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- 1. Sec on 1.4 Calcula ng Limits V63.0121.001: Calculus I Professor Ma hew Leingang New York University February 2, 2011 Announcements First wri en HW due today.
- 2. Announcements First wri en HW due today Get-to-know-you survey and photo deadline is February 11
- 3. Objectives Know basic limits like lim x = a x→a and lim c = c. 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.
- 4. Limit.
- 5. Yoda on teaching course concepts You must unlearn what you have learned. In other words, we are building up concepts and allowing ourselves only to speak in terms of what we personally have produced.
- 6. Outline Recall: The concept of limit Basic Limits Limit Laws The direct subs tu on property Limits with Algebra Two more limit theorems Two important trigonometric limits
- 7. Heuristic Deﬁnition of a Limit 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.
- 8. The error-tolerance game A game between two players (Dana and Emerson) to decide if a limit lim f(x) exists. x→a Step 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 coun ng 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.
- 9. The error-tolerance game L . a To be legit, the part of the graph inside the blue (ver cal) strip must also be inside the green (horizontal) strip. Even if Emerson shrinks the error, Dana can s ll move.
- 10. Limit FAIL: Jump y 1 . x Part of graph −1 inside blue is not inside green
- 11. Limit FAIL: Jump yPart of graphinside blueis not inside 1green . x −1
- 12. Limit FAIL: Jump yPart of graph |x| So lim does notinside blue x→0 xis not inside exist. 1green . x −1
- 13. Limit FAIL: unboundedness y 1 lim+ does not exist be- x→0 x cause the func on is un- bounded near 0 L? . x 0
- 14. Limit EPIC FAIL (π ) Here is a graph of the func on f(x) = sin : x y 1 . x −1 For every y in [−1, 1], there are inﬁnitely many points x arbitrarily close to zero where f(x) = y. So lim f(x) cannot exist. x→0
- 15. Outline Recall: The concept of limit Basic Limits Limit Laws The direct subs tu on property Limits with Algebra Two more limit theorems Two important trigonometric limits
- 16. 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
- 17. 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.
- 18. ET game for f(x) = x y . x
- 19. ET game for f(x) = x y . x
- 20. ET game for f(x) = x y a . x a
- 21. ET game for f(x) = x y a . x a
- 22. ET game for f(x) = x y a . x a
- 23. ET game for f(x) = x y a . x a
- 24. ET game for f(x) = x y a Se ng error equal to tolerance works! . x a
- 25. ET game for f(x) = c .
- 26. ET game for f(x) = c y . x
- 27. ET game for f(x) = c y . x
- 28. ET game for f(x) = c y c . x a
- 29. ET game for f(x) = c y c . x a
- 30. ET game for f(x) = c y c . x a
- 31. ET game for f(x) = c y c any tolerance works! . x a
- 32. 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.
- 33. Outline Recall: The concept of limit Basic Limits Limit Laws The direct subs tu on property Limits with Algebra Two more limit theorems Two important trigonometric limits
- 34. 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
- 35. 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
- 36. 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
- 37. 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
- 38. 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
- 39. 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.
- 40. 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 (combina on of adding and scaling) x→a 3. lim [cf(x)] = cL (error scales) x→a
- 41. 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 (combina on 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
- 42. 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 (combina on 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
- 43. Limits and arithmetic II Fact (Con nued) f(x) L 5. lim = , if M ̸= 0. x→a g(x) M
- 44. Caution! The quo ent 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 quo ents where numerator and denominator both tend to 0 are exactly where the magic happens.
- 45. Limits and arithmetic II Fact (Con nued) f(x) L 5. lim = , if M ̸= 0. x→a g(x) M [ ]n n 6. lim [f(x)] = lim f(x) x→a x→a
- 46. Limits and arithmetic II Fact (Con nued) f(x) L 5. lim = , if M ̸= 0. x→a g(x) M [ ]n n 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly) x→a x→a
- 47. Limits and arithmetic II Fact (Con nued) f(x) L 5. lim = , if M ̸= 0. x→a g(x) 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
- 48. Limits and arithmetic II Fact (Con nued) f(x) L 5. lim = , if M ̸= 0. x→a g(x) 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
- 49. Limits and arithmetic II Fact (Con nued) f(x) L 5. lim = , if M ̸= 0. x→a g(x) 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
- 50. Limits and arithmetic II Fact (Con nued) f(x) L 5. lim = , if M ̸= 0. x→a g(x) 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 √ √ n 9. lim f(x) = n lim f(x) (If n is even, we must addi onally x→a x→a assume that lim f(x) > 0) x→a
- 51. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3
- 52. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solu on By applying the limit laws repeatedly: ( ) lim x2 + 2x + 4 x→3
- 53. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solu on By applying the limit laws repeatedly: ( ) ( ) lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4) x→3 x→3 x→3 x→3
- 54. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solu on 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
- 55. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solu on 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 2 = (3) + 2 · 3 + 4
- 56. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solu on 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 2 = (3) + 2 · 3 + 4 = 9 + 6 + 4 = 19.
- 57. Your turn Example x2 + 2x + 4 Find lim x→3 x3 + 11
- 58. Your turn Example x2 + 2x + 4 Find lim x→3 x3 + 11 Solu on 19 1 The answer is = . 38 2
- 59. Direct Substitution Property As a direct consequence of the limit laws and the really basic limits we have: 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
- 60. Outline Recall: The concept of limit Basic Limits Limit Laws The direct subs tu on property Limits with Algebra Two more limit theorems Two important trigonometric limits
- 61. 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
- 62. Example of the MTP principle Example x2 + 2x + 1 Find lim , if it exists. x→−1 x+1
- 63. Example of the MTP principle Example x2 + 2x + 1 Find lim , if it exists. x→−1 x+1 Solu on 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
- 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.
- 65. 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.
- 66. Limit of a piecewise function Example { x2 x ≥ 0 Let f(x) = . Does lim f(x) exist? −x x < 0 x→0 Solu on .
- 67. Limit of a piecewise function Example { x2 x ≥ 0 Let f(x) = . Does lim f(x) exist? −x x < 0 x→0 Solu on MTP DSP We have lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 .
- 68. Limit of a piecewise function Example { x2 x ≥ 0 Let f(x) = . Does lim f(x) exist? −x x < 0 x→0 Solu on MTP DSP We have lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 .
- 69. Limit of a piecewise function Example { x2 x ≥ 0 Let f(x) = . Does lim f(x) exist? −x x < 0 x→0 Solu on MTP DSP We have lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 .
- 70. Limit of a piecewise function Example { x2 x ≥ 0 Let f(x) = . Does lim f(x) exist? −x x < 0 x→0 Solu on MTP DSP We have lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 Likewise: lim− f(x) = lim− −x = −0 = 0 . x→0 x→0
- 71. Limit of a piecewise function Example { x2 x ≥ 0 Let f(x) = . Does lim f(x) exist? −x x < 0 x→0 Solu on MTP DSP We have lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 Likewise: lim− f(x) = lim− −x = −0 = 0 . x→0 x→0
- 72. Limit of a piecewise function Example { x2 x ≥ 0 Let f(x) = . Does lim f(x) exist? −x x < 0 x→0 Solu on MTP DSP We have lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 Likewise: lim− f(x) = lim− −x = −0 = 0 . x→0 x→0 So lim f(x) = 0. x→0
- 73. Finding limits by algebra Example √ x−2 Find lim . x→4 x−4
- 74. Finding limits by algebra Example √ x−2 Find lim . x→4 x−4 Solu on √ 2 √ √ Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2).
- 75. Finding limits by algebra Example √ x−2 Find lim . x→4 x−4 Solu on √ 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
- 76. Your turn Example { 1 − x2 x≥1 Let f(x) = . Find lim f(x) if it exists. 2x x<1 x→1 Solu on
- 77. Your turn Example { 1 − x2 x≥1 Let f(x) = . Find lim f(x) if it exists. 2x x<1 x→1 Solu on ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1
- 78. Your turn Example { 1 − x2 x≥1 Let f(x) = . Find lim f(x) if it exists. 2x x<1 x→1 Solu on ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 . 1
- 79. Your turn Example { 1 − x2 x≥1 Let f(x) = . Find lim f(x) if it exists. 2x x<1 x→1 Solu on ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 DSP lim− f(x) = lim− (2x) = 2 x→1 x→1 . 1
- 80. Your turn Example { 1 − x2 x≥1 Let f(x) = . Find lim f(x) if it exists. 2x x<1 x→1 Solu on ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 DSP lim− f(x) = lim− (2x) = 2 x→1 x→1 . 1
- 81. Your turn Example { 1 − x2 x≥1 Let f(x) = . Find lim f(x) if it exists. 2x x<1 x→1 Solu on ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 DSP lim− f(x) = lim− (2x) = 2 x→1 x→1 The le - and right-hand limits disagree, so the . limit does not exist. 1
- 82. A message fromthe Mathematical Grammar Police Please do not say “lim f(x) = DNE.” Does not compute. x→a
- 83. A message fromthe Mathematical Grammar Police Please do not say “lim f(x) = DNE.” Does not compute. x→a Too many verbs
- 84. A message fromthe 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
- 85. Two Important Limit Theorems Theorem Theorem (The Squeeze/ If f(x) ≤ g(x) when x is near a Sandwich/ Pinching Theorem) (except possibly at a), then If f(x) ≤ g(x) ≤ h(x) when x is near a (as usual, except lim f(x) ≤ lim g(x) possibly at a), and x→a x→a (as usual, provided these limits lim f(x) = lim h(x) = L, x→a x→a exist). then lim g(x) = L. x→a
- 86. Using the Squeeze Theorem We can use the Squeeze Theorem to replace complicated expressions with simple ones when taking the limit.
- 87. Using the Squeeze Theorem We can use the Squeeze Theorem to replace complicated expressions with simple ones when taking the limit. Example (π ) 2 Show that lim x sin = 0. x→0 x
- 88. Using the Squeeze Theorem We can use the Squeeze Theorem to replace complicated expressions with simple ones when taking the limit. Example (π ) 2 Show that lim x sin = 0. x→0 x Solu on We have for all x, (π ) (π ) −1 ≤ sin ≤ 1 =⇒ −x ≤ x sin 2 2 ≤ x2 x x The le and right sides go to zero as x → 0.
- 89. Illustrating the Squeeze Theorem y h(x) = x2 . x
- 90. Illustrating the Squeeze Theorem y h(x) = x2 . x f(x) = −x2
- 91. Illustrating the Squeeze Theorem y h(x) = x2 (π ) 2 g(x) = x sin x . x f(x) = −x2
- 92. Outline Recall: The concept of limit Basic Limits Limit Laws The direct subs tu on property Limits with Algebra Two more limit theorems Two important trigonometric limits
- 93. Two trigonometric limits Theorem The following two limits hold: sin θ lim =1 θ→0 θ cos θ − 1 lim =0 θ→0 θ
- 94. Proof of the Sine Limit Proof. No ce θ θ . θ 1
- 95. Proof of the Sine Limit Proof. No ce sin θ ≤ θ sin θ θ . θ cos θ 1
- 96. Proof of the Sine Limit Proof. No ce sin θ ≤ θ tan θ sin θ θ tan θ . θ cos θ 1
- 97. Proof of the Sine Limit Proof. θ No ce sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 sin θ θ tan θ . θ cos θ 1
- 98. Proof of the Sine Limit Proof. θ No ce sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 θ 1 Divide by sin θ: 1 ≤ ≤ sin θ cos θ sin θ θ tan θ . θ cos θ 1
- 99. Proof of the Sine Limit Proof. θ No ce sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 θ 1 Divide by sin θ: 1 ≤ ≤ sin θ cos θ sin θ θ tan θ sin θ . θ Take reciprocals: 1 ≥ ≥ cos θ θ cos θ 1
- 100. Proof of the Sine Limit Proof. θ No ce sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 θ 1 Divide by sin θ: 1 ≤ ≤ sin θ cos θ sin θ θ tan θ sin θ . θ Take reciprocals: 1 ≥ ≥ cos θ θ cos θ 1 As θ → 0, the le and right sides tend to 1. So, then, must the middle expression.
- 101. Proof of the Cosine Limit Proof. 1 − cos θ 1 − cos θ 1 + cos θ = · θ θ 1 + cos θ
- 102. 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)
- 103. 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) 2 sin θ = θ(1 + cos θ)
- 104. 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) 2 sin θ sin θ sin θ = = · θ(1 + cos θ) θ 1 + cos θ
- 105. 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) 2 sin 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
- 106. 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) 2 sin θ sin θ sin θ = = · θ(1 + cos θ) θ 1 + cos θ So ( ) ( ) 1 − cos θ sin θ sin θ 0 lim = lim · lim =1· = 0. θ→0 θ θ→0 θ θ→0 1 + cos θ 2
- 107. Try these Example tan θ 1. lim θ→0 θ sin 2θ 2. lim θ→0 θ
- 108. Try these Example tan θ 1. lim θ→0 θ sin 2θ 2. lim θ→0 θ Answer 1. 1 2. 2
- 109. Solutions 1. Use the basic trigonometric limit and the deﬁni on 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
- 110. Solutions 1. Use the basic trigonometric limit and the deﬁni on 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
- 111. Solutions 1. Use the basic trigonometric limit and the deﬁni on 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 iden ty: 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
- 112. Summary y The limit laws allow us to compute limits reasonably. BUT we cannot make up . x extra laws otherwise we get into trouble.

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