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# Lesson 4: Calculating Limits (Section 41 slides)

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Basic limits, limit laws, algebraic limits, and limits of some trigonometric quotients.

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### Lesson 4: Calculating Limits (Section 41 slides)

1. 1. Section 1.4 Calculating Limits V63.0121.041, Calculus I New York University September 15, 2010Announcements First written homework due today (put it in the envelope) Remember to put your lecture and recitation section numbers on your paper . . . . . .
2. 2. Announcements First written homework due today (put it in the envelope) Remember to put your lecture and recitation section numbers on your paper . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 2 / 45
3. 3. Yoda on teaching a concepts course“You must unlearn what you have learned.”In other words, we are building up concepts and allowing ourselvesonly to speak in terms of what we personally have produced. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 3 / 45
4. 4. 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 4 / 45
5. 5. OutlineRecall: The concept of limitBasic LimitsLimit Laws The direct substitution propertyLimits with Algebra Two more limit theoremsTwo important trigonometric limits . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 5 / 45
6. 6. Heuristic Definition of a LimitDefinitionWe write lim f(x) = L x→aand 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 aswe like) by taking x to be sufficiently close to a (on either side of a) butnot equal to a. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 6 / 45
7. 7. The error-tolerance gameA game between two players (Dana and Emerson) to decide if a limitlim f(x) exists.x→aStep 1 Dana proposes L to be the limit.Step 2 Emerson challenges with an “error” level around L.Step 3 Dana chooses a “tolerance” level around a so that points x within that tolerance of a (not counting a itself) are taken to values y within the error level of L. If Dana cannot, Emerson wins and the limit cannot be L.Step 4 If Dana’s move is a good one, Emerson can challenge again or give up. If Emerson gives up, Dana wins and the limit is L. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 7 / 45
8. 8. The error-tolerance game L . . a . To be legit, the part of the graph inside the blue (vertical) strip must also be inside the green (horizontal) strip. If Emerson shrinks the error, Dana can still win. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 8 / 45
9. 9. Limit FAIL: Jump y . . . 1 . x . . Part of graph . 1. − inside blue is not inside green . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 9 / 45
10. 10. Limit FAIL: Jump y . . Part of graph inside blue is not . . 1 inside green . x . . 1. − . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 9 / 45
11. 11. Limit FAIL: Jump y . . Part of graph inside blue is not . . 1 inside green . x . . 1. − |x| So lim does not exist. x→0 x . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 9 / 45
12. 12. Limit FAIL: unboundedness y . . 1 lim+ does not exist x→0 x because the function is unbounded near 0 .?. L . x . 0 . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 10 / 45
13. 13. Limit EPIC FAIL (π )Here is a graph of the function f(x) = sin : x y . . . 1 . x . . 1. −For every y in [−1, 1], there are infinitely many points x arbitrarily closeto zero where f(x) = y. So lim f(x) cannot exist. x→0 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 11 / 45
14. 14. OutlineRecall: The concept of limitBasic LimitsLimit Laws The direct substitution propertyLimits with Algebra Two more limit theoremsTwo important trigonometric limits . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 12 / 45
15. 15. Really basic limitsFactLet c be a constant and a a real number. (i) lim x = a x→a (ii) lim c = c x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 13 / 45
16. 16. Really basic limitsFactLet c be a constant and a a real number. (i) lim x = a x→a (ii) lim c = c x→aProof.The first is tautological, the second is trivial. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 13 / 45
17. 17. ET game for f(x) = x y . . x . . . . . . .
18. 18. ET game for f(x) = x y . . x . . . . . . .
19. 19. ET game for f(x) = x y . . . a . . x . a . . . . . . .
20. 20. ET game for f(x) = x y . . . a . . x . a . . . . . . .
21. 21. ET game for f(x) = x y . . . a . . x . a . . . . . . .
22. 22. ET game for f(x) = x y . . . a . . x . a . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 14 / 45
23. 23. ET game for f(x) = x y . . . a . . x . a . Setting error equal to tolerance works! . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 14 / 45
24. 24. ET game for f(x) = c . . . . . . .
25. 25. ET game for f(x) = c y . . x . . . . . . .
26. 26. ET game for f(x) = c y . . x . . . . . . .
27. 27. ET game for f(x) = c y . . c . . . x . a . . . . . . .
28. 28. ET game for f(x) = c y . . c . . . x . a . . . . . . .
29. 29. ET game for f(x) = c y . . c . . . x . a . . . . . . .
30. 30. ET game for f(x) = c y . . c . . . x . a . any tolerance works! . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 15 / 45
31. 31. Really basic limitsFactLet c be a constant and a a real number. (i) lim x = a x→a (ii) lim c = c x→aProof.The first is tautological, the second is trivial. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 16 / 45
32. 32. OutlineRecall: The concept of limitBasic LimitsLimit Laws The direct substitution propertyLimits with Algebra Two more limit theoremsTwo important trigonometric limits . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 17 / 45
33. 33. Limits and arithmeticFactSuppose 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 18 / 45
34. 34. Limits and arithmeticFactSuppose 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 18 / 45
35. 35. Limits and arithmeticFactSuppose 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 19 / 45
36. 36. Limits and arithmeticFactSuppose 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 19 / 45
37. 37. Limits and arithmeticFactSuppose 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 19 / 45
38. 38. 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 Emerson gives us an error of 1 (for instance), Dana 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. Dana wins the round. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 20 / 45
39. 39. Limits and arithmeticFactSuppose 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 21 / 45
40. 40. Limits and arithmeticFactSuppose 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 22 / 45
41. 41. Limits and arithmeticFactSuppose 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 22 / 45
42. 42. Limits and arithmetic IIFact (Continued) f(x) L 5. lim = , if M ̸= 0. x→a g(x) M . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 23 / 45
43. 43. 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, limits of quotients where numerator and denominator both tend to 0 are exactly where the magic happens. more about this later . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 24 / 45
44. 44. Limits and arithmetic IIFact (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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 25 / 45
45. 45. Limits and arithmetic IIFact (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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 25 / 45
46. 46. Limits and arithmetic IIFact (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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 25 / 45
47. 47. Limits and arithmetic IIFact (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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 25 / 45
48. 48. Limits and arithmetic IIFact (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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 25 / 45
49. 49. Limits and arithmetic IIFact (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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 25 / 45
50. 50. Applying the limit lawsExample ( )Find lim x2 + 2x + 4 . x→3 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 26 / 45
51. 51. Applying the limit lawsExample ( )Find lim x2 + 2x + 4 . x→3SolutionBy applying the limit laws repeatedly: ( ) lim x2 + 2x + 4 x→3 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 26 / 45
52. 52. Applying the limit lawsExample ( )Find lim x2 + 2x + 4 . x→3SolutionBy applying the limit laws repeatedly: ( ) ( ) lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4) x→3 x→3 x→3 x→3 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 26 / 45
53. 53. Applying the limit lawsExample ( )Find lim x2 + 2x + 4 . x→3SolutionBy 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 26 / 45
54. 54. Applying the limit lawsExample ( )Find lim x2 + 2x + 4 . x→3SolutionBy 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 26 / 45
55. 55. Applying the limit lawsExample ( )Find lim x2 + 2x + 4 . x→3SolutionBy 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 26 / 45
56. 56. Your turnExample x2 + 2x + 4Find lim x→3 x3 + 11 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 27 / 45
57. 57. Your turnExample x2 + 2x + 4Find lim x→3 x3 + 11Solution 19 1The answer is = . 38 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 27 / 45
58. 58. Direct Substitution PropertyTheorem (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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 28 / 45
59. 59. OutlineRecall: The concept of limitBasic LimitsLimit Laws The direct substitution propertyLimits with Algebra Two more limit theoremsTwo important trigonometric limits . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 29 / 45
60. 60. Limits do not see the point! (in a good way)TheoremIf f(x) = g(x) when x ̸= a, and lim g(x) = L, then lim f(x) = L. x→a x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 30 / 45
61. 61. Limits do not see the point! (in a good way)TheoremIf f(x) = g(x) when x ̸= a, and lim g(x) = L, then lim f(x) = L. x→a x→aExample x2 + 2x + 1Find lim , if it exists. x→−1 x+1 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 30 / 45
62. 62. Limits do not see the point! (in a good way)TheoremIf f(x) = g(x) when x ̸= a, and lim g(x) = L, then lim f(x) = L. x→a x→aExample x2 + 2x + 1Find lim , if it exists. x→−1 x+1Solution x2 + 2x + 1Since = 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 30 / 45
63. 63. x2 + 2x + 1ET game for f(x) = x+1 y . . . x . − . 1 Even if f(−1) were something else, it would not effect the limit. . . . . . .
64. 64. x2 + 2x + 1ET game for f(x) = x+1 y . . . x . − . 1 Even if f(−1) were something else, it would not effect the limit. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 31 / 45
65. 65. Limit of a function defined piecewise at a boundarypointExampleLet { x2 x ≥ 0 . f(x) = −x x < 0Does lim f(x) exist? x→0 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45
66. 66. Limit of a function defined piecewise at a boundarypointExampleLet { x2 x ≥ 0 . f(x) = −x x < 0Does lim f(x) exist? x→0SolutionWe have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45
67. 67. Limit of a function defined piecewise at a boundarypointExampleLet { x2 x ≥ 0 . f(x) = −x x < 0Does lim f(x) exist? x→0SolutionWe have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45
68. 68. Limit of a function defined piecewise at a boundarypointExampleLet { x2 x ≥ 0 . f(x) = −x x < 0Does lim f(x) exist? x→0SolutionWe have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45
69. 69. Limit of a function defined piecewise at a boundarypointExampleLet { x2 x ≥ 0 . f(x) = −x x < 0Does lim f(x) exist? x→0SolutionWe have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0Likewise: lim f(x) = lim −x = −0 = 0 x→0− x→0− . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45
70. 70. Limit of a function defined piecewise at a boundarypointExampleLet { x2 x ≥ 0 . f(x) = −x x < 0Does lim f(x) exist? x→0SolutionWe have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0Likewise: lim f(x) = lim −x = −0 = 0 x→0− x→0− . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45
71. 71. Limit of a function defined piecewise at a boundarypointExampleLet { x2 x ≥ 0 . f(x) = −x x < 0Does lim f(x) exist? x→0SolutionWe have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0Likewise: lim f(x) = lim −x = −0 = 0 x→0− x→0− . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45
72. 72. Finding limits by algebraic manipulationsExample √ x−2Find lim . x→4 x − 4 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 33 / 45
73. 73. Finding limits by algebraic manipulationsExample √ x−2Find lim . x→4 x − 4Solution √ 2 √ √Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 33 / 45
74. 74. Finding limits by algebraic manipulationsExample √ x−2Find lim . x→4 x − 4Solution √ 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 33 / 45
75. 75. Your turnExampleLet { 1 − x2 x≥1 f(x) = 2x x<1Find lim f(x) if it exists. x→1 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 34 / 45
76. 76. Your turnExampleLet { 1 − x2 x≥1 f(x) = 2x x<1Find lim f(x) if it exists. x→1SolutionWe have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 34 / 45
77. 77. Your turnExampleLet { 1 − x2 x≥1 f(x) = 2x x<1 . .Find lim f(x) if it exists. 1 . x→1SolutionWe have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 34 / 45
78. 78. Your turnExampleLet { 1 − x2 x≥1 f(x) = 2x x<1 . .Find lim f(x) if it exists. 1 . x→1SolutionWe have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 DSP lim f(x) = lim (2x) = 2 − − x→1 x→1 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 34 / 45
79. 79. Your turnExample .Let { 1 − x2 x≥1 f(x) = 2x x<1 . .Find lim f(x) if it exists. 1 . x→1SolutionWe have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 DSP lim f(x) = lim (2x) = 2 − − x→1 x→1 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 34 / 45
80. 80. Your turnExample .Let { 1 − x2 x≥1 f(x) = 2x x<1 . .Find lim f(x) if it exists. 1 . x→1SolutionWe have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 DSP lim f(x) = lim (2x) = 2 − − x→1 x→1The left- and right-hand limits disagree, so the limit does not exist. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 34 / 45
81. 81. A message from the Mathematical Grammar PolicePlease do not say “ lim f(x) = DNE.” Does not compute. x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 35 / 45
82. 82. A message from the Mathematical Grammar PolicePlease do not say “ lim f(x) = DNE.” Does not compute. x→a Too many verbs . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 35 / 45
83. 83. A message from the Mathematical Grammar PolicePlease 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 35 / 45
84. 84. Two More Important Limit TheoremsTheoremIf 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→athen lim g(x) = L. x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 36 / 45
85. 85. Using the Squeeze TheoremWe can use the Squeeze Theorem to replace complicated expressionswith simple ones when taking the limit. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 37 / 45
86. 86. Using the Squeeze TheoremWe can use the Squeeze Theorem to replace complicated expressionswith simple ones when taking the limit.Example (π )Show that lim x2 sin = 0. x→0 x . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 37 / 45
87. 87. Using the Squeeze TheoremWe can use the Squeeze Theorem to replace complicated expressionswith simple ones when taking the limit.Example (π )Show that lim x2 sin = 0. x→0 xSolutionWe have for all x, (π ) (π ) −1 ≤ sin ≤ 1 =⇒ −x2 ≤ x2 sin ≤ x2 x xThe left and right sides go to zero as x → 0. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 37 / 45
88. 88. Illustration of the Squeeze Theorem y . . (x) = x2 h . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 38 / 45
89. 89. Illustration of the Squeeze Theorem y . . (x) = x2 h . x . .(x) = −x2 f . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 38 / 45
90. 90. Illustration of the Squeeze Theorem y . . (x) = x2 h (π ) . (x) = x2 sin g x . x . .(x) = −x2 f . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 38 / 45
91. 91. OutlineRecall: The concept of limitBasic LimitsLimit Laws The direct substitution propertyLimits with Algebra Two more limit theoremsTwo important trigonometric limits . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 39 / 45
92. 92. Two important trigonometric limitsTheoremThe following two limits hold: sin θ lim =1 θ→0 θ cos θ − 1 lim =0 θ→0 θ . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 40 / 45
93. 93. Proof of the Sine LimitProof. Notice θ . θ . θ . 1 . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 41 / 45
94. 94. Proof of the Sine LimitProof. Notice sin θ ≤ θ . in θ . s θ . θ . c . os θ 1 . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 41 / 45
95. 95. Proof of the Sine LimitProof. Notice sin θ ≤ θ tan θ . in θ . .an θ s θ t . θ . c . os θ 1 . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 41 / 45
96. 96. Proof of the Sine LimitProof. Notice θ sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 . in θ . .an θ s θ t . θ . c . os θ 1 . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 41 / 45
97. 97. Proof of the Sine LimitProof. 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 . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 41 / 45
98. 98. Proof of the Sine LimitProof. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 41 / 45
99. 99. Proof of the Sine LimitProof. 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 θAs θ → 0, the left and right sides tend to 1. So, then, must the middleexpression. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 41 / 45
100. 100. Proof of the Cosine LimitProof. 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 θSo ( ) ( ) 1 − cos θ sin θ sin θ lim = lim · lim θ→0 θ θ→0 θ θ→0 1 + cos θ = 1 · 0 = 0. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 42 / 45
101. 101. Try theseExample tan θ 1. lim θ→0 θ sin 2θ 2. lim θ→0 θ . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 43 / 45
102. 102. Try theseExample tan θ 1. lim θ→0 θ sin 2θ 2. lim θ→0 θAnswer 1. 1 2. 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 43 / 45
103. 103. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 44 / 45
104. 104. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 44 / 45
105. 105. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 44 / 45
106. 106. Summary The limit laws allow us to compute limits reasonably. BUT we cannot make up extra laws otherwise we get into trouble. . . . . . .V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 45 / 45
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