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

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- 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. 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. Recall Recall the limit laws from Chapter 2. Limit of a sum is the sum of the limits
- 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. 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. 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. 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. 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. Experiments sin2 x lim+ x x→0
- 10. Experiments sin2 x lim+ =0 x x→0
- 11. Experiments sin2 x lim+ =0 x x→0 x lim x→0 sin2 x
- 12. Experiments sin2 x lim+ =0 x x→0 x lim does not exist x→0 sin2 x
- 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. 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. 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. 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. 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. 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. Indeterminate forms are like Tug Of War Which side wins depends on which side is stronger.
- 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. Question If f and g are lines and f (a) = g (a) = 0, what is f (x) lim ? x→a g (x)
- 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. 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. 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. 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. 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. How does this aﬀect our examples above? Example sin2 x lim x x→0
- 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. 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. 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. 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. 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. 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. 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. 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. Beware of Red Herrings Example Find x lim x→0 cos x
- 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. Theorem Let r be any positive number. Then ex = ∞. lim x→∞ x r
- 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. 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. Indeterminate products Example Find √ lim x ln x x→0+
- 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. Indeterminate diﬀerences Example 1 − cot 2x lim x x→0
- 44. Indeterminate diﬀerences Example 1 − cot 2x lim x x→0 This limit is of the form ∞ − ∞, which is indeterminate.
- 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. Indeterminate powers Example lim (1 − 2x)1/x x→0+
- 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. Example lim (3x)4x x→0
- 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. 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. 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. 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

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