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# Lesson 25: Indeterminate Forms and L'Hôpital's Rule

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Recognizing indeterminate forms and resolving them with L'Hôpital's Rule

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### Lesson 25: Indeterminate Forms and L'Hôpital's Rule

1. 1. Section 4.5 Indeterminate Forms and L’Hˆpital’s Rule o Math 1a November 26, 2007 Announcements Special review session on optimization problems: Tues 11/27 (tomorrow) 7:00–9:00 (SC 507) my next oﬃce hours: today 1–2, tomorrow 3–4 (SC 323) MT II Review session: Sunday, 11/2, 7:30–9:00 (SC Hall D) Midterm II: Tues 12/4 7:00-9:00pm (SC Hall B)
2. 2. Outline Indeterminate Forms L’Hˆpital’s Rule o Application to Indeterminate Products Application to Indeterminate Diﬀerences Application to Indeterminate Powers Summary The Cauchy Mean Value Theorem (Bonus)
3. 3. Recall Recall the limit laws from Chapter 2. Limit of a sum is the sum of the limits
4. 4. 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
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 Limit of a product is the product 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 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.
7. 7. We know dividing by zero is bad. Most of the time, if you have a numerator which approaches a ﬁnite number and a denominator which approaches zero, the quotient approaches some kind of inﬁnity. An exception would be something like 1 lim = lim x sec x. x→∞ 1 sin x x→∞ x which doesn’t exist.
8. 8. We know dividing by zero is bad. Most of the time, if you have a numerator which approaches a ﬁnite number and a denominator which approaches zero, the quotient approaches some kind of inﬁnity. An exception would be something like 1 lim = lim x sec x. x→∞ 1 sin x x→∞ x which doesn’t exist. Even worse is the situation where the numerator and denominator both go to zero.
9. 9. Experiments sin2 x lim+ x x→0
10. 10. Experiments sin2 x lim+ =0 x x→0
11. 11. Experiments sin2 x lim+ =0 x x→0 x lim x→0 sin2 x
12. 12. Experiments sin2 x lim+ =0 x x→0 x lim does not exist x→0 sin2 x
13. 13. Experiments sin2 x lim+ =0 x x→0 x lim does not exist x→0 sin2 x sin2 x lim x→0 sin x 2
14. 14. Experiments sin2 x lim+ =0 x x→0 x lim does not exist x→0 sin2 x sin2 x lim =1 x→0 sin x 2
15. 15. Experiments sin2 x lim+ =0 x x→0 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
16. 16. Experiments sin2 x lim+ =0 x x→0 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
17. 17. Experiments sin2 x lim+ =0 x x→0 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 0 answers in diﬀerent cases, we say this form is indeterminate.
18. 18. 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 0 limit in each case “is” , and therefore nonexistent because 0 this expression is undeﬁned. 0 The limit is of the form , which means we cannot evaluate it 0 with our limit laws.
19. 19. Indeterminate forms are like Tug Of War Which side wins depends on which side is stronger.
20. 20. Outline Indeterminate Forms L’Hˆpital’s Rule o Application to Indeterminate Products Application to Indeterminate Diﬀerences Application to Indeterminate Powers Summary The Cauchy Mean Value Theorem (Bonus)
21. 21. Question If f and g are lines and f (a) = g (a) = 0, what is f (x) lim ? x→a g (x)
22. 22. 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 f (x) = = . g (x) m2 g (x)
23. 23. 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 f (x) = = . g (x) m2 g (x) But what if the functions aren’t linear? If only there were a way to deal with functions which were only approximately linear!
24. 24. 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) = ±∞ lim g (x) = ±∞ and 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 −∞.
25. 25. 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) = ±∞ lim g (x) = ±∞ and 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 −∞. ∞ L’Hˆpital’s rule also applies for limits of the form o . ∞
26. 26. 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 stipend to Johann Bernoulli, who proved this theorem and named it after him! Guillaume Fran¸is Antoine, o Marquis de L’Hˆpital o (1661–1704)
27. 27. How does this aﬀect our examples above? Example sin2 x lim x x→0
28. 28. How does this aﬀect our examples above? Example sin2 x H 2 sin x cos x lim = lim = 0. x 1 x→0 x→0
29. 29. How does this aﬀect our examples above? Example sin2 x H 2 sin x cos x lim = lim = 0. x 1 x→0 x→0 Example sin2 x lim x→0 sin x 2
30. 30. How does this aﬀect our examples above? Example sin2 x H 2 sin x cos x lim = lim = 0. x 1 x→0 x→0 Example sin2 x H 2 sin x cos x ¡ lim = lim 2 x→0 (cos x 2 ) (2x) x→0 sin x ¡
31. 31. How does this aﬀect our examples above? Example sin2 x H 2 sin x cos x lim = lim = 0. x 1 x→0 x→0 Example sin2 x H cos2 x − sin2 x 2 sin x cos x H ¡ lim = lim = lim x→0 sin x 2 x→0 (cos x 2 ) (2x) x→0 cos x 2 − x 2 sin(x 2 ) ¡
32. 32. How does this aﬀect our examples above? Example sin2 x H 2 sin x cos x lim = lim = 0. x 1 x→0 x→0 Example sin2 x H cos2 x − sin2 x 2 sin x cos x H ¡ lim = lim = lim =1 x→0 sin x 2 x→0 (cos x 2 ) (2x) x→0 cos x 2 − x 2 sin(x 2 ) ¡
33. 33. How does this aﬀect our examples above? Example sin2 x H 2 sin x cos x lim = lim = 0. x 1 x→0 x→0 Example sin2 x H cos2 x − sin2 x 2 sin x cos x H ¡ lim = lim = lim =1 x→0 sin x 2 x→0 (cos x 2 ) (2x) x→0 cos x 2 − x 2 sin(x 2 ) ¡ Example sin 3x lim x→0 sin x
34. 34. How does this aﬀect our examples above? Example sin2 x H 2 sin x cos x lim = lim = 0. x 1 x→0 x→0 Example sin2 x H cos2 x − sin2 x 2 sin x cos x H ¡ lim = lim = lim =1 x→0 sin x 2 x→0 (cos x 2 ) (2x) x→0 cos x 2 − x 2 sin(x 2 ) ¡ Example sin 3x H 3 cos 3x lim = lim = 3. x→0 sin x x→0 cos x
35. 35. Sketch of Proof of L’Hˆpital’s Rule o f (x) − f (a) Let x be a number close to a. We know that = f (c), x −a g (x) − g (a) for some c ∈ (a, x); also = g (d), for some x −a d ∈ (a, x). This means f (x) f (c) ≈ . g (x) g (d) The miracle of the MVT is that a tweaking of it allows us to assume c = d, so that f (x) f (c) = . g (x) g (c) The number c depends on x and since it is between a and x, we must have lim c(x) = a. x→a
36. 36. Beware of Red Herrings Example Find x lim x→0 cos x
37. 37. 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 o not apply. The limit is 0.
38. 38. Theorem Let r be any positive number. Then ex = ∞. lim x→∞ x r
39. 39. Theorem Let r be any positive number. Then ex = ∞. lim x→∞ x r Proof. If r is a positive integer, then apply L’Hˆpital’s rule r times to the o fraction. You get ex H ex H = ∞. lim = . . . = lim x→∞ x r x→∞ r !
40. 40. Theorem Let r be any positive number. Then ex = ∞. lim x→∞ x r Proof. If r is a positive integer, then apply L’Hˆpital’s rule r times to the o fraction. You get ex H ex H = ∞. lim = . . . = lim x→∞ x r x→∞ r ! If r is not an integer, let n = [[x]] and m = n + 1. Then if x > 1, x n < x r < x m , so ex ex ex > r > m. xn x x Now apply the Squeeze Theorem.
41. 41. Indeterminate products Example Find √ lim x ln x x→0+
42. 42. Indeterminate products Example Find √ lim x ln x x→0+ Solution Jury-rig the expression to make an indeterminate quotient. Then apply L’Hˆpital’s Rule: o x −1 √ ln x H lim x ln x = lim √ = lim x→0+ − 1 x −3/2 x→0+ 1/ x x→0+ 2 √ = lim −2 x = 0 x→0+
43. 43. Indeterminate diﬀerences Example 1 − cot 2x lim x x→0
44. 44. Indeterminate diﬀerences Example 1 − cot 2x lim x x→0 This limit is of the form ∞ − ∞, which is indeterminate.
45. 45. Indeterminate diﬀerences Example 1 − cot 2x lim x x→0 This limit is of the form ∞ − ∞, which is indeterminate. Solution Again, rig it to make an indeterminate quotient. 2x csc2 (2x) − cot(2x) 1 − x cot 2x H 2x − cos 2x lim+ = lim+ = lim+ 2 x 1 x→0 sin x sin 2x x→0 x→0 2 + 2 sin 2x H = lim+ x→0 2 cos 2x sin2 x + 2 cos x sin x sin 2x =∞
46. 46. Indeterminate powers Example lim (1 − 2x)1/x x→0+
47. 47. Indeterminate powers Example lim (1 − 2x)1/x x→0+ Take the logarithm: 1 ln lim (1 − 2x)1/x = lim ln (1 − 2x)1/x = lim ln(1 − 2x) x→0 x x→0 x→0 0 This limit is of the form , so we can use L’Hˆpital: o 0 −2 1 H 1−2x ln(1 − 2x) = lim = −2 lim x→0 x 1 x→0 This is not the answer, it’s the log of the answer! So the answer we want is e −2 .
48. 48. Example lim (3x)4x x→0
49. 49. 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 = lim+ −1/4x 2 /4x x→0 x→0 = lim+ (−4x) = 0 x→0 So the answer is e 0 = 1.
50. 50. Summary Form Method 0 L’Hˆpital’s rule directly o 0 ∞ L’Hˆpital’s rule directly o ∞ ∞ 0 0·∞ jiggle to make or ∞. 0 ∞−∞ factor to make an indeterminate product 00 take ln to make an indeterminate product ∞0 ditto 1∞ ditto
51. 51. Outline Indeterminate Forms L’Hˆpital’s Rule o Application to Indeterminate Products Application to Indeterminate Diﬀerences Application to Indeterminate Powers Summary The Cauchy Mean Value Theorem (Bonus)
52. 52. The Cauchy Mean Value Theorem (Bonus) Apply the MVT to the function h(x) = (f (b) − f (a))g (x) − (g (b) − g (a))f (x). We have h(a) = h(b). So there exists a c in (a, b) such that h (c) = 0. Thus (f (b) − f (a))g (c) = (g (b) − g (a))f (c) This is how L’Hˆpitalis proved. o