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Lesson 3: Limit Laws

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  • 1. Section 1.4 Calculating Limits V63.0121.002.2010Su, Calculus I New York University May 18, 2010 Announcements WebAssign Class Key: nyu 0127 7953 Office Hours: MR 5:00–5:45, TW 7:50–8:30, CIWW 102 (here) Quiz 1 Thursday on 1.1–1.4 . . . . . .
  • 2. Announcements WebAssign Class Key: nyu 0127 7953 Office Hours: MR 5:00–5:45, TW 7:50–8:30, CIWW 102 (here) Quiz 1 Thursday on 1.1–1.4 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 2 / 37
  • 3. Objectives Know basic limits like lim x = a and lim c = c. x→a x→a Use the limit laws to compute elementary limits. Use algebra to simplify limits. Understand and state the Squeeze Theorem. Use the Squeeze Theorem to demonstrate a limit. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 3 / 37
  • 4. Outline Basic Limits Limit Laws The direct substitution property Limits with Algebra Two more limit theorems Two important trigonometric limits . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 4 / 37
  • 5. Really basic limits Fact Let c be a constant and a a real number. (i) lim x = a x→a (ii) lim c = c x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 5 / 37
  • 6. Really basic limits Fact Let c be a constant and a a real number. (i) lim x = a x→a (ii) lim c = c x→a Proof. The first is tautological, the second is trivial. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 5 / 37
  • 7. ET game for f(x) = x y . . x . . . . . . .
  • 8. ET game for f(x) = x y . . x . . . . . . .
  • 9. ET game for f(x) = x y . . . a . . x . a . . . . . . .
  • 10. ET game for f(x) = x y . . . a . . x . a . . . . . . .
  • 11. ET game for f(x) = x y . . . a . . x . a . . . . . . .
  • 12. ET game for f(x) = x y . . . a . . x . a . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 6 / 37
  • 13. ET game for f(x) = x y . . . a . . x . a . Setting error equal to tolerance works! . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 6 / 37
  • 14. ET game for f(x) = c . . . . . . .
  • 15. ET game for f(x) = c y . . x . . . . . . .
  • 16. ET game for f(x) = c y . . x . . . . . . .
  • 17. ET game for f(x) = c y . . c . . . x . a . . . . . . .
  • 18. ET game for f(x) = c y . . c . . . x . a . . . . . . .
  • 19. ET game for f(x) = c y . . c . . . x . a . . . . . . .
  • 20. ET game for f(x) = c y . . c . . . x . a . any tolerance works! . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 7 / 37
  • 21. Really basic limits Fact Let c be a constant and a a real number. (i) lim x = a x→a (ii) lim c = c x→a Proof. The first is tautological, the second is trivial. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 8 / 37
  • 22. Outline Basic Limits Limit Laws The direct substitution property Limits with Algebra Two more limit theorems Two important trigonometric limits . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 9 / 37
  • 23. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 10 / 37
  • 24. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M (errors add) x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 10 / 37
  • 25. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M (errors add) x→a 2. lim [f(x) − g(x)] = L − M x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 11 / 37
  • 26. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M (errors add) x→a 2. lim [f(x) − g(x)] = L − M x→a 3. lim [cf(x)] = cL x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 11 / 37
  • 27. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M (errors add) x→a 2. lim [f(x) − g(x)] = L − M x→a 3. lim [cf(x)] = cL (error scales) x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 11 / 37
  • 28. Justification of the scaling law errors scale: If f(x) is e away from L, then (c · f(x) − c · L) = c · (f(x) − L) = c · e That is, (c · f)(x) is c · e away from cL, So if Player 2 gives us an error of 1 (for instance), Player 1 can use the fact that lim f(x) = L to find a tolerance for f and g x→a corresponding to the error 1/c. Player 1 wins the round. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 12 / 37
  • 29. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M (errors add) x→a 2. lim [f(x) − g(x)] = L − M (combination of adding and scaling) x→a 3. lim [cf(x)] = cL (error scales) x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 13 / 37
  • 30. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M (errors add) x→a 2. lim [f(x) − g(x)] = L − M (combination of adding and scaling) x→a 3. lim [cf(x)] = cL (error scales) x→a 4. lim [f(x)g(x)] = L · M x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 14 / 37
  • 31. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M (errors add) x→a 2. lim [f(x) − g(x)] = L − M (combination of adding and scaling) x→a 3. lim [cf(x)] = cL (error scales) x→a 4. lim [f(x)g(x)] = L · M (more complicated, but doable) x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 14 / 37
  • 32. Limits and arithmetic II Fact (Continued) f(x) L 5. lim = , if M ̸= 0. x→a g(x) M . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 15 / 37
  • 33. Caution! The quotient rule for limits says that if lim g(x) ̸= 0, then x→a f(x) limx→a f(x) lim = x→a g(x) limx→a g(x) It does NOT say that if lim g(x) = 0, then x→a f(x) lim does not exist x→a g(x) In fact, it can. more about this later . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 16 / 37
  • 34. Limits and arithmetic II Fact (Continued) f(x) L 5. lim = , if M ̸= 0. g(x) x→a M [ ]n n 6. lim [f(x)] = lim f(x) x→a x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 17 / 37
  • 35. Limits and arithmetic II Fact (Continued) f(x) L 5. lim = , if M ̸= 0. g(x) x→a M [ ]n n 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly) x→a x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 17 / 37
  • 36. Limits and arithmetic II Fact (Continued) f(x) L 5. lim = , if M ̸= 0. g(x) x→a M [ ]n n 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly) x→a x→a n n 7. lim x = a x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 17 / 37
  • 37. Limits and arithmetic II Fact (Continued) f(x) L 5. lim = , if M ̸= 0. g(x) x→a M [ ]n n 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly) x→a x→a n n 7. lim x = a x→a √ √ 8. lim n x = n a x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 17 / 37
  • 38. Limits and arithmetic II Fact (Continued) f(x) L 5. lim = , if M ̸= 0. g(x) x→a M [ ]n n 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly) x→a x→a n n 7. lim x = a (follows from 6) x→a √ √ 8. lim n x = n a x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 17 / 37
  • 39. Limits and arithmetic II Fact (Continued) f(x) L 5. lim = , if M ̸= 0. g(x) x→a M [ ]n n 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly) x→a x→a n n 7. lim x = a (follows from 6) x→a √ √ 8. lim n x = n a x→a √ √ 9. lim n f(x) = n lim f(x) (If n is even, we must additionally assume x→a x→a that lim f(x) > 0) x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 17 / 37
  • 40. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 18 / 37
  • 41. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: ( ) lim x2 + 2x + 4 x→3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 18 / 37
  • 42. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: ( ) ( ) lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4) x→3 x→3 x→3 x→3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 18 / 37
  • 43. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: ( ) ( ) lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4) x→3 x→3 x→3 x→3 ( )2 = lim x + 2 · lim (x) + 4 x→3 x→3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 18 / 37
  • 44. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: ( ) ( ) lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4) x→3 x→3 x→3 x→3 ( )2 = lim x + 2 · lim (x) + 4 x→3 x→3 = (3)2 + 2 · 3 + 4 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 18 / 37
  • 45. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: ( ) ( ) lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4) x→3 x→3 x→3 x→3 ( )2 = lim x + 2 · lim (x) + 4 x→3 x→3 = (3)2 + 2 · 3 + 4 = 9 + 6 + 4 = 19. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 18 / 37
  • 46. Your turn Example x2 + 2x + 4 Find lim x→3 x3 + 11 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 19 / 37
  • 47. Your turn Example x2 + 2x + 4 Find lim x→3 x3 + 11 Solution 19 1 The answer is = . 38 2 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 19 / 37
  • 48. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 20 / 37
  • 49. Outline Basic Limits Limit Laws The direct substitution property Limits with Algebra Two more limit theorems Two important trigonometric limits . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 21 / 37
  • 50. Limits do not see the point! (in a good way) Theorem If f(x) = g(x) when x ̸= a, and lim g(x) = L, then lim f(x) = L. x→a x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 22 / 37
  • 51. Limits do not see the point! (in a good way) Theorem If f(x) = g(x) when x ̸= a, and lim g(x) = L, then lim f(x) = L. x→a x→a Example x2 + 2x + 1 Find lim , if it exists. x→−1 x+1 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 22 / 37
  • 52. Limits do not see the point! (in a good way) Theorem If f(x) = g(x) when x ̸= a, and lim g(x) = L, then lim f(x) = L. x→a x→a Example x2 + 2x + 1 Find lim , if it exists. x→−1 x+1 Solution x2 + 2x + 1 Since = x + 1 whenever x ̸= −1, and since x+1 x2 + 2x + 1 lim x + 1 = 0, we have lim = 0. x→−1 x→−1 x+1 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 22 / 37
  • 53. x2 + 2x + 1 ET game for f(x) = x+1 y . . . x . − . 1 Even if f(−1) were something else, it would not effect the limit. . . . . . .
  • 54. x2 + 2x + 1 ET game for f(x) = x+1 y . . . x . − . 1 Even if f(−1) were something else, it would not effect the limit. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 23 / 37
  • 55. Limit of a function defined piecewise at a boundary point Example Let { x2 x ≥ 0 . f(x) = −x x < 0 Does lim f(x) exist? x→0 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37
  • 56. Limit of a function defined piecewise at a boundary point Example Let { x2 x ≥ 0 . f(x) = −x x < 0 Does lim f(x) exist? x→0 Solution We have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37
  • 57. Limit of a function defined piecewise at a boundary point Example Let { x2 x ≥ 0 . f(x) = −x x < 0 Does lim f(x) exist? x→0 Solution We have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37
  • 58. Limit of a function defined piecewise at a boundary point Example Let { x2 x ≥ 0 . f(x) = −x x < 0 Does lim f(x) exist? x→0 Solution We have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37
  • 59. Limit of a function defined piecewise at a boundary point Example Let { x2 x ≥ 0 . f(x) = −x x < 0 Does lim f(x) exist? x→0 Solution We have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 Likewise: lim f(x) = lim −x = −0 = 0 x→0− x→0− . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37
  • 60. Limit of a function defined piecewise at a boundary point Example Let { x2 x ≥ 0 . f(x) = −x x < 0 Does lim f(x) exist? x→0 Solution We have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 Likewise: lim f(x) = lim −x = −0 = 0 x→0− x→0− . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37
  • 61. Limit of a function defined piecewise at a boundary point Example Let { x2 x ≥ 0 . f(x) = −x x < 0 Does lim f(x) exist? x→0 Solution We have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 Likewise: lim f(x) = lim −x = −0 = 0 x→0− x→0− . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37
  • 62. Finding limits by algebraic manipulations Example √ x−2 Find lim . x→4 x − 4 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 25 / 37
  • 63. Finding limits by algebraic manipulations Example √ x−2 Find lim . x→4 x − 4 Solution √ 2 √ √ Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 25 / 37
  • 64. Finding limits by algebraic manipulations Example √ x−2 Find lim . x→4 x − 4 Solution √ 2 √ √ Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). So √ √ x−2 x−2 lim = lim √ √ x→4 x − 4 x→4 ( x − 2)( x + 2) 1 1 = lim √ = x→4 x+2 4 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 25 / 37
  • 65. Your turn Example Let { 1 − x2 x≥1 f(x) = 2x x<1 Find lim f(x) if it exists. x→1 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 26 / 37
  • 66. Your turn Example Let { 1 − x2 x≥1 f(x) = 2x x<1 Find lim f(x) if it exists. x→1 Solution We have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 26 / 37
  • 67. Your turn Example Let { 1 − x2 x≥1 f(x) = 2x x<1 . . Find lim f(x) if it exists. 1 . x→1 Solution We have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 26 / 37
  • 68. Your turn Example Let { 1 − x2 x≥1 f(x) = 2x x<1 . . Find lim f(x) if it exists. 1 . x→1 Solution We have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 DSP lim f(x) = lim (2x) = 2 − − x→1 x→1 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 26 / 37
  • 69. Your turn Example . Let { 1 − x2 x≥1 f(x) = 2x x<1 . . Find lim f(x) if it exists. 1 . x→1 Solution We have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 DSP lim f(x) = lim (2x) = 2 − − x→1 x→1 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 26 / 37
  • 70. Your turn Example . Let { 1 − x2 x≥1 f(x) = 2x x<1 . . Find lim f(x) if it exists. 1 . x→1 Solution We have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 DSP lim f(x) = lim (2x) = 2 − − x→1 x→1 The left- and right-hand limits disagree, so the limit does not exist. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 26 / 37
  • 71. A message from the Mathematical Grammar Police Please do not say “ lim f(x) = DNE.” Does not compute. x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 27 / 37
  • 72. A message from the Mathematical Grammar Police Please do not say “ lim f(x) = DNE.” Does not compute. x→a Too many verbs . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 27 / 37
  • 73. A message from the Mathematical Grammar Police Please do not say “ lim f(x) = DNE.” Does not compute. x→a Too many verbs Leads to FALSE limit laws like “If lim f(x) DNE and lim g(x) DNE, x→a x→a then lim (f(x) + g(x)) DNE.” x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 27 / 37
  • 74. Two More Important Limit Theorems Theorem If f(x) ≤ g(x) when x is near a (except possibly at a), then lim f(x) ≤ lim g(x) x→a x→a (as usual, provided these limits exist). Theorem (The Squeeze/Sandwich/Pinching Theorem) If f(x) ≤ g(x) ≤ h(x) when x is near a (as usual, except possibly at a), and lim f(x) = lim h(x) = L, x→a x→a then lim g(x) = L. x→a . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 28 / 37
  • 75. Using the Squeeze Theorem We can use the Squeeze Theorem to replace complicated expressions with simple ones when taking the limit. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 29 / 37
  • 76. Using the Squeeze Theorem We can use the Squeeze Theorem to replace complicated expressions with simple ones when taking the limit. Example (π ) Show that lim x2 sin = 0. x→0 x . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 29 / 37
  • 77. Using the Squeeze Theorem We can use the Squeeze Theorem to replace complicated expressions with simple ones when taking the limit. Example (π ) Show that lim x2 sin = 0. x→0 x Solution We have for all x, (π ) (π ) −1 ≤ sin ≤ 1 =⇒ −x2 ≤ x2 sin ≤ x2 x x The left and right sides go to zero as x → 0. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 29 / 37
  • 78. Illustration of the Squeeze Theorem y . . (x) = x2 h . x . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 30 / 37
  • 79. Illustration of the Squeeze Theorem y . . (x) = x2 h . x . .(x) = −x2 f . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 30 / 37
  • 80. Illustration of the Squeeze Theorem y . . (x) = x2 h (π ) . (x) = x2 sin g x . x . .(x) = −x2 f . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 30 / 37
  • 81. Outline Basic Limits Limit Laws The direct substitution property Limits with Algebra Two more limit theorems Two important trigonometric limits . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 31 / 37
  • 82. Two important trigonometric limits Theorem The following two limits hold: sin θ lim =1 θ→0 θ cos θ − 1 lim =0 θ→0 θ . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 32 / 37
  • 83. Proof of the Sine Limit Proof. Notice θ . θ . θ . 1 . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 33 / 37
  • 84. Proof of the Sine Limit Proof. Notice sin θ ≤ θ . in θ . s θ . θ . c . os θ 1 . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 33 / 37
  • 85. Proof of the Sine Limit Proof. Notice sin θ ≤ θ tan θ . in θ . .an θ s θ t . θ . c . os θ 1 . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 33 / 37
  • 86. Proof of the Sine Limit Proof. Notice θ sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 . in θ . .an θ s θ t . θ . c . os θ 1 . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 33 / 37
  • 87. Proof of the Sine Limit Proof. Notice θ sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 Divide by sin θ: θ 1 1≤ ≤ sin θ cos θ . in θ . .an θ s θ t . θ . c . os θ 1 . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 33 / 37
  • 88. Proof of the Sine Limit Proof. Notice θ sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 Divide by sin θ: θ 1 1≤ ≤ sin θ cos θ . in θ . .an θ s θ t . θ Take reciprocals: . c . os θ 1 . sin θ 1≥ ≥ cos θ θ . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 33 / 37
  • 89. Proof of the Sine Limit Proof. Notice θ sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 Divide by sin θ: θ 1 1≤ ≤ sin θ cos θ . in θ . .an θ s θ t . θ Take reciprocals: . c . os θ 1 . sin θ 1≥ ≥ cos θ θ As θ → 0, the left and right sides tend to 1. So, then, must the middle expression. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 33 / 37
  • 90. Proof of the Cosine Limit Proof. 1 − cos θ 1 − cos θ 1 + cos θ 1 − cos2 θ = · = θ θ 1 + cos θ θ(1 + cos θ) sin2 θ sin θ sin θ = = · θ(1 + cos θ) θ 1 + cos θ So ( ) ( ) 1 − cos θ sin θ sin θ lim = lim · lim θ→0 θ θ→0 θ θ→0 1 + cos θ = 1 · 0 = 0. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 34 / 37
  • 91. Try these Example tan θ 1. lim θ→0 θ sin 2θ 2. lim θ→0 θ . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 35 / 37
  • 92. Try these Example tan θ 1. lim θ→0 θ sin 2θ 2. lim θ→0 θ Answer 1. 1 2. 2 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 35 / 37
  • 93. Solutions 1. Use the basic trigonometric limit and the definition of tangent. tan θ sin θ sin θ 1 1 lim = lim = lim · lim = 1 · = 1. θ→0 θ θ→0 θ cos θ θ→0 θ θ→0 cos θ 1 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 36 / 37
  • 94. Solutions 1. Use the basic trigonometric limit and the definition of tangent. tan θ sin θ sin θ 1 1 lim = lim = lim · lim = 1 · = 1. θ→0 θ θ→0 θ cos θ θ→0 θ θ→0 cos θ 1 2. Change the variable: sin 2θ sin 2θ sin 2θ lim = lim = 2 · lim =2·1=2 θ→0 θ 2θ→0 2θ · 1 2θ→0 2θ 2 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 36 / 37
  • 95. Solutions 1. Use the basic trigonometric limit and the definition of tangent. tan θ sin θ sin θ 1 1 lim = lim = lim · lim = 1 · = 1. θ→0 θ θ→0 θ cos θ θ→0 θ θ→0 cos θ 1 2. Change the variable: sin 2θ sin 2θ sin 2θ lim = lim = 2 · lim =2·1=2 θ→0 θ 2θ→0 2θ · 1 2θ→0 2θ 2 OR use a trigonometric identity: sin 2θ 2 sin θ cos θ sin θ lim = lim = 2· lim · lim cos θ = 2·1·1 = 2 θ→0 θ θ→0 θ θ→0 θ θ→0 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 36 / 37
  • 96. Summary Limits laws say limits play well with the rules of arithmetic When limit laws do not work we can be algebraically creative When algebra does not work we can try Squeezing. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 37 / 37

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