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- 1. Section 3.7 Indeterminate Forms and L’Hˆpital’s o Rule V63.0121.002.2010Su, Calculus I New York University June 7, 2010 Announcements
- 2. Announcements V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 2 / 26
- 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 o June 7, 2010 3 / 26
- 4. Experiments with funny limits sin2 x lim x→0 x V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 4 / 26
- 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 o June 7, 2010 4 / 26
- 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 o June 7, 2010 4 / 26
- 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 o June 7, 2010 4 / 26
- 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 o June 7, 2010 4 / 26
- 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 o June 7, 2010 5 / 26
- 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 o June 7, 2010 5 / 26
- 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 o June 7, 2010 5 / 26
- 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 o June 7, 2010 5 / 26
- 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 o June 7, 2010 6 / 26
- 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 o June 7, 2010 6 / 26
- 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 o June 7, 2010 7 / 26
- 21. Indeterminate forms are like Tug Of War Which side wins depends on which side is stronger. V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 8 / 26
- 22. Outline L’Hˆpital’s Rule o Other Indeterminate Limits Indeterminate Products Indeterminate Diﬀerences Indeterminate Powers Summary V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 9 / 26
- 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 o June 7, 2010 10 / 26
- 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 o June 7, 2010 10 / 26
- 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 o June 7, 2010 11 / 26
- 26. What then? But what if the functions aren’t linear? V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 12 / 26
- 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 o June 7, 2010 12 / 26
- 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 o June 7, 2010 12 / 26
- 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 o June 7, 2010 12 / 26
- 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 −∞. V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 13 / 26
- 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) V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 14 / 26
- 32. Revisiting the previous examples Example sin2 x lim x→0 x V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
- 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 o June 7, 2010 15 / 26
- 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 o June 7, 2010 15 / 26
- 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 o June 7, 2010 15 / 26
- 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 o June 7, 2010 15 / 26
- 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 o June 7, 2010 15 / 26
- 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 o June 7, 2010 15 / 26
- 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 o June 7, 2010 15 / 26
- 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 o June 7, 2010 15 / 26
- 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 o June 7, 2010 15 / 26
- 47. Another Example Example Find x lim x→0 cos x V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 16 / 26
- 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 o June 7, 2010 16 / 26
- 49. Outline L’Hˆpital’s Rule o Other Indeterminate Limits Indeterminate Products Indeterminate Diﬀerences Indeterminate Powers Summary V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 17 / 26
- 50. Indeterminate products Example Find √ lim+ x ln x x→0 This limit is of the form 0 · (−∞). V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 18 / 26
- 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 o June 7, 2010 18 / 26
- 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 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 18 / 26
- 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 o June 7, 2010 18 / 26
- 56. Indeterminate diﬀerences Example 1 lim+ − cot 2x x→0 x This limit is of the form ∞ − ∞. V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 19 / 26
- 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 o June 7, 2010 19 / 26
- 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 o June 7, 2010 19 / 26
- 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 o June 7, 2010 19 / 26
- 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 o June 7, 2010 20 / 26
- 62. Indeterminate powers Example Find lim+ (1 − 2x)1/x x→0 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 21 / 26
- 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 o June 7, 2010 21 / 26
- 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 o June 7, 2010 21 / 26
- 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 o June 7, 2010 21 / 26
- 66. Another indeterminate power limit Example lim (3x)4x x→0 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 22 / 26
- 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 o June 7, 2010 22 / 26
- 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 o June 7, 2010 23 / 26
- 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

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