Lesson 5: Continuity (slides)

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Continuity says that the limit of a function at a point equals the value of the function at that point, or, that small changes in the input give only small changes in output. This has important implications, such as the Intermediate Value Theorem.

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Lesson 5: Continuity (slides)

  1. 1. Sec on 1.5 Con nuity V63.0121.011: Calculus I Professor Ma hew Leingang New York University February 7, 2011.
  2. 2. Announcements Get-to-know-you extra credit due Friday February 11 Quiz 1 February 17/18 in recita on
  3. 3. Objectives Understand and apply the defini on of con nuity for a func on at a point or on an interval. Given a piecewise defined func on, decide where it is con nuous or discon nuous. State and understand the Intermediate Value Theorem. Use the IVT to show that a func on takes a certain value, or that an equa on has a solu on
  4. 4. Last time Defini on We write lim f(x) = L x→a and say “the limit of f(x), as x approaches a, equals L” if we can make the values of f(x) arbitrarily close to L (as close to L as we like) by taking x to be sufficiently close to a (on either side of a) but not equal to a.
  5. 5. Basic Limits Theorem (Basic Limits) lim x = a x→a lim c = c x→a
  6. 6. Limit Laws for arithmetic Theorem (Limit Laws) Let f and g be func ons with limits at a point a. Then lim (f(x) + g(x)) = lim f(x) + lim g(x) x→a x→a x→a lim (f(x) − g(x)) = lim f(x) − lim g(x) x→a x→a x→a lim (f(x) · g(x)) = lim f(x) · lim g(x) x→a x→a x→a f(x) limx→a f(x) lim = if lim g(x) ̸= 0 x→a g(x) limx→a g(x) x→a
  7. 7. Hatsumon Here are some discussion ques ons to start. True or False At some point in your life you were exactly three feet tall.
  8. 8. Hatsumon Here are some discussion ques ons to start. True or False At some point in your life you were exactly three feet tall. True or False At some point in your life your height (in inches) was equal to your weight (in pounds).
  9. 9. Hatsumon Here are some discussion ques ons to start. True or False At some point in your life you were exactly three feet tall. True or False At some point in your life your height (in inches) was equal to your weight (in pounds). True or False Right now there are a pair of points on opposite sides of the world measuring the exact same temperature.
  10. 10. Outline Con nuity The Intermediate Value Theorem Back to the Ques ons
  11. 11. Recall: Direct SubstitutionProperty Theorem (The Direct Subs tu on Property) If f is a polynomial or a ra onal func on and a is in the domain of f, then lim f(x) = f(a) x→a This property is so useful it’s worth naming.
  12. 12. Definition of Continuity y Defini on Let f be a func on defined near a. We say that f is con nuous at f(a) a if lim f(x) = f(a). x→a . x a
  13. 13. Definition of Continuity y Defini on Let f be a func on defined near a. We say that f is con nuous at f(a) a if lim f(x) = f(a). x→a A func on f is con nuous if it is con nuous at every point in its domain. . x a
  14. 14. Scholium Defini on Let f be a func on defined near a. We say that f is con nuous at a if lim f(x) = f(a). x→a There are three important parts to this defini on. The func on has to have a limit at a, the func on has to have a value at a, and these values have to agree.
  15. 15. Free Theorems Theorem (a) Any polynomial is con nuous everywhere; that is, it is con nuous on R = (−∞, ∞). (b) Any ra onal func on is con nuous wherever it is defined; that is, it is con nuous on its domain.
  16. 16. Showing a function is continuous . Example √ Let f(x) = 4x + 1. Show that f is con nuous at 2.
  17. 17. Showing a function is continuous . Example √ Let f(x) = 4x + 1. Show that f is con nuous at 2. Solu on 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.
  18. 18. At which other points? Ques on √ As before, let f(x) = 4x + 1. At which points is f con nuous?
  19. 19. At which other points? Ques on √ As before, let f(x) = 4x + 1. At which points is f con nuous? Solu on If a > −1/4, then lim (4x + 1) = 4a + 1 > 0, so x→a √ √ √ lim f(x) = lim 4x + 1 = lim (4x + 1) = 4a + 1 = f(a) x→a x→a x→a and f is con nuous at a.
  20. 20. At which other points? Ques on √ As before, let f(x) = 4x + 1. At which points is f con nuous? Solu on If a = −1/4, then 4x + 1 < 0 to the le of a, which means √ 4x + 1 is undefined. S ll, √ √ √ lim+ f(x) = lim+ 4x + 1 = lim+ (4x + 1) = 0 = 0 = f(a) x→a x→a x→a so f is con nuous on the right at a = −1/4.
  21. 21. Limit Laws give Continuity Laws Theorem If f(x) and g(x) are con nuous at a and c is a constant, then the following func ons are also con nuous at a: (f + g)(x) (fg)(x) (f − g)(x) f (x) (if g(a) ̸= 0) (cf)(x) g
  22. 22. Sum of continuous is continuous We want to show that lim (f + g)(x) = (f + g)(a). x→a
  23. 23. Sum of continuous is continuous We want to show that lim (f + g)(x) = (f + g)(a). x→a We just follow our nose.
  24. 24. Sum of continuous is continuous We want to show that lim (f + g)(x) = (f + g)(a). x→a We just follow our nose. (def of f + g) lim (f + g)(x) x→a
  25. 25. Sum of continuous is continuous We want to show that lim (f + g)(x) = (f + g)(a). x→a We just follow our nose. (def of f + g) lim (f + g)(x) = lim [f(x) + g(x)] x→a x→a
  26. 26. Sum of continuous is continuous We want to show that lim (f + g)(x) = (f + g)(a). x→a We just follow our nose. (def of f + g) lim (f + g)(x) = lim [f(x) + g(x)] x→a x→a (if these limits exist)
  27. 27. Sum of continuous is continuous We want to show that lim (f + g)(x) = (f + g)(a). x→a We just follow our nose. (def of f + g) lim (f + g)(x) = lim [f(x) + g(x)] x→a x→a (if these limits exist) = lim f(x) + lim g(x) x→a x→a
  28. 28. Sum of continuous is continuous We want to show that lim (f + g)(x) = (f + g)(a). x→a We just follow our nose. (def of f + g) lim (f + g)(x) = lim [f(x) + g(x)] x→a x→a (if these limits exist) = lim f(x) + lim g(x) x→a x→a (they do; f and g are cts.)
  29. 29. Sum of continuous is continuous We want to show that lim (f + g)(x) = (f + g)(a). x→a We just follow our nose. (def of f + g) lim (f + g)(x) = lim [f(x) + g(x)] x→a x→a (if these limits exist) = lim f(x) + lim g(x) x→a x→a (they do; f and g are cts.) = f(a) + g(a)
  30. 30. Sum of continuous is continuous We want to show that lim (f + g)(x) = (f + g)(a). x→a We just follow our nose. (def of f + g) lim (f + g)(x) = lim [f(x) + g(x)] x→a x→a (if these limits exist) = lim f(x) + lim g(x) x→a x→a (they do; f and g are cts.) = f(a) + g(a) (def of f + g again)
  31. 31. Sum of continuous is continuous We want to show that lim (f + g)(x) = (f + g)(a). x→a We just follow our nose. (def of f + g) lim (f + g)(x) = lim [f(x) + g(x)] x→a x→a (if these limits exist) = lim f(x) + lim g(x) x→a x→a (they do; f and g are cts.) = f(a) + g(a) (def of f + g again) = (f + g)(a)
  32. 32. Sum of continuous is continuous We want to show that lim (f + g)(x) = (f + g)(a). x→a We just follow our nose. (def of f + g) lim (f + g)(x) = lim [f(x) + g(x)] x→a x→a (if these limits exist) = lim f(x) + lim g(x) x→a x→a (they do; f and g are cts.) = f(a) + g(a) (def of f + g again) = (f + g)(a)
  33. 33. Trig functions are continuous sin and cos are con nuous on R. . sin
  34. 34. Trig functions are continuous sin and cos are con nuous on R. cos . sin
  35. 35. Trig functions are continuous tan sin and cos are con nuous on R. sin 1 tan = and sec = are cos cos cos con nuous on their domain, {π which is } . sin R + kπ k ∈ Z . 2
  36. 36. Trig functions are continuous tan sec sin and cos are con nuous on R. sin 1 tan = and sec = are cos cos cos con nuous on their domain, {π which is } . sin R + kπ k ∈ Z . 2
  37. 37. Trig functions are continuous tan sec sin and cos are con nuous on R. sin 1 tan = and sec = are cos cos cos con nuous on their domain, {π which is } . sin R + kπ k ∈ Z . 2 cos 1 cot = and csc = are sin sin con nuous on their domain, which is R { kπ | k ∈ Z }. cot
  38. 38. Trig functions are continuous tan sec sin and cos are con nuous on R. sin 1 tan = and sec = are cos cos cos con nuous on their domain, {π which is } . sin R + kπ k ∈ Z . 2 cos 1 cot = and csc = are sin sin con nuous on their domain, which is R { kπ | k ∈ Z }. cot csc
  39. 39. Exp and Log are continuous For any base a 1, the func on x → ax is ax con nuous on R .
  40. 40. Exp and Log are continuous For any base a 1, the func on x → ax is ax loga x con nuous on R the func on loga is con nuous on its . domain: (0, ∞)
  41. 41. Exp and Log are continuous For any base a 1, the func on x → ax is ax loga x con nuous on R the func on loga is con nuous on its . domain: (0, ∞) In par cular ex and ln = loge are con nuous on their domains
  42. 42. Inverse trigonometric functionsare mostly continuous sin−1 and cos−1 are con nuous on (−1, 1), le con nuous at 1, and right con nuous at −1. π π/2 −1 . sin
  43. 43. Inverse trigonometric functionsare mostly continuous sin−1 and cos−1 are con nuous on (−1, 1), le con nuous at 1, and right con nuous at −1. π cos−1 π/2 −1 . sin
  44. 44. Inverse trigonometric functionsare mostly continuous sin−1 and cos−1 are con nuous on (−1, 1), le con nuous at 1, and right con nuous at −1. sec−1 and csc−1 are con nuous on (−∞, −1) ∪ (1, ∞), le con nuous at −1, and right con nuous at 1. π cos−1 sec−1 π/2 −1 . sin
  45. 45. Inverse trigonometric functionsare mostly continuous sin−1 and cos−1 are con nuous on (−1, 1), le con nuous at 1, and right con nuous at −1. sec−1 and csc−1 are con nuous on (−∞, −1) ∪ (1, ∞), le con nuous at −1, and right con nuous at 1. π cos−1 sec−1 π/2 −1 csc−1 sin .
  46. 46. Inverse trigonometric functionsare mostly continuous sin−1 and cos−1 are con nuous on (−1, 1), le con nuous at 1, and right con nuous at −1. sec−1 and csc−1 are con nuous on (−∞, −1) ∪ (1, ∞), le con nuous at −1, and right con nuous at 1. tan−1 and cot−1 are con nuous on R. π cos−1 sec−1 π/2 −1 tan−1 csc−1 sin .
  47. 47. Inverse trigonometric functionsare mostly continuous sin−1 and cos−1 are con nuous on (−1, 1), le con nuous at 1, and right con nuous at −1. sec−1 and csc−1 are con nuous on (−∞, −1) ∪ (1, ∞), le con nuous at −1, and right con nuous at 1. tan−1 and cot−1 are con nuous on R. π cot−1 cos−1 sec−1 π/2 −1 tan−1 csc−1 sin .
  48. 48. What could go wrong? In what ways could a func on f fail to be con nuous at a point a? Look again at the equa on from the defini on: lim f(x) = f(a) x→a
  49. 49. Continuity FAIL . Example { x2 if 0 ≤ x ≤ 1 Let f(x) = . At which points is f con nuous? 2x if 1 x ≤ 2
  50. 50. Continuity FAIL: no limit . Example { x2 if 0 ≤ x ≤ 1 Let f(x) = . At which points is f con nuous? 2x if 1 x ≤ 2 Solu on At any point a besides 1, lim f(x) = f(a) because f is represented by a x→a polynomial near a, and polynomials have the direct subs tu on property. lim− f(x) = lim− x2 = 12 = 1 and lim+ f(x) = lim+ 2x = 2(1) = 2 x→1 x→1 x→1 x→1 So f has no limit at 1. Therefore f is not con nuous at 1.
  51. 51. Graphical Illustration of Pitfall #1 y 4 3 The func on cannot be con nuous at a point if the 2 func on has no limit at that 1 point. . x −1 1 2 −1
  52. 52. Graphical Illustration of Pitfall #1 y 4 3 The func on cannot be con nuous at a point if the 2 1 FAIL func on has no limit at that point. . x −1 1 2 −1
  53. 53. Continuity FAIL Example Let x2 + 2x + 1 f(x) = x+1 At which points is f con nuous?
  54. 54. Continuity FAIL: no value Example Let x2 + 2x + 1 f(x) = x+1 At which points is f con nuous? Solu on Because f is ra onal, it is con nuous on its whole domain. Note that −1 is not in the domain of f, so f is not con nuous there.
  55. 55. Graphical Illustration of Pitfall #2 y 1 The func on cannot be con nuous at a point outside . x its domain (that is, a point −1 where it has no value).
  56. 56. Graphical Illustration of Pitfall #2 y FAIL 1 The func on cannot be con nuous at a point outside . x its domain (that is, a point −1 where it has no value).
  57. 57. Continuity FAIL Example Let { 7 if x ̸= 1 f(x) = π if x = 1 At which points is f con nuous?
  58. 58. Continuity FAIL: value ̸= limit Example Let { 7 if x ̸= 1 f(x) = π if x = 1 At which points is f con nuous? Solu on f is not con nuous at 1 because f(1) = π but lim f(x) = 7. x→1
  59. 59. Graphical Illustration of Pitfall #3 y 7 If the func on has a limit and a value at a point the two π must s ll agree. . x 1
  60. 60. Graphical Illustration of Pitfall #3 y 7 If the func on has a limit and a value at a point the two FAIL π . x must s ll agree. 1
  61. 61. Special types of discontinuities removable discon nuity 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 discon nuity The limits lim− f(x) and lim+ f(x) exist, but are x→a x→a different.
  62. 62. Special discontinuities graphically y y 7 2 π 1 . x . x 1 1 removable jump
  63. 63. Special discontinuities graphically y y Presto! con nuous! 7 2 π 1 . x . x 1 1 removable jump
  64. 64. Special discontinuities graphically y y Presto! con nuous! 7 2 π 1 con nuous? . x . x 1 1 removable jump
  65. 65. Special discontinuities graphically y y Presto! con nuous! 7 2 con nuous? π 1 . x . x 1 1 removable jump
  66. 66. Special discontinuities graphically y y Presto! con nuous! 7 2 con nuous? π 1 . x . x 1 1 removable jump
  67. 67. Special types of discontinuities removable discon nuity 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. By re-defining f(a) = lim f(x), f can be made con nuous x→a at a jump discon nuity The limits lim− f(x) and lim+ f(x) exist, but are x→a x→a different.
  68. 68. Special types of discontinuities removable discon nuity 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. By re-defining f(a) = lim f(x), f can be made con nuous x→a at a jump discon nuity The limits lim− f(x) and lim+ f(x) exist, but are x→a x→a different. The func on cannot be made con nuous by changing a single value.
  69. 69. The greatest integer function [[x]] is the greatest integer ≤ x. y 3 x [[x]] y = [[x]] 0 0 2 1 1 1.5 1 1 1.9 1 . x 2.1 2 −2 −1 1 2 3 −0.5 −1 −1 −0.9 −1 −1.1 −2 −2
  70. 70. The greatest integer function [[x]] is the greatest integer ≤ x. y 3 x [[x]] y = [[x]] 0 0 2 1 1 1.5 1 1 1.9 1 . x 2.1 2 −2 −1 1 2 3 −0.5 −1 −1 −0.9 −1 −1.1 −2 −2 This func on has a jump discon nuity at each integer.
  71. 71. Outline Con nuity The Intermediate Value Theorem Back to the Ques ons
  72. 72. A Big Time Theorem Theorem (The Intermediate Value Theorem) Suppose that f is con nuous 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.
  73. 73. Illustrating the IVT f(x) Theorem . x
  74. 74. Illustrating the IVT f(x) Theorem Suppose that f is con nuous on the closed interval [a, b] . x
  75. 75. Illustrating the IVT f(x) Theorem Suppose that f is con nuous f(b) on the closed interval [a, b] f(a) . a b x
  76. 76. Illustrating the IVT f(x) Theorem Suppose that f is con nuous f(b) on the closed interval [a, b] N and let N be any number between f(a) and f(b), where f(a) f(a) ̸= f(b). . a b x
  77. 77. Illustrating the IVT f(x) Theorem Suppose that f is con nuous f(b) on the closed interval [a, b] N and let N be any number between f(a) and f(b), where f(a) f(a) ̸= f(b). Then there exists a number c in (a, b) such that f(c) = N. . a c b x
  78. 78. Illustrating the IVT f(x) Theorem Suppose that f is con nuous f(b) on the closed interval [a, b] N and let N be any number between f(a) and f(b), where f(a) f(a) ̸= f(b). Then there exists a number c in (a, b) such that f(c) = N. . a b x
  79. 79. Illustrating the IVT f(x) Theorem Suppose that f is con nuous f(b) on the closed interval [a, b] N and let N be any number between f(a) and f(b), where f(a) f(a) ̸= f(b). Then there exists a number c in (a, b) such that f(c) = N. . ac1 c2 c3b x
  80. 80. 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. S ll, it can be used in itera on or in conjunc on with other theorems to answer these ques ons.
  81. 81. Using the IVT to find zeroes Example Let f(x) = x3 − x − 1. Show that there is a zero for f on the interval [1, 2].
  82. 82. Using the IVT to find zeroes Example Let f(x) = x3 − x − 1. Show that there is a zero for f on the interval [1, 2]. Solu on f(1) = −1 and f(2) = 5. So there is a zero between 1 and 2.
  83. 83. Using the IVT to find zeroes Example Let f(x) = x3 − x − 1. Show that there is a zero for f on the interval [1, 2]. Solu on f(1) = −1 and f(2) = 5. So there is a zero between 1 and 2. In fact, we can “narrow in” on the zero by the method of bisec ons.
  84. 84. Finding a zero by bisection y x f(x) . x
  85. 85. Finding a zero by bisection y x f(x) 1 −1 . x
  86. 86. Finding a zero by bisection y x f(x) 1 −1 2 5 . x
  87. 87. Finding a zero by bisection y x f(x) 1 −1 1.5 0.875 2 5 . x
  88. 88. Finding a zero by bisection y x f(x) 1 −1 1.25 − 0.296875 1.5 0.875 2 5 . x
  89. 89. Finding a zero by bisection y x f(x) 1 −1 1.25 − 0.296875 1.375 0.224609 1.5 0.875 2 5 . x
  90. 90. Finding a zero by bisection y x f(x) 1 −1 1.25 − 0.296875 1.3125 − 0.0515137 1.375 0.224609 1.5 0.875 2 5 . x
  91. 91. Finding a zero by bisection y x f(x) 1 −1 1.25 − 0.296875 1.3125 − 0.0515137 1.375 0.224609 1.5 0.875 2 5 . x (More careful analysis yields 1.32472.)
  92. 92. Using the IVT to assert existenceof numbers Example Suppose we are unaware of the square root func on and that it’s con nuous. Prove that the square root of two exists.
  93. 93. Using the IVT to assert existenceof numbers Example Suppose we are unaware of the square root func on and that it’s con nuous. Prove that the square root of two exists. Proof. Let f(x) = x2 , a con nuous func on on [1, 2].
  94. 94. Using the IVT to assert existenceof numbers Example Suppose we are unaware of the square root func on and that it’s con nuous. Prove that the square root of two exists. Proof. Let f(x) = x2 , a con nuous func on 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.
  95. 95. Outline Con nuity The Intermediate Value Theorem Back to the Ques ons
  96. 96. Back to the Questions True or False At one point in your life you were exactly three feet tall.
  97. 97. Question 1 Answer The answer is TRUE. Let h(t) be height, which varies con nuously over me. Then h(birth) 3 ft and h(now) 3 ft. So by the IVT there is a point c in (birth, now) where h(c) = 3.
  98. 98. 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.
  99. 99. Question 2 Answer The answer is TRUE. Let h(t) be height in inches and w(t) be weight in pounds, both varying con nuously over me. Let f(t) = h(t) − w(t). For most of us (call your mom), f(birth) 0 and f(now) 0. So by the IVT there is a point c in (birth, now) where f(c) = 0. In other words, h(c) − w(c) = 0 ⇐⇒ h(c) = w(c).
  100. 100. 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.
  101. 101. Question 3 Answer The answer is TRUE. 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?
  102. 102. Question 3 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!
  103. 103. SummaryWhat have we learned today? Defini on: a func on is con nuous at a point if the limit of the func on at that point agrees with the value of the func on at that point. We o en make a fundamental assump on that func ons we meet in nature are con nuous. The Intermediate Value Theorem is a basic property of real numbers that we need and use a lot.
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