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# Lesson 5: Continuity

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A function is continuous at a point if the limit of the function at the point equals the value of the function at that point. Another way to say it, f is continuous at a if values of f(x) are close to f(a) if x is close to a. This property has deep implications, such as this: right now there are two points on opposites sides of the world with the same temperature!

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### Lesson 5: Continuity

1. 1. Section 1.5 Continuity V63.0121.027, Calculus I September 17, 2009 Announcements Please put your homework in the envelope by last name Quiz next week in Recitation on §§1.1–1.3 WebAssignments 1 and 2 due Tuesday . . . . . .
2. 2. Hatsumon Here are some discussion questions to start. Were you ever exactly three feet tall? . . . . . .
3. 3. Hatsumon Here are some discussion questions to start. Were you ever exactly three feet tall? Was your height (in inches) ever equal to your weight (in pounds)? . . . . . .
4. 4. Hatsumon Here are some discussion questions to start. Were you ever exactly three feet tall? Was your height (in inches) ever equal to your weight (in pounds)? Is there a pair of points on opposite sides of the world at the same temperature at the same time? . . . . . .
5. 5. Outline Continuity The Intermediate Value Theorem Back to the Questions . . . . . .
6. 6. Recall: Direct Substitution Property Theorem (The Direct Substitution Property) If f is a polynomial or a rational function and a is in the domain of f, then lim f(x) = f(a) x→a . . . . . .
7. 7. Deﬁnition of Continuity Deﬁnition Let f be a function deﬁned near a. We say that f is continuous at a if lim f(x) = f(a). x→a . . . . . .
8. 8. Deﬁnition of Continuity Deﬁnition y . Let f be a function deﬁned near a. We say that f is continuous at a f . (a ) . if lim f(x) = f(a). x→a A function f is continuous if it is continuous at every . x . point in its domain. a . There are three important parts to this deﬁnition. The limit has to exist the function has to be deﬁned and these values have to agree. . . . . . .
9. 9. Free Theorems Theorem (a) Any polynomial is continuous everywhere; that is, it is continuous on R = (−∞, ∞). (b) Any rational function is continuous wherever it is deﬁned; that is, it is continuous on its domain. . . . . . .
10. 10. Showing a function is continuous Example √ Let f(x) = 4x + 1. Show that f is continuous at 2. . . . . . .
11. 11. Showing a function is continuous Example √ Let f(x) = 4x + 1. Show that f is continuous at 2. Solution We want to show that lim f(x) = f(2). We have x→2 √ √ √ lim f(x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f(2). x→a x→2 x→2 Each step comes from the limit laws. . . . . . .
12. 12. Showing a function is continuous Example √ Let f(x) = 4x + 1. Show that f is continuous at 2. Solution We want to show that lim f(x) = f(2). We have x→2 √ √ √ lim f(x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f(2). x→a x→2 x→2 Each step comes from the limit laws. Question At which other points is f continuous? . . . . . .
13. 13. Showing a function is continuous Example √ Let f(x) = 4x + 1. Show that f is continuous at 2. Solution We want to show that lim f(x) = f(2). We have x→2 √ √ √ lim f(x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f(2). x→a x→2 x→2 Each step comes from the limit laws. Question At which other points is f continuous? Answer The function f is continuous on (−1/4, ∞). . . . . . .
14. 14. Showing a function is continuous Example √ Let f(x) = 4x + 1. Show that f is continuous at 2. Solution We want to show that lim f(x) = f(2). We have x→2 √ √ √ lim f(x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f(2). x→a x→2 x→2 Each step comes from the limit laws. Question At which other points is f continuous? Answer The function f is continuous on (−1/4, ∞). It is right continuous at the point −1/4 since lim f(x) = f(−1/4). x→−1/4+ . . . . . .
15. 15. The Limit Laws give Continuity Laws Theorem If f and g are continuous at a and c is a constant, then the following functions are also continuous at a: f+g f−g cf fg f (if g(a) ̸= 0) g . . . . . .
16. 16. Why a sum of continuous functions is continuous We want to show that lim (f + g)(x) = (f + g)(a). x→a We just follow our nose: lim (f + g)(x) = lim [f(x) + g(x)] x→a x→a = lim f(x) + lim g(x) (if these limits exist) x→a x→a = f(a) + g(a) (they do; f and g are cts) = (f + g)(a) . . . . . .
17. 17. Trigonometric functions are continuous sin and cos are continuous on R. . s . in . . . . . .
18. 18. Trigonometric functions are continuous sin and cos are continuous on R. c . os . s . in . . . . . .
19. 19. Trigonometric functions are continuous t .an sin and cos are continuous on R. sin 1 tan = and sec = cos cos c . os are continuous on their domain, which is {π } . R + kπ k ∈ Z . s . in 2 . . . . . .
20. 20. Trigonometric functions are continuous t .an s . ec sin and cos are continuous on R. sin 1 tan = and sec = cos cos c . os are continuous on their domain, which is {π } . R + kπ k ∈ Z . s . in 2 . . . . . .
21. 21. Trigonometric functions are continuous t .an s . ec sin and cos are continuous on R. sin 1 tan = and sec = cos cos c . os are continuous on their domain, which is {π } . R + kπ k ∈ Z . s . in 2 cos 1 cot = and csc = sin sin are continuous on their domain, which is R { kπ | k ∈ Z }. c . ot . . . . . .
22. 22. Trigonometric functions are continuous t .an s . ec sin and cos are continuous on R. sin 1 tan = and sec = cos cos c . os are continuous on their domain, which is {π } . R + kπ k ∈ Z . s . in 2 cos 1 cot = and csc = sin sin are continuous on their domain, which is R { kπ | k ∈ Z }. c . ot . sc c . . . . . .
23. 23. Exponential and Logarithmic functions are continuous For any base a > 1, .x a the function x → ax is continuous on R . . . . . . .
24. 24. Exponential and Logarithmic functions are continuous For any base a > 1, .x a the function x → ax is .oga x l continuous on R the function loga is continuous on its . domain: (0, ∞) . . . . . .
25. 25. Exponential and Logarithmic functions are continuous For any base a > 1, .x a the function x → ax is .oga x l continuous on R the function loga is continuous on its . domain: (0, ∞) In particular ex and ln = loge are continuous on their domains . . . . . .
26. 26. Inverse trigonometric functions are mostly continuous sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and right continuous at −1. . π . . /2 π . . in−1 s . − . π/2 . . . . . .
27. 27. Inverse trigonometric functions are mostly continuous sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and right continuous at −1. . . π . os−1 c . . /2 π . . . in−1 s . − . π/2 . . . . . .
28. 28. Inverse trigonometric functions are mostly continuous sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and right continuous at −1. sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left continuous at −1, and right continuous at 1. . . π . os−1 c . . ec−1 s . /2 π . . . in−1 s . − . π/2 . . . . . .
29. 29. Inverse trigonometric functions are mostly continuous sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and right continuous at −1. sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left continuous at −1, and right continuous at 1. . . π . os−1 c . . ec−1 s . /2 π . sc−1 c . . . in−1 s . − . π/2 . . . . . .
30. 30. Inverse trigonometric functions are mostly continuous sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and right continuous at −1. sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left continuous at −1, and right continuous at 1. tan−1 and cot−1 are continuous on R. . . π . os−1 c . . ec−1 s . /2 π .an−1 t . sc−1 c . . . in−1 s . − . π/2 . . . . . .
31. 31. Inverse trigonometric functions are mostly continuous sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and right continuous at −1. sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left continuous at −1, and right continuous at 1. tan−1 and cot−1 are continuous on R. . . π . ot−1 c . os−1 c . . ec−1 s . /2 π .an−1 t . sc−1 c . . . in−1 s . − . π/2 . . . . . .
32. 32. What could go wrong? In what ways could a function f fail to be continuous at a point a? Look again at the deﬁnition: lim f(x) = f(a) x→a . . . . . .
33. 33. Pitfall #1 Example Let { x2 if 0 ≤ x ≤ 1 f (x ) = 2x if 1 < x ≤ 2 At which points is f continuous? . . . . . .
34. 34. Pitfall #1: The limit does not exist Example Let { x2 if 0 ≤ x ≤ 1 f (x ) = 2x if 1 < x ≤ 2 At which points is f continuous? Solution At any point a in [0, 2] besides 1, lim f(x) = f(a) because f is x→a represented by a polynomial near a, and polynomials have the direct substitution property. However, lim f(x) = lim x2 = 12 = 1 x→1− x→1− lim f(x) = lim 2x = 2(1) = 2 x→1+ x→1+ So f has no limit at 1. Therefore f is not continuous at 1. . . . . . .
35. 35. Graphical Illustration of Pitfall #1 y . . . 4 . . . 3 . . 2 . . . 1 . . . . . x . − . 1 1 . 2 . . 1 . − . . . . . .
36. 36. Pitfall #2 Example Let x2 + 2x + 1 f (x ) = x+1 At which points is f continuous? . . . . . .
37. 37. Pitfall #2: The function has no value Example Let x2 + 2x + 1 f (x ) = x+1 At which points is f continuous? Solution Because f is rational, it is continuous on its whole domain. Note that −1 is not in the domain of f, so f is not continuous there. . . . . . .
38. 38. Graphical Illustration of Pitfall #2 y . . . 1 . . x . − . 1 f cannot be continuous where it has no value. . . . . . .
39. 39. Pitfall #3 Example Let { 7 if x ̸= 1 f(x) = π if x = 1 At which points is f continuous? . . . . . .
40. 40. Pitfall #3: function value ̸= limit Example Let { 7 if x ̸= 1 f(x) = π if x = 1 At which points is f continuous? Solution f is not continuous at 1 because f(1) = π but lim f(x) = 7. x→1 . . . . . .
41. 41. Graphical Illustration of Pitfall #3 y . . . 7 . . . π . . . x . 1 . . . . . . .
42. 42. Special types of discontinuites removable discontinuity The limit lim f(x) exists, but f is not x→a deﬁned at a or its value at a is not equal to the limit at a. jump discontinuity The limits lim f(x) and lim f(x) exist, but x→a− x→a+ are different. f(a) is one of these limits. . . . . . .
43. 43. Graphical representations of discontinuities y . . . 4 . y . . . 3 . . 7 . . . 2 . . . π . . . 1 . . . x . 1 . . . . . x . − . 1 1 . 2 . removable . 1 . − jump . . . . . .
44. 44. The greatest integer function [[x]] is the greatest integer ≤ x. y . . . 3 y . = [[x]] . . 2 . . . . 1 . . . . . . . . x . − . 2 − . 1 1 . 2 . 3 . .. 1 . − . .. 2 . − . . . . . .
45. 45. The greatest integer function [[x]] is the greatest integer ≤ x. y . . . 3 y . = [[x]] . . 2 . . . . 1 . . . . . . . . x . − . 2 − . 1 1 . 2 . 3 . .. 1 . − . .. 2 . − This function has a jump discontinuity at each integer. . . . . . .
46. 46. Outline Continuity The Intermediate Value Theorem Back to the Questions . . . . . .
47. 47. A Big Time Theorem Theorem (The Intermediate Value Theorem) Suppose that f is continuous on the closed interval [a, b] and let N be any number between f(a) and f(b), where f(a) ̸= f(b). Then there exists a number c in (a, b) such that f(c) = N. . . . . . .
48. 48. Illustrating the IVT f . (x ) . x . . . . . . .
49. 49. Illustrating the IVT Suppose that f is continuous on the closed interval [a, b] f . (x ) . . . x . . . . . . .
50. 50. Illustrating the IVT Suppose that f is continuous on the closed interval [a, b] f . (x ) f . (b ) . f . (a ) . . a . x . b . . . . . . .
51. 51. Illustrating the IVT Suppose that f is continuous on the closed interval [a, b] and let N be any number between f(a) and f(b), where f(a) ̸= f(b). f . (x ) f . (b ) . N . f . (a ) . . a . x . b . . . . . . .
52. 52. Illustrating the IVT Suppose that f is continuous on the closed interval [a, b] and let N be any number between f(a) and f(b), where f(a) ̸= f(b). Then there exists a number c in (a, b) such that f(c) = N. f . (x ) f . (b ) . N . . f . (a ) . . a . c . x . b . . . . . . .
53. 53. Illustrating the IVT Suppose that f is continuous on the closed interval [a, b] and let N be any number between f(a) and f(b), where f(a) ̸= f(b). Then there exists a number c in (a, b) such that f(c) = N. f . (x ) f . (b ) . N . f . (a ) . . a . x . b . . . . . . .
54. 54. Illustrating the IVT Suppose that f is continuous on the closed interval [a, b] and let N be any number between f(a) and f(b), where f(a) ̸= f(b). Then there exists a number c in (a, b) such that f(c) = N. f . (x ) f . (b ) . N . . . . f . (a ) . . a c . .1 x . c .2 c b .3 . . . . . . .
55. 55. What the IVT does not say The Intermediate Value Theorem is an “existence” theorem. It does not say how many such c exist. It also does not say how to ﬁnd c. Still, it can be used in iteration or in conjunction with other theorems to answer these questions. . . . . . .
56. 56. Using the IVT Example Suppose we are unaware of the square root function and that it’s continuous. Prove that the square root of two exists. . . . . . .
57. 57. Using the IVT Example Suppose we are unaware of the square root function and that it’s continuous. Prove that the square root of two exists. Proof. Let f(x) = x2 , a continuous function on [1, 2]. . . . . . .
58. 58. Using the IVT Example Suppose we are unaware of the square root function and that it’s continuous. Prove that the square root of two exists. Proof. Let f(x) = x2 , a continuous function on [1, 2]. Note f(1) = 1 and f(2) = 4. Since 2 is between 1 and 4, there exists a point c in (1, 2) such that f(c) = c2 = 2. . . . . . .
59. 59. Using the IVT Example Suppose we are unaware of the square root function and that it’s continuous. Prove that the square root of two exists. Proof. Let f(x) = x2 , a continuous function on [1, 2]. Note f(1) = 1 and f(2) = 4. Since 2 is between 1 and 4, there exists a point c in (1, 2) such that f(c) = c2 = 2. In fact, we can “narrow in” on the square root of 2 by the method of bisections. . . . . . .
60. 60. √ Finding 2 by bisections . .(x) = x2 x f . .. 2 4 . .. 1 1 . . . . . .
61. 61. √ Finding 2 by bisections . .(x) = x2 x f . .. 2 4 . .. 1 1 . . . . . .
62. 62. √ Finding 2 by bisections . .(x) = x2 x f . .. 2 4 . .5 . . .25 1 2 . .. 1 1 . . . . . .
63. 63. √ Finding 2 by bisections . .(x) = x2 x f . .. 2 4 . .5 . . .25 1 2 . .25 . . .5625 1 1 . .. 1 1 . . . . . .
64. 64. √ Finding 2 by bisections . .(x) = x2 x f . .. 2 4 1 . .5 . . .25 2 1 . .375 . . .890625 1 1 . .25 . . .5625 1 . .. 1 1 . . . . . .
65. 65. √ Finding 2 by bisections . .(x) = x2 x f . .. 2 4 . . .25 1 . .5 . 2.06640625 1 . .4375 1 . .375 . . .890625 2 1 . 1 . .25 . . .5625 1 . .. 1 1 . . . . . .
66. 66. Using the IVT Example Let f(x) = x3 − x − 1. Show that there is a zero for f. Solution f(1) = −1 and f(2) = 5. So there is a zero between 1 and 2. . . . . . .
67. 67. Using the IVT Example Let f(x) = x3 − x − 1. Show that there is a zero for f. Solution f(1) = −1 and f(2) = 5. So there is a zero between 1 and 2. (More careful analysis yields 1.32472.) . . . . . .
68. 68. Outline Continuity The Intermediate Value Theorem Back to the Questions . . . . . .
69. 69. Back to the Questions True or False At one point in your life you were exactly three feet tall. . . . . . .
70. 70. Question 1: True! Let h(t) be height, which varies continuously over time. Then h(birth) < 3 ft and h(now) > 3 ft. So there is a point c in (birth, now) where h(c) = 3. . . . . . .
71. 71. Back to the Questions True or False At one point in your life you were exactly three feet tall. True or False At one point in your life your height in inches equaled your weight in pounds. . . . . . .
72. 72. Question 2: True! Let h(t) be height in inches and w(t) be weight in pounds, both varying continuously over time. Let f(t) = h(t) − w(t). For most of us (call your mom), f(birth) > 0 and f(now) < 0. So there is a point c in (birth, now) where f(c) = 0. In other words, h(c) − w(c) = 0 ⇐⇒ h(c) = w(c). . . . . . .
73. 73. Back to the Questions True or False At one point in your life you were exactly three feet tall. True or False At one point in your life your height in inches equaled your weight in pounds. True or False Right now there are two points on opposite sides of the Earth with exactly the same temperature. . . . . . .
74. 74. Question 3 Let T(θ) be the temperature at the point on the equator at longitude θ. How can you express the statement that the temperature on opposite sides is the same? How can you ensure this is true? . . . . . .
75. 75. Question 3: True! Let f(θ) = T(θ) − T(θ + 180◦ ) Then f(0) = T(0) − T(180) while f(180) = T(180) − T(360) = −f(0) So somewhere between 0 and 180 there is a point θ where f(θ) = 0! . . . . . .