x→∞ xrProof.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. 76. Any exponential versus any powerTheoremLet a 1 and r 0. Then ax lim = ∞. x→∞ xrProof.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)xSo even lim = ∞! x→∞ x100000000 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 25 / 34
77. 77. Logarithmic versus power growthTheoremLet 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. 78. Logarithmic versus power growthTheoremLet r be any positive number. Then ln x lim = 0. x→∞ xrProof.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. 79. OutlineL’Hôpital’s RuleRelative Rates of GrowthOther 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. 80. Indeterminate productsExampleFind √ lim+ x ln x x→0This 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. 81. Indeterminate productsExampleFind √ lim+ x ln x x→0This limit is of the form 0 · (−∞).SolutionJury-rig the expression to make an indeterminate quotient. Then applyL’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. 82. Indeterminate productsExampleFind √ lim+ x ln x x→0This limit is of the form 0 · (−∞).SolutionJury-rig the expression to make an indeterminate quotient. Then applyL’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. 83. Indeterminate productsExampleFind √ lim+ x ln x x→0This limit is of the form 0 · (−∞).SolutionJury-rig the expression to make an indeterminate quotient. Then applyL’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. 84. Indeterminate productsExampleFind √ lim+ x ln x x→0This limit is of the form 0 · (−∞).SolutionJury-rig the expression to make an indeterminate quotient. Then applyL’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. 85. Indeterminate productsExampleFind √ lim+ x ln x x→0This limit is of the form 0 · (−∞).SolutionJury-rig the expression to make an indeterminate quotient. Then applyL’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. 86. Indeterminate differencesExample ( ) 1 lim+ − cot 2x x→0 xThis limit is of the form ∞ − ∞. . . . . . . V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 29 / 34
87. 87. Indeterminate differencesExample ( ) 1 lim+ − cot 2x x→0 xThis limit is of the form ∞ − ∞.SolutionAgain, 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. 88. Indeterminate differencesExample ( ) 1 lim+ − cot 2x x→0 xThis limit is of the form ∞ − ∞.SolutionAgain, 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. 89. Indeterminate differencesExample ( ) 1 lim+ − cot 2x x→0 xThis limit is of the form ∞ − ∞.SolutionAgain, 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. 90. Indeterminate differencesExample ( ) 1 lim+ − cot 2x x→0 xThis limit is of the form ∞ − ∞.SolutionAgain, 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 thedenominator 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. 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. 92. Indeterminate powersExampleFind 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. 93. Indeterminate powersExampleFind lim+ (1 − 2x)1/x x→0Take 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. 94. Indeterminate powersExampleFind lim+ (1 − 2x)1/x x→0Take 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 0This 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. 95. Indeterminate powersExampleFind lim+ (1 − 2x)1/x x→0Take 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 0This 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 1This is not the answer, it’s the log of the answer! So the answer wewant is e−2 . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34
96. 96. Another indeterminate power limitExample lim (3x)4x x→0 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 32 / 34
97. 97. Another indeterminate power limitExample lim (3x)4x x→0Solution 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→0So the answer is e0 = 1. . . . . . . V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 32 / 34
98. 98. Summary Form Method 0 0 L’Hôpital’s rule directly ∞ ∞ L’Hôpital’s rule directly0·∞ 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. 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