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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, 2010 Announcements 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 lim x→0 sin2 x 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 lim x→0 sin2 x x = 0 . V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34
6. 6. . . . . . . Experiments with funny limits lim x→0 sin2 x x = 0 lim x→0 x 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 lim x→0 sin2 x x = 0 lim x→0 x sin2 x does not exist . V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34
8. 8. . . . . . . Experiments with funny limits lim x→0 sin2 x x = 0 lim x→0 x sin2 x does not exist lim x→0 sin2 x 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 lim x→0 sin2 x x = 0 lim x→0 x sin2 x does not exist lim x→0 sin2 x sin(x2) = 1 . V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34
10. 10. . . . . . . Experiments with funny limits lim x→0 sin2 x x = 0 lim x→0 x sin2 x does not exist lim x→0 sin2 x sin(x2) = 1 lim x→0 sin 3x 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 lim x→0 sin2 x x = 0 lim x→0 x sin2 x does not exist lim x→0 sin2 x sin(x2) = 1 lim x→0 sin 3x sin x = 3 . V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34
12. 12. . . . . . . Experiments with funny limits lim x→0 sin2 x x = 0 lim x→0 x sin2 x does not exist lim x→0 sin2 x sin(x2) = 1 lim x→0 sin 3x sin x = 3 . All of these are of the form 0 0 , and since we can get different answers 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. 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. 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. 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. 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. 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: lim x→0+ 1 x = +∞ lim x→0− cos x 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: lim x→0+ 1 x = +∞ lim x→0− cos x x3 = −∞ An exception would be something like lim x→∞ 1 1 x sin x = lim x→∞ x csc 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: lim x→0+ 1 x = +∞ lim x→0− cos x x3 = −∞ An exception would be something like lim x→∞ 1 1 x sin x = lim x→∞ x csc 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 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 0 , and therefore nonexistent because this expression is undefined. The limit is of the form 0 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 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. 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. 23. . . . . . . The Linear Case Question If f and g are lines and f(a) = g(a) = 0, what is lim x→a f(x) g(x) ? V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 10 / 34
24. 24. . . . . . . The Linear Case Question If f and g are lines and f(a) = g(a) = 0, what is lim x→a f(x) 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) g(x) = m1 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 . .x .y .y = f(x) .y = g(x) . .a . .x .f(x) .g(x) f(x) g(x) = f(x) − f(a) g(x) − g(a) = (f(x) − f(a))/(x − a) (g(x) − g(a))/(x − a) = m1 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 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 x→a f(x) = 0 and lim x→a g(x) = 0 or lim x→a f(x) = ±∞ and lim x→a g(x) = ±∞ Then lim x→a f(x) g(x) = lim x→a f′ (x) 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: 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. 32. . . . . . . Revisiting the previous examples Example lim x→0 sin2 x x V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
33. 33. . . . . . . Revisiting the previous examples Example lim x→0 sin2 x x H = lim x→0 2 sin x cos x 1 V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
34. 34. . . . . . . Revisiting the previous examples Example lim x→0 sin2 x x H = lim x→0 2 sin x . .sin x → 0 cos x 1 V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
35. 35. . . . . . . Revisiting the previous examples Example lim x→0 sin2 x x H = lim x→0 2 sin x . .sin x → 0 cos x 1 = 0 V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
36. 36. . . . . . . Revisiting the previous examples Example lim x→0 sin2 x x H = lim x→0 2 sin x . .sin x → 0 cos x 1 = 0 Example lim x→0 sin2 x 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 examples Example lim x→0 sin2 x x H = lim x→0 2 sin x cos x 1 = 0 Example lim x→0 sin2 x . .numerator → 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 examples Example lim x→0 sin2 x x H = lim x→0 2 sin x cos x 1 = 0 Example lim x→0 sin2 x . .numerator → 0 sin x2. .denominator → 0 V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
39. 39. . . . . . . Revisiting the previous examples Example lim x→0 sin2 x x H = lim x→0 2 sin x cos x 1 = 0 Example lim x→0 sin2 x . .numerator → 0 sin x2. .denominator → 0 H = lim x→0 2 sin x cos x ( cos x2 ) (2x) V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
40. 40. . . . . . . Revisiting the previous examples Example lim x→0 sin2 x x H = lim x→0 2 sin x cos x 1 = 0 Example lim x→0 sin2 x sin x2 H = lim x→0 2 sin x cos x . .numerator → 0 ( cos x2 ) (2x) V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
41. 41. . . . . . . Revisiting the previous examples Example lim x→0 sin2 x x H = lim x→0 2 sin x cos x 1 = 0 Example lim x→0 sin2 x sin x2 H = lim x→0 2 sin x cos x . .numerator → 0 ( cos x2 ) (2x . .denominator → 0 ) V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
42. 42. . . . . . . Revisiting the previous examples Example lim x→0 sin2 x x H = lim x→0 2 sin x cos x 1 = 0 Example lim x→0 sin2 x sin x2 H = lim x→0 2 sin x cos x . .numerator → 0 ( cos x2 ) (2x . .denominator → 0 ) H = lim x→0 cos2 x − sin2 x cos x2 − 2x2 sin(x2) V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
43. 43. . . . . . . Revisiting the previous examples Example lim x→0 sin2 x x H = lim x→0 2 sin x cos x 1 = 0 Example lim x→0 sin2 x sin x2 H = lim x→0 2 sin x cos x ( cos x2 ) (2x) H = lim x→0 cos2 x − sin2 x . .numerator → 1 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 examples Example lim x→0 sin2 x x H = lim x→0 2 sin x cos x 1 = 0 Example lim x→0 sin2 x sin x2 H = lim x→0 2 sin x cos x ( cos x2 ) (2x) H = lim x→0 cos2 x − sin2 x . .numerator → 1 cos x2 − 2x2 sin(x2) . .denominator → 1 V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
45. 45. . . . . . . Revisiting the previous examples Example lim x→0 sin2 x x H = lim x→0 2 sin x cos x 1 = 0 Example lim x→0 sin2 x sin x2 H = lim x→0 2 sin x cos x ( cos x2 ) (2x) H = lim x→0 cos2 x − sin2 x cos x2 − 2x2 sin(x2) = 1 V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
46. 46. . . . . . . Revisiting the previous examples Example lim x→0 sin2 x x H = lim x→0 2 sin x cos x 1 = 0 Example lim x→0 sin2 x sin x2 H = lim x→0 2 sin x cos x ( cos x2 ) (2x) H = lim x→0 cos2 x − sin2 x cos x2 − 2x2 sin(x2) = 1 Example lim x→0 sin 3x 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 examples Example lim x→0 sin2 x x H = lim x→0 2 sin x cos x 1 = 0 Example lim x→0 sin2 x sin x2 H = lim x→0 2 sin x cos x ( cos x2 ) (2x) H = lim x→0 cos2 x − sin2 x cos x2 − 2x2 sin(x2) = 1 Example lim x→0 sin 3x sin x H = lim x→0 3 cos 3x cos x = 3. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
48. 48. . . . . . . Another Example Example Find lim x→0 x 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 Herrings Example Find lim x→0 x 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. 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. 51. . . . . . . Limits of Rational Functions revisited Example Find lim x→∞ 5x2 + 3x − 1 3x2 + 7x + 27 if it exists. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34
52. 52. . . . . . . Limits of Rational Functions revisited Example Find lim x→∞ 5x2 + 3x − 1 3x2 + 7x + 27 if it exists. Solution Using L’Hôpital: lim x→∞ 5x2 + 3x − 1 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 revisited Example Find lim x→∞ 5x2 + 3x − 1 3x2 + 7x + 27 if it exists. Solution Using L’Hôpital: lim x→∞ 5x2 + 3x − 1 3x2 + 7x + 27 H = lim x→∞ 10x + 3 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 revisited Example Find lim x→∞ 5x2 + 3x − 1 3x2 + 7x + 27 if it exists. Solution Using L’Hôpital: lim x→∞ 5x2 + 3x − 1 3x2 + 7x + 27 H = lim x→∞ 10x + 3 6x + 7 H = lim x→∞ 10 6 = 5 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 revisited Example Find lim x→∞ 5x2 + 3x − 1 3x2 + 7x + 27 if it exists. Solution Using L’Hôpital: lim x→∞ 5x2 + 3x − 1 3x2 + 7x + 27 H = lim x→∞ 10x + 3 6x + 7 H = lim x→∞ 10 6 = 5 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 II Example Find lim x→∞ 5x2 + 3x − 1 7x + 27 if it exists. 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 II Example Find lim x→∞ 5x2 + 3x − 1 7x + 27 if it exists. Solution Using L’Hôpital: lim x→∞ 5x2 + 3x − 1 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 II Example Find lim x→∞ 5x2 + 3x − 1 7x + 27 if it exists. Solution Using L’Hôpital: lim x→∞ 5x2 + 3x − 1 7x + 27 H = lim x→∞ 10x + 3 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 II Example Find lim x→∞ 5x2 + 3x − 1 7x + 27 if it exists. Solution Using L’Hôpital: lim x→∞ 5x2 + 3x − 1 7x + 27 H = lim x→∞ 10x + 3 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 III Example Find lim x→∞ 4x + 7 3x2 + 7x + 27 if it exists. 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 III Example Find lim x→∞ 4x + 7 3x2 + 7x + 27 if it exists. Solution Using L’Hôpital: lim x→∞ 4x + 7 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 III Example Find lim x→∞ 4x + 7 3x2 + 7x + 27 if it exists. Solution Using L’Hôpital: lim x→∞ 4x + 7 3x2 + 7x + 27 H = lim x→∞ 4 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 III Example Find lim x→∞ 4x + 7 3x2 + 7x + 27 if it exists. Solution Using L’Hôpital: lim x→∞ 4x + 7 3x2 + 7x + 27 H = lim x→∞ 4 6x + 7 = 0 V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 20 / 34
64. 64. . . . . . . Limits of Rational Functions Fact Let f(x) and g(x) be polynomials of degree p and q. If p q, then lim x→∞ f(x) g(x) = ∞ If p q, then lim x→∞ f(x) g(x) = 0 If p = q, then lim x→∞ f(x) g(x) is the ratio of the leading coefficients of f 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 growth Example Find lim x→∞ ex x2 , if it exists. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34
66. 66. . . . . . . Exponential versus geometric growth Example Find lim x→∞ ex x2 , if it exists. Solution We have lim x→∞ ex x2 H = lim x→∞ ex 2x H = lim x→∞ ex 2 = ∞. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34
67. 67. . . . . . . Exponential versus geometric growth Example Find lim x→∞ ex x2 , if it exists. Solution We have lim x→∞ ex x2 H = lim x→∞ ex 2x H = lim x→∞ ex 2 = ∞. Example What about lim x→∞ ex x3 ? V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34
68. 68. . . . . . . Exponential versus geometric growth Example Find lim x→∞ ex x2 , if it exists. Solution We have lim x→∞ ex x2 H = lim x→∞ ex 2x H = lim x→∞ ex 2 = ∞. Example What about lim x→∞ ex x3 ? Answer Still ∞. (Why?) V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34
69. 69. . . . . . . Exponential versus fractional powers Example Find lim x→∞ ex √ x , if it exists. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 23 / 34
70. 70. . . . . . . Exponential versus fractional powers Example Find lim x→∞ ex √ x , if it exists. Solution (without L’Hôpital) We have for all x 1, x1/2 x1 , so ex x1/2 ex 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. 71. . . . . . . Exponential versus fractional powers Example Find lim x→∞ ex √ x , if it exists. Solution (without L’Hôpital) We have for all x 1, x1/2 x1 , so ex x1/2 ex x The right hand side tends to ∞, so the left-hand side must, too. Solution (with L’Hôpital) lim x→∞ ex √ x = lim x→∞ ex 1 2 x−1/2 = lim x→∞ 2 √ xex = ∞ V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 23 / 34
72. 72. . . . . . . Exponential versus any power Theorem Let r be any positive number. Then lim x→∞ ex xr = ∞. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 24 / 34
73. 73. . . . . . . Exponential versus any power Theorem Let r be any positive number. Then lim x→∞ ex xr = ∞. Proof. If r is a positive integer, then apply L’Hôpital’s rule r times to the fraction. You get lim x→∞ ex xr H = . . . H = lim x→∞ ex r! = ∞. If r is not an integer, let m be the smallest integer greater than r. Then if x 1, xr xm , so ex xr ex xm . The right-hand side tends to ∞ by the previous 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 power Theorem Let a 1 and r 0. Then lim x→∞ ax 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 power Theorem Let a 1 and r 0. Then lim x→∞ ax xr = ∞. Proof. If r is a positive integer, we have lim x→∞ ax xr H = . . . H = lim x→∞ (ln a)rax 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 power Theorem Let a 1 and r 0. Then lim x→∞ ax xr = ∞. Proof. If r is a positive integer, we have lim x→∞ ax xr H = . . . H = lim x→∞ (ln a)rax r! = ∞. If r isn’t an integer, we can compare it as before. So even lim x→∞ (1.00000001)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 growth Theorem Let r be any positive number. Then lim x→∞ ln x xr = 0. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 26 / 34
78. 78. . . . . . . Logarithmic versus power growth Theorem Let r be any positive number. Then lim x→∞ ln x xr = 0. Proof. One application of L’Hôpital’s Rule here suffices: lim x→∞ ln x xr H = lim x→∞ 1/x rxr−1 = lim x→∞ 1 rxr = 0. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 26 / 34
79. 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. 80. . . . . . . Indeterminate products Example Find lim x→0+ √ x ln x 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. 81. . . . . . . Indeterminate products Example Find lim x→0+ √ x ln x 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→0+ √ x ln x V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34
82. 82. . . . . . . Indeterminate products Example Find lim x→0+ √ x ln x 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→0+ √ x ln x = lim x→0+ ln x 1/ √ x V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34
83. 83. . . . . . . Indeterminate products Example Find lim x→0+ √ x ln x 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→0+ √ x ln x = lim x→0+ ln x 1/ √ x H = lim x→0+ x−1 −1 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 products Example Find lim x→0+ √ x ln x 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→0+ √ x ln x = lim x→0+ ln x 1/ √ x H = lim x→0+ x−1 −1 2 x−3/2 = lim x→0+ −2 √ x V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34
85. 85. . . . . . . Indeterminate products Example Find lim x→0+ √ x ln x 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→0+ √ x ln x = lim x→0+ ln x 1/ √ x H = lim x→0+ x−1 −1 2 x−3/2 = lim x→0+ −2 √ x = 0 V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34
86. 86. . . . . . . Indeterminate differences Example lim x→0+ ( 1 x − cot 2x ) 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. 87. . . . . . . Indeterminate differences Example lim x→0+ ( 1 x − cot 2x ) This limit is of the form ∞ − ∞. Solution Again, rig it to make an indeterminate quotient. lim x→0+ sin(2x) − x cos(2x) x sin(2x) V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 29 / 34
88. 88. . . . . . . Indeterminate differences Example lim x→0+ ( 1 x − cot 2x ) This limit is of the form ∞ − ∞. Solution Again, rig it to make an indeterminate quotient. lim x→0+ sin(2x) − x cos(2x) x sin(2x) H = lim x→0+ cos(2x) + 2x sin(2x) 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 differences Example lim x→0+ ( 1 x − cot 2x ) This limit is of the form ∞ − ∞. Solution Again, rig it to make an indeterminate quotient. lim x→0+ sin(2x) − x cos(2x) x sin(2x) H = lim x→0+ cos(2x) + 2x sin(2x) 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 differences Example lim x→0+ ( 1 x − cot 2x ) This limit is of the form ∞ − ∞. Solution Again, rig it to make an indeterminate quotient. lim x→0+ sin(2x) − x cos(2x) x sin(2x) H = lim x→0+ cos(2x) + 2x sin(2x) 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. 91. . . . . . . Checking your work . . lim x→0 tan 2x 2x = 1, so for small x, tan 2x ≈ 2x. So cot 2x ≈ 1 2x and 1 x − cot 2x ≈ 1 x − 1 2x = 1 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 powers Example Find lim x→0+ (1 − 2x)1/x V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34
93. 93. . . . . . . Indeterminate powers Example Find lim x→0+ (1 − 2x)1/x Take the logarithm: ln ( lim x→0+ (1 − 2x)1/x ) = lim x→0+ ln ( (1 − 2x)1/x ) = lim x→0+ ln(1 − 2x) x V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34
94. 94. . . . . . . Indeterminate powers Example Find lim x→0+ (1 − 2x)1/x Take the logarithm: ln ( lim x→0+ (1 − 2x)1/x ) = lim x→0+ ln ( (1 − 2x)1/x ) = lim x→0+ ln(1 − 2x) x This limit is of the form 0 0 , so we can use L’Hôpital: lim x→0+ ln(1 − 2x) x H = lim x→0+ −2 1−2x 1 = −2 V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34
95. 95. . . . . . . Indeterminate powers Example Find lim x→0+ (1 − 2x)1/x Take the logarithm: ln ( lim x→0+ (1 − 2x)1/x ) = lim x→0+ ln ( (1 − 2x)1/x ) = lim x→0+ ln(1 − 2x) x This limit is of the form 0 0 , so we can use L’Hôpital: lim x→0+ ln(1 − 2x) x H = lim x→0+ −2 1−2x 1 = −2 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. 96. . . . . . . Another indeterminate power limit Example lim x→0 (3x)4x V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 32 / 34
97. 97. . . . . . . Another indeterminate power limit Example lim x→0 (3x)4x Solution ln lim x→0+ (3x)4x = lim x→0+ ln(3x)4x = lim x→0+ 4x ln(3x) = lim x→0+ ln(3x) 1/4x H = lim x→0+ 3/3x −1/4x2 = lim x→0+ (−4x) = 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. 98. . . . . . . Summary Form Method 0 0 L’Hôpital’s rule directly ∞ ∞ L’Hôpital’s rule directly 0 · ∞ jiggle to make 0 0 or ∞ ∞. ∞ − ∞ 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