Test bank and solution of calculus 10e (2)

C H A P T E R Limits and Their Properties Section 2.1 A Preview of Calculus 81 Section 2.2 Finding Limits Graphically and Numerically .82 Section 2.3 Evaluating Limits Analytically 93 Section 2.4 Continuity and One-Sided Limits 105 Section 2.5 Infinite Limits .117 Review Exercises 125 Problem Solving 133 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part C H A P T E R Limits and Their Properties Section 2.1 A Preview of Calculus Precalculus: ( 20 ft/sec)(15 sec) = 300 ft f ( x ) = x − x 2 Calculus required: Velocity is not constant (a) y 10 Distance ≈ (20 ft/sec)(15 sec) = 300 ft P Calculus required: Slope of the tangent line at x = is the rate of change, and equals about 0.16 Precalculus: rate of change = slope = 0.08 (a) Precalculus: Area = bh 2 = (5)( 4) = 10 sq units ≈ 2( 2.5) = sq units (a) (b) slope = m = (6 x − x ) − x−2 = ( − x), x ≠ (b) Calculus required: Area = bh f ( x) = x −2 x = (x For x = 3, m = − = For x = 2.5, m = − 2.5 = 1.5 = For x = 1.5, m = − 1.5 = 2.5 = y P(4, 2) − 2)( − x) x−2 (c) At P( 2, 8), the slope is You can improve your approximation by considering values of x close to Answers will vary Sample answer: x = The instantaneous rate of change of an automobile’s position is the velocity of the automobile, and can be determined by the speedometer x −2 x −4 (b) slope = m = = ( x −2 x + )( x −2 ) ,x ≠ x + 1 = + x = 3: m = ≈ 0.2679 + x = 5: m = ≈ 0.2361 + 1 = = 0.25 (c) At P( 4, 2) the slope is 4 + x = 1: m = You can improve your approximation of the slope at x = by considering x-values very close to © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 81 82 Chapter Limits and Their Properties (a) Area ≈ + Area ≈ + 5 + ≈ 10.417 (5 + 1.55 + 52 + 2.55 + 53 + 3.55 + 54 + 4.55 ) ≈ 9.145 (b) You could improve the approximation by using more rectangles 10 (a) D1 = (5 − 1) (b) D2 = 1+ ( 52 ) + (1 − 5) + 1+ = 16 + 16 ≈ 5.66 ( 52 − 53 ) + ( 53 − 54 ) 1+ + 1+ ( 54 − 1) ≈ 2.693 + 1.302 + 1.083 + 1.031 ≈ 6.11 (c) Increase the number of line segments Section 2.2 Finding Limits Graphically and Numerically x 3.9 3.99 3.999 4.001 4.01 4.1 f (x) 0.2041 0.2004 0.2000 0.2000 0.1996 0.1961 lim x → x2 –0.1 –0.01 –0.001 0.001 0.01 0.1 f (x) 0.5132 0.5013 0.5001 ? 0.4999 0.4988 0.4881 x +1 −1 1⎞ ⎛ ≈ 0.5000 ⎜ Actual limit is ⎟ x 2⎠ ⎝ lim x –0.1 –0.01 –0.001 0.001 0.01 0.1 f (x) 0.9983 0.99998 1.0000 1.0000 0.99998 0.9983 lim x→0 ( Actual limit is 1.) ( Make sure you use radian mode.) –0.1 –0.01 –0.001 0.001 0.01 0.1 f (x) 0.0500 0.0050 0.0005 –0.0005 –0.0050 –0.0500 lim cos x − ≈ 0.0000 x ( Actual limit is 0.) ( Make sure you use radian mode.) x –0.1 –0.01 –0.001 0.001 0.01 0.1 f (x) 0.9516 0.9950 0.9995 1.0005 1.0050 1.0517 lim x→0 sin x ≈ 1.0000 x x x→0 1⎞ ⎛ ⎜ Actual limit is ⎟ 5⎠ ⎝ x x→0 x − ≈ 0.2000 − 3x − ex − ≈ 1.0000 x (Actual limit is 1.) x –0.1 –0.01 –0.001 0.001 0.01 0.1 f (x) 1.0536 1.0050 1.0005 0.9995 0.9950 0.9531 lim x→0 ln ( x + 1) x ≈ 1.0000 (Actual limit is 1.) © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Section 2.2 x 0.9 0.99 0.999 1.001 1.01 1.1 f (x) 0.2564 0.2506 0.2501 0.2499 0.2494 0.2439 lim x →1 x − ≈ 0.2500 x + x −6 x –4.1 –4.01 –4.001 –4 –3.999 –3.99 –3.9 f (x) 1.1111 1.0101 1.0010 ? 0.9990 0.9901 0.9091 x + ≈ 1.0000 x + x + 20 lim 0.9 0.99 0.999 1.001 1.01 1.1 f (x) 0.7340 0.6733 0.6673 0.6660 0.6600 0.6015 x →1 10 x4 − ≈ 0.6666 x6 − –3.1 –3.01 –3.001 –3 –2.999 –2.99 –2.9 f (x) 27.91 27.0901 27.0090 ? 26.9910 26.9101 26.11 x + 27 ≈ 27.0000 x + x → −3 –6.1 –6.01 –6.001 –6 –5.999 –5.99 –5.9 f (x) –0.1248 –0.1250 –0.1250 ? –0.1250 –0.1250 –0.1252 10 − x − ≈ − 0.1250 x + lim x f (x) lim x→2 13 1⎞ ⎛ ⎜ Actual limit is − ⎟ 8⎠ ⎝ 1.9 1.99 1.999 2.001 2.01 2.1 0.1149 0.115 0.1111 ? 0.1111 0.1107 0.1075 x ( x + 1) − ≈ 0.1111 x − 1⎞ ⎛ ⎜ Actual limit is ⎟ 9⎠ ⎝ x –0.1 –0.01 –0.001 0.001 0.01 0.1 f (x) 1.9867 1.9999 2.0000 2.0000 1.9999 1.9867 lim x→0 14 ( Actual limit is 27.) x x → −6 12 2⎞ ⎛ ⎜ Actual limit is ⎟ 3⎠ ⎝ x lim 11 (Actual limit is 1.) x lim sin x ≈ 2.0000 x (Actual limit is 2.) (Make sure you use radian mode.) x –0.1 –0.01 –0.001 0.001 0.01 0.1 f (x) 0.4950 0.5000 0.5000 0.5000 0.5000 0.4950 lim x→0 tan x ≈ 0.5000 tan x 83 1⎞ ⎛ ⎜ Actual limit is ⎟ 4⎠ ⎝ x → −4 Finding Limits Graphically and Numerically 1⎞ ⎛ ⎜ Actual limit is ⎟ 2⎠ ⎝ © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 84 15 Chapter x 1.9 1.99 1.999 2.001 2.01 2.1 f (x) 0.5129 0.5013 0.5001 0.4999 0.4988 0.4879 lim x→2 16 Limits and Their Properties ln x − ln ≈ 0.5000 x − 1⎞ ⎛ ⎜ Actual limit is ⎟ 2⎠ ⎝ x –0.1 –0.01 –0.001 0.001 0.01 0.1 f (x) 3.99982 4 0 0.00018 lim x→0 does not exist + e1 x 25 (a) f (1) exists The black dot at (1, 2) indicates that 17 lim ( − x) = x →3 f (1) = 18 lim sec x = (b) lim f ( x) does not exist As x approaches from the x→0 x →1 19 lim f ( x) = lim ( − x ) = x→2 left, f (x) approaches 3.5, whereas as x approaches from the right, f (x) approaches x→2 20 lim f ( x) = lim ( x + 3) = x →1 21 lim x→2 (c) f ( 4) does not exist The hollow circle at x →1 x − x − (4, 2) indicates that (d) lim f ( x) exists As x approaches 4, f ( x) approaches does not exist For values of x to the left of 2, for values of x to the right of 2, x→4 x − (x − 2) 2: lim f ( x) = = −1, whereas x − (x − 2) = 23 lim cos(1 x) does not exist because the function x→0 oscillates between –1 and as x approaches lim tan x does not exist because the function increases x →π without bound as x approaches π from the left and decreases without bound as x approaches the right x→4 26 (a) f ( −2) does not exist The vertical dotted line 22 lim does not exist The function approaches x → + e1 x from the left side of by it approaches from the left side of 24 f is not defined at π from indicates that f is not defined at –2 (b) lim f ( x) does not exist As x approaches –2, the x → −2 values of f ( x) not approach a specific number (c) f (0) exists The black dot at (0, 4) indicates that f (0) = (d) lim f ( x ) does not exist As x approaches from the x→0 left, f ( x) approaches 12 , whereas as x approaches from the right, f ( x) approaches (e) f ( 2) does not exist The hollow circle at (2, 12 ) indicates that f ( 2) is not defined (f ) lim f ( x) exists As x approaches 2, f ( x) approaches x→2 : lim f x→2 ( x) = 12 (g) f ( 4) exists The black dot at ( 4, 2) indicates that f ( 4) = (h) lim f ( x) does not exist As x approaches 4, the x→4 values of f ( x) not approach a specific number © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Section 2.2 27 y Finding Limits Graphically and Numerically 32 You need f ( x) − = 85 − x −1 = < 0.01 x −1 x −1 Let δ = f − x −2 −1 −1 −2 lim f ( x ) exists for all values of c ≠ x→c 28 y f ( x) − = π −π 33 You need to find δ such that < x − < δ implies lim f ( x ) exists for all values of c ≠ π x→c 29 One possible answer is f x 30 One possible answer is y So take δ = 1 −1 31 You need f ( x ) − = (x + 1) − = x − < 0.4 So, take δ = 0.4 If < x − < 0.4, then x − = (x Then < x − < δ implies 11 1 < x −1< 11 11 1 − < x −1< 11 − x −1 − < 0.1 That is, x − < 0.1 x 1 − 0.1 < < + 0.1 x 11 < < 10 10 x 10 10 x > > 11 10 10 −1 > x −1 > −1 11 1 > x −1> − 11 f ( x) − = −0.1 < y −2 2− x 101 −1 = < = x −1 x −1 100 101 100 = 0.01 x π −1 −3 1 1 < x − < ⇒1− < x −1 101 and you have −2 −1 −1 1 If < x − < , then 101 101 + 1) − = f ( x) − < 0.4, as desired Using the first series of equivalent inequalities, you obtain f ( x) − = − < 0.1 x © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 86 Chapter Limits and Their Properties 34 You need to find δ such that < x − < δ implies f ( x) − = x − − = x − < 0.2 That is, 37 lim ( x − 3) = 22 − = = L x→2 ( x2 −0.2 < x − < 0.2 − 0.2 < x2 < + 0.2 3.8 < x2 < 4.2 3.8 < x < So take δ = ( (x + 2)( x − 2) < 0.01 4.2 − 4.2 − ≈ 0.0494 Then < x − < δ implies − x − < 0.01 x + x − < 0.01 4.2 3.8 − < x − < ) x − < 4.2 − 3.8 − < x − < 4.2 − Using the first series of equivalent inequalities, you obtain If you assume < x < 3, then δ ≈ 0.01 = 0.002 x − < 0.002 = 35 lim (3 x + 2) = 3(2) + = = L x + x − < 0.01 ( x2 + 2) − < 0.01 38 lim ( x + 6) = 42 + = 22 = L x − < 0.01 x→4 x − < 0.01 ≈ 0.0033 = δ So, if < x − < δ = 0.01 , you have ( x2 + 6) − 22 < 0.01 x − 16 < 0.01 (x + 4)( x − 4) < 0.01 x − < 0.01 x − < 3x − < 0.01 (3 x − 3) − < 0.01 f ( x) − L < 0.01 x→2 < x − < 1 (0.01) < (0.01) x + x − < 0.01 f ( x ) − = x − < 0.2 0.01 0.01 x + So, if < x − < δ ≈ 0.002, you have 4.2 − < x − < (3 x − 3) − < 0.01 0.01 x + + 2) − < 0.01 If you assume < x < 5, then δ = f ( x) − L < 0.01 So, if < x − < δ ≈ 2− x < 0.01 − 13 ( x − 6) < 0.01 ≈ 0.00111 0.01 , you have 0.01 0.01 x − < < x + x⎞ ⎛ 36 lim ⎜ − ⎟ = − = = L x → 6⎝ 3⎠ x⎞ ⎛ ⎜ − ⎟ − < 0.01 3⎠ ⎝ 0.01 ( x + 4)( x − 4) < 0.01 x − 16 < 0.01 ( x2 + 6) − 22 < 0.01 f ( x) − L < 0.01 x − < 0.03 < x − < 0.03 = δ So, if < x − < δ = 0.03, you have − 13 ( x − 6) < 0.01 2− x < 0.01 x⎞ ⎛ ⎜ − ⎟ − < 0.01 3⎠ ⎝ f ( x) − L < 0.01 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Section 2.2 39 lim ( x + 2) = + = 42 lim x→4 ( x + 1) = x →3 Given ε > 0: (x Finding Limits Graphically and Numerically + 2) − < ε ( 34 x + 1) − 134 x So, let δ = ε So, if < x − < δ = ε , you have x −4 < ε 40 lim ( x + 5) = 4(− 2) + = − 3 + 5) − (− 3) < ε x + < ε x + < ε 4ε 4ε , you have ε x −3 < ε < ε ( 34 x + 1) − 134 < ε − f ( x) − L < ε 43 lim = x→6 Given ε > 0: x + < ε , you have ε 3−3 < ε f ( x) − L < ε f ( x) − L < ε ( x − 1) = ( − 4) − = −3 44 lim ( −1) = −1 x→2 Given ε > 0: −1 − ( −1) < ε Given ε > 0: ( 12 x − 1) − (−3) < ε So, for any δ > 0, you have (4 x + 5) − ( −3) < ε x → −4 3−3 < ε So, any δ > will work 4x + < ε 4ε = δ So, if < x + < δ = 1x x −3 < ε x 4x + < ε 41 lim 13 < ε x −3 < Given ε > 0: So, if < x − < δ = x → −2 ε − So, let δ = f ( x) − L < ε So, let δ = = < ε x −3 < + 2) − < ε (4 x (3) + Given ε > 0: x −4 < ε = δ (x 87 < ε + < ε x − ( −4) < ε x − ( −4) < 2ε < ε So, any δ > will work So, for any δ > 0, you have (−1) − (−1) f ( x) − L < ε < ε So, let δ = 2ε So, if < x − ( −4) < δ = 2ε , you have x − ( −4) < 2ε 1x + < ε ( 12 x − 1) + < ε f ( x) − L < ε © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 88 Chapter 45 lim x→0 Limits and Their Properties 48 lim x − = − = x = x →3 Given ε > 0: Given ε > 0: x −0 < ε x −3 < ε x < ε So, let δ = ε x < ε3 = δ So, for < x − < δ = ε , you have So, let δ = ε x −3 < ε So, for x − δ = ε , you have x −3 −0 < ε x < ε3 3 f ( x) − L < ε x < ε x →1 x→4 ( x2 + 1) − < ε x2 − < ε = (x Given ε > 0: + 1)( x − 1) < ε x − < ε x − x + < ε x + x − < ε x + Assuming < x < 9, you can choose δ = 3ε Then, < x − < δ = 3ε ⇒ x − < ε ⇒ x −1 < So for < x − < δ = x2 − < ε ( x2 + 1) − < ε f ( x) − < ε x − − 10 < ε −( x − 5) − 10 < ε (x − < 0) −x − < ε 50 lim ( x + x) = (− 4) + 4(− 4) = x → −4 Given ε > 0: x − ( −5) < ε ( x2 + x) − < ε So, let δ = ε x( x + 4) < ε So for x − ( −5) < δ = ε , you have x + < −( x + 5) < ε −( x − 5) − 10 < ε x − − 10 < ε f ( x) − L < ε ε , you have 1 ε x −1 < ε < x +1 x − < ε x → −5 ε x +1 If you assume < x < 2, then δ = ε x + 47 lim x − = ( −5) − = −10 = 10 Given ε > 0: ) Given ε > 0: f ( x) − L < ε x = ( 49 lim x + = 12 + = x −0 < ε 46 lim x −3 −0 < ε ε x If you assume − < x < − 3, then δ = (because x − < 0) So for < x − (− 4) < δ = x + < ε < ε ε , you have ε x x ( x + 4) < ε (x + x) − < ε f ( x) − L < ε 51 lim f ( x) = lim = x →π x →π 52 lim f ( x) = lim x = π x →π x →π © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Section 2.2 Finding Limits Graphically and Numerically x +5 −3 x − lim f ( x) = x→4 55 f ( x ) = 53 f ( x) = lim f ( x) = 10 The domain is [−5, 4) ∪ (4, ∞) The graphing utility does not show the hole ⎛ 1⎞ at ⎜ 4, ⎟ ⎝ 6⎠ − 0.1667 The domain is all x ≥ except x = The graphing utility does not show the hole at (9, 6) ex − x lim f ( x) = x→0 56 f ( x) = The domain is all x ≠ 1, The graphing −3 10 x −3 x − 4x + lim f ( x) = x →3 54 f ( x) = x −9 x −3 x →9 0.5 −6 89 utility does not show the ⎛ 1⎞ hole at ⎜ 3, ⎟ ⎝ 2⎠ −4 −2 −1 The domain is all x ≠ The graphing utility does not ⎛ 1⎞ show the hole at ⎜ 0, ⎟ ⎝ 2⎠ 57 C (t ) = 9.99 − 0.79 ⎡⎣⎡− ⎣ (t − 1)⎤⎦ ⎤⎦ (a) 16 (b) t 3.3 3.4 3.5 3.6 3.7 C 11.57 12.36 12.36 12.36 12.36 12.36 12.36 lim C (t ) = 12.36 t → 3.5 (c) t 2.5 2.9 3.1 3.5 C 10.78 11.57 11.57 11.57 12.36 12.36 12.36 The lim C (t ) does not exist because the values of C approach different values as t approaches from both sides t →3 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 122 Chapter (b) Limits and Their Properties x 0.5 0.2 0.1 0.01 0.001 0.0001 f (x) 0.1585 0.0823 0.0333 0.0167 0.0017 ≈ ≈ 0.25 lim x → 0+ −1.5 x − sin x = x2 1.5 −0.25 (c) x 0.5 0.2 0.1 0.01 0.001 0.0001 f (x) 0.1585 0.1646 0.1663 0.1666 0.1667 0.1667 0.1667 0.25 lim x → 0+ − 1.5 x − sin x = 0.1667 (1 6) x3 1.5 − 0.25 (d) x 0.5 0.2 0.1 0.01 0.001 0.0001 f (x) 0.1585 0.3292 0.8317 1.6658 16.67 166.7 1667.0 1.5 lim x → 0+ −1.5 x − sin x x − sin x = ∞ = ∞ or n > 3, lim xn x4 x → 0+ 1.5 −1.5 66 lim P = ∞ Total distance Total time 2d 50 = (d x) + (d y) 68 (a) Average speed = V → 0+ As the volume of the gas decreases, the pressure increases 67 (a) r = (b) r = (c) lim x → 25− 2(7) 625 − 49 2(15) 625 − 225 2x 625 − x = ft sec 12 = 50 = xy y + x 50 y + 50 x = xy ft sec 50 x = xy − 50 y 50 x = y ( x − 25) = ∞ 25 x = y x − 25 Domain: x > 25 (b) (c) x 30 40 50 60 y 150 66.667 50 42.857 lim x → 25+ 25 x = ∞ x − 25 As x gets close to 25 mi/h, y becomes larger and larger © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Section 2.5 Infinite Limits 123 1 1 bh − r 2θ = (10)(10 tan θ ) − (10) θ = 50 tan θ − 50 θ 2 2 69 (a) A = ⎛ π⎞ Domain: ⎜ 0, ⎟ ⎝ 2⎠ (b) θ 0.3 0.6 0.9 1.2 1.5 f (θ ) 0.47 4.21 18.0 68.6 630.1 100 1.5 (c) lim A = ∞ θ → π 2− 70 (a) Because the circumference of the motor is half that of the saw arbor, the saw makes 1700 = 850 revolutions per minute (b) The direction of rotation is reversed ⎛ ⎛π ⎞⎞ (c) 2( 20 cot φ ) + 2(10 cot φ ): straight sections The angle subtended in each circle is 2π − ⎜ 2⎜ − φ ⎟ ⎟ = π + 2φ ⎠⎠ ⎝ ⎝ So, the length of the belt around the pulleys is 20(π + 2φ ) + 10(π + 2φ ) = 30(π + 2φ ) Total length = 60 cot φ + 30(π + 2φ ) ⎛ π⎞ Domain: ⎜ 0, ⎟ ⎝ 2⎠ (d) (e) φ 0.3 0.6 0.9 1.2 1.5 L 306.2 217.9 195.9 189.6 188.5 450 ␲ 0 (f ) lim L = 60π ≈ 188.5 φ → (π 2)− (All the belts are around pulleys.) (g) lim L = ∞ φ → 0+ 71 False For instance, let f ( x) = x −1 or x −1 g ( x) = x x2 + 72 True 73 False The graphs of y = tan x, y = cot x, y = sec x and y = csc x have vertical asymptotes 74 False Let ⎧1 ⎪ , x ≠ f ( x) = ⎨ x ⎪3, x = ⎩ The graph of f has a vertical asymptote at x = 0, but f (0) = © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 124 Chapter Limits and Their Properties 1 and g ( x ) = , and c = x x 75 Let f ( x ) = ⎛ x2 − 1⎞ 1⎞ 1 ⎛1 and but − = lim lim = ∞ = ∞ lim , ⎜ ⎟ = −∞ ≠ ⎜ ⎟ x → 0⎝ x x→0 x → x2 x → x4 x4 ⎠ ⎝ x ⎠ lim 76 Given lim f ( x) = ∞ and lim g ( x) = L : x→c x→c (1) Difference: Let h( x) = − g ( x) Then lim h( x ) = − L, and lim ⎡⎣ f ( x) − g ( x)⎤⎦ = lim ⎡⎣ f ( x) + h( x)⎤⎦ = ∞, by the Sum Property x→c x→c x→c (2) Product: If L > 0, then for ε = L > there exists δ1 > such that g ( x) − L < L whenever < x − c < δ1 So, L < g ( x) < 3L Because lim f ( x) = ∞ then for M > 0, there exists δ > such that x→c f ( x) > M ( L) whenever x − c < δ Let δ be the smaller of δ1 and δ Then for < x − c < δ , you have f ( x) g ( x) > M ( L)( L 2) = M Therefore lim f ( x) g ( x) = ∞ The proof is similar for L < x→c (3) Quotient: Let ε > be given There exists δ1 > such that f ( x) > 3L 2ε whenever < x − c < δ1 and there exists δ > such that g ( x) − L < L whenever < x − c < δ This inequality gives us L < g ( x) < 3L Let δ be the smaller of δ1 and δ Then for < x − c < δ , you have g ( x) f ( x) < 3L = ε 3L 2ε Therefore, lim x→c g ( x) f ( x) = 77 Given lim f ( x) = ∞, let g ( x) = Then x→c lim x→c g ( x) f ( x) 78 Given lim x →c Then, lim x→c = by Theorem 1.15 = Suppose lim f ( x ) exists and equals L x→c f ( x) lim 1 = x→c = = f ( x) L lim f ( x) x→c This is not possible So, lim f ( x ) does not exist x→c is defined for all x > x −3 Let M > be given You need δ > such that f ( x) = > M whenever < x < + δ x −3 79 f ( x) = Equivalently, x − < whenever M x − < δ , x > Then for x > and M 1 x − < δ, > = M and so f ( x ) > M x −3 So take δ = 1 is defined for all x < Let N < be given You need δ > such that f ( x ) = < N whenever x −5 x −5 1 − δ < x < Equivalently, x − > whenever x − < δ , x < Equivalently, < − whenever N x −5 N 80 f ( x ) = Note that δ > because N < For x − < δ and N 1 1 = − < N x < 5, > = − N , and δ x −5 x −5 x −5 x − < δ , x < So take δ = − © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Review Exercises for Chapter 125 Review Exercises for Chapter Calculus required Using a graphing utility, you can estimate the length to be 8.3 Or, the length is slightly longer than the distance between the two points, approximately 8.25 11 −9 −1 Precalculus L = (9 − 1) + (3 − 1) 2 ≈ 8.25 x −3 x − x + 12 f ( x) = x 2.9 2.99 2.999 3.001 3.01 3.1 f (x) –0.9091 –0.9901 –0.9990 ? –1.0010 –1.0101 –1.1111 lim f ( x ) ≈ −1.0000 (Actual limit is −1.) x →3 −6 12 −6 x + − x f ( x ) = x –0.1 –0.01 –0.001 0.001 0.01 0.1 f (x) 0.2516 0.2502 0.2500 ? 0.2500 0.2498 0.2485 lim f ( x ) ≈ 0.2500 x→0 (Actual limit is 14 ) 0.5 −5 5 h( x) = x( − x ) 4x − x2 = = − x, x ≠ x x (a) lim h( x) = − = x→0 (b) lim h( x) = − ( −1) = x → −1 f (t ) = ln (t + 2) t (a) lim f (t ) does not exist because lim f (t ) = − ∞ t →0− t →0 and lim f (t ) = ∞ t →0+ (b) lim f (t ) = t → −1 ln = −1 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 126 Chapter Limits and Their Properties lim( x + 4) = + = x →1 Let ε > be given Choose δ = ε Then for < x − < δ = ε , you have x −1 < ε (x + 4) − < ε f ( x) − L < ε x = lim x →9 = Let ε > be given You need x −3 < ε ⇒ x −3 < ε x +3 x +3 ⇒ x −9 < ε x + Assuming < x < 16, you can choose δ = 5ε So, for < x − < δ = 5ε , you have x − < 5ε < x +3ε x −3 < ε f ( x) − L < ε lim (1 − x ) = − 22 = −3 x→2 Let ε > be given You need − x − ( −3) < ε ⇒ x − = x − x + < ε ⇒ x − < Assuming < x < 3, you can choose δ = So, for < x − < δ = x − < ε < ε ε ε x + , you have ε x + x − x + < ε x2 − < ε − x2 < ε (1 − x ) − (−3) < ε f ( x) − L < ε 10 lim = Let ε > be given δ can be any positive x →5 number So, for < x − < δ , you have 9−9 < ε f ( x) − L < ε 11 lim x = (− 6) = 36 13 lim ( x − 2) = (6 − 2) = 16 14 lim x −3 = x → −5 (− 5) − = −8 = − 15 lim 4 = = −1 −1 16 lim x 2 = = = +1 +1 4+1 x→4 x x → −6 12 lim (5 x − 3) = 5(0) − = − x→6 x → x2 x→0 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Review Exercises for Chapter t + 1 = lim = − t → − − t − 17 lim 19 lim t → −2 t x→4 x −3 −1 = lim x→4 x − t − 16 (t − 4)(t + 4) = lim t →4 t − t →4 t − = lim(t + 4) = + = = lim 18 lim x→4 4+ x − ⋅ x 4+ x + = lim x→0 4+ x + x −3 +1 x −3 +1 − 3) − ( ) x −3 +1 1 = x −3 +1 x→4 4+ x − = lim x→0 x x→0 (x ( x − 4) = lim t →4 20 lim x −3 −1 ⋅ x − 127 1 = 4+ x + ⎡1 ( x + 1)⎤⎦ − 1 − ( x + 1) −1 = lim = lim = −1 21 lim ⎣ x→0 x → x( x + 1) x→0 x + x 22 lim (1 ) 1+ s −1 s s →0 ( ⎡1 = lim ⎢ s → 0⎢ ⎣ ) 1+ s −1 s ⋅ (1 (1 ) ) + s + 1⎤ ⎥ + s + 1⎥ ⎦ ⎡1 (1 + s )⎤⎦ − 1 −1 = lim ⎣ = lim = − s →0 ⎡ s → s 1 + s + 1⎤ (1 + s)⎡⎣ 1 + s + 1⎤⎦ ⎣ ⎦ ( 23 lim x→0 24 ) ( ⎛ x ⎞⎛ − cos x ⎞ − cos x = lim ⎜ ⎟⎜ ⎟ = (1)(0) = x → sin x ⎝ sin x x ⎠ ⎝ ⎠ 25 lim e x −1 sin x →1 4(π 4) 4x = = π x → (π 4) tan x 26 lim lim 27 lim ∆x → x→2 ∆x → ∆x → πx ln ( x − 1) = e sin ln ( x − 1) = lim x→2 π =1 ln ( x − 1) ln ( x − 1) = lim = x→2 sin ⎡⎣(π 6) + ∆x⎤⎦ − (1 2) sin (π 6)cos ∆x + cos(π 6)sin ∆x − (1 2) = lim ∆ x → ∆x ∆x = lim 28 lim ) cos(π + ∆x) + ∆x = lim ∆x → ⋅ (cos ∆x ∆x − 1) + lim ∆x → sin ∆x ⋅ = 0+ (1) = ∆x cos π cos ∆x − sin π sin ∆x + ∆x ⎡ (cos ∆x − 1) ⎤ sin ∆x ⎤ ⎡ sin π = lim ⎢− ⎥ − ∆lim ∆x → x → 0⎢ ∆x ∆x ⎥⎦ ⎣ ⎣ ⎦ = −0 − (0)(1) = 29 lim ⎡⎣ f ( x) g ( x )⎤⎦ = ⎡⎢ lim f ( x )⎤⎡ g ( x)⎤⎥ ⎥⎢xlim x→c →c ⎣x → c ⎦⎣ ⎦ = (− 6) ( 12 ) = − ( 12 ) = − lim f ( x) f ( x) −6 30 lim = x→c = = −12 x → c g ( x) lim g ( x ) x→c 31 lim ⎡⎣ f ( x) + g ( x)⎤⎦ = lim f ( x) + lim g ( x) x→c x→c x→c ( 2) = −6 + 2 32 lim ⎡⎣ f ( x)⎤⎦ = ⎡⎢ lim f ( x )⎤⎥ x→c ⎣x → c ⎦ = ( − 6) = 36 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 128 Chapter 33 f ( x ) = Limits and Their Properties 2x + − x The limit appears to be −1 x –0.01 –0.001 0.001 0.01 f (x) 0.3335 0.3333 ? 0.3333 0.331 lim f ( x) ≈ 0.3333 x→0 2x + − ⋅ x lim x→0 2x + + (2 x + 9) − = lim = lim x → x ⎡ x + + 3⎤ x→0 2x + + ⎣ ⎦ = 2x + + = + ⎡1 ( x + 4)⎤⎦ − (1 4) 34 f ( x) = ⎣ x The limit appears to be − −8 1 16 −3 x –0.01 –0.001 0.001 0.01 f (x) –0.0627 –0.0625 ? –0.0625 –0.0623 lim f ( x) ≈ − 0.0625 = − x→0 16 1 − −1 + x 4 = lim − ( x + 4) = lim = − lim x→0 x → ( x + 4)4( x ) x → ( x + 4)4 16 x 35 f ( x ) = lim 20(e x − 1) x→0 x −1 The limit appears to be −3 −3 x –0.1 –0.01 –0.001 0.001 0.01 0.1 f (x) 0.8867 0.0988 0.0100 −0.0100 −0.1013 −1.1394 lim f ( x) ≈ 0.0000 x→0 lim x→0 20(e x − 1) x −1 = 20(e0 − 1) −1 = = −1 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Review Exercises for Chapter 129 ln ( x + 1) 36 f ( x ) = x +1 The limit appears to be −1 −1 x –0.1 –0.01 –0.001 0.001 0.01 0.1 f (x) −0.1171 −0.0102 −0.0010 ? 0.0010 0.0099 0.0866 lim f ( x) ≈ 0.0000 x→0 ln ( x + 1) lim x +1 x→0 37 v = lim s ( 4) − s ( t ) −t t →4 = lim t →4 = lim t →4 = lim ln = = 1 = ⎣⎡−4.9(16) + 250⎤⎦ − ⎡⎣−4.9t + 250⎤⎦ −t 4.9(t − 16) −t 4.9(t − 4)(t + 4) 4−t t →4 = lim ⎡− 4.9(t + 4)⎤⎦ = −39.2 m/sec t →4 ⎣ The object is falling at about 39.2 m/sec 38 −4.9t + 250 = ⇒ t = When a = lim 50 , the velocity is s ( a ) − s (t ) t →a 50 sec a −t ⎡−4.9a + 250⎤⎦ − ⎡⎣−4.9t + 250⎤⎦ = lim ⎣ t →a a −t 2 4.9(t − a ) = lim t →a a −t 4.9(t − a )(t + a ) = lim t →a a −t 4.9 t + a )⎤⎦ = lim ⎡− ( ⎣ t →a = −4.9( 2a ) 50 ⎞ ⎛ ⎜a = ⎟ 7⎠ ⎝ = −70 m/sec The velocity of the object when it hits the ground is about 70 m/sec 39 lim x → 3+ 1 = = 3+ x + 40 lim x → 6− x −6 x −6 = lim x − 36 x → 6− ( x − 6)( x + 6) = lim x → 6− = x + 12 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 130 Chapter 41 lim x → 4− Limits and Their Properties has a nonremovable discontinuity at x −5 x = because lim f ( x) does not exist x − x − x + = lim ⋅ x − x → 4− x − x + x − = lim − x → ( x − 4) x + ( 51 f ( x ) = x →5 ) 52 f ( x ) = = lim x + x → 4− has nonremovable discontinuities at x = ± = x −3 42 lim x → 3− x −3 because lim f ( x ) and lim f ( x) not exist x →3 −( x − 3) = lim x → 3− x −3 53 f ( x ) = = −1 x → 2− x x ,x ≠ = = x3 − x x ( x − 1) ( x − 1)( x + 1) x → −1 x→4 at x = 45 lim f ( x) = x→2 x +3 x − x − 18 x +3 = ( x + 3)( x − 6) 54 f ( x ) = 46 lim g ( x) = + = x →1+ 47 lim h(t ) does not exist because lim h(t ) = + = t →1− t →1 (1 + 1) x →1 and has a removable discontinuity at x = because lim f ( x) = lim = −1 x→0 x → ( x − 1)( x + 1) 44 lim x − does not exist There is a break in the graph t →1+ x → −3 has nonremovable discontinuities at x = ±1 because lim f ( x) and lim f ( x) not exist, 43 lim ( x + 1) = 2(1) + = and lim h(t ) = 1 = x2 − ( x − 3)( x + 3) = = , x ≠ −3 x −6 has a nonremovable discontinuity at x = because lim f ( x) does not exist, and has a 48 lim f ( s ) = x→6 s → −2 removable discontinuity at x = − because 49 f ( x ) = x − is continuous for all real x lim f ( x ) = lim x → −3 50 f ( x) = x − x + 20 is continuous for all real x x → −3 1 = − x −6 55 f ( 2) = Find c so that lim (cx + 6) = x → 2+ c ( 2) + = 2c = −1 c = − 56 lim ( x + 1) = x →1+ lim ( x + 1) = x → 3− Find b and c so that lim ( x + bx + c ) = and lim ( x + bx + c ) = x →1− Consequently you get Solving simultaneously, 57 f ( x ) = −3 x + Continuous on ( −∞, ∞) x → 3+ 1+ b + c = b and + 3b + c = = −3 and c = (4 x − 1)( x + 2) x2 + x − = x + x + Continuous on (− ∞, − 2) ∪ (− 2, ∞) There is a 58 f ( x ) = removable discontinuity at x = − © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Review Exercises for Chapter 59 f ( x ) = x − (a) 60 f ( x) = x + lim x + = k + where k is an integer x → k+ lim x + = k + where k is an integer (b) Nonremovable discontinuity at each integer k Continuous on ( k , k + 1) for all integers k 61 g ( x) = 2e is continuous on all intervals ( n, n + 1), where n is an integer g has nonremovable discontinuities at each n 62 h( x) = −2 ln − x Because − x > except for x = 5, h is continuous on ( −∞, 5) ∪ (5, ∞) lim f ( x) = x → 2+ x→2 68 f ( x ) = (x − 1) x (a) Domain: ( −∞, 0] ∪ [1, ∞) (b) lim f ( x) = x → 0− (c) lim f ( x) = x →1+ 69 f ( x) = x = −∞ x lim = ∞ x → 0+ x lim x → 0− (3x + 2)( x − 1) 3x − x − = x −1 x −1 lim f ( x) = lim (3 x + 2) = 63 f ( x ) = x →1 lim f ( x ) = −4 x → 2− (c) lim f ( x) does not exist x → k− x ⎡ x − 2⎤ x2 − = ( x + 2) ⎢ ⎥ x − ⎢⎣ x − ⎦⎥ 67 f ( x) = Continuous on [4, ∞ ) 131 Therefore, x = is a vertical asymptote x →1 Removable discontinuity at x = Continuous on ( −∞, 1) ∪ (1, ∞) ⎧5 − x, x ≤ 64 f ( x) = ⎨ ⎩2 x − 3, x > lim (5 − x) = x → 2− lim ( x − 3) = Nonremovable discontinuity at x = Continuous on ( −∞, 2) ∪ ( 2, ∞) 65 f is continuous on [1, 2] f (1) = −1 < and f ( 2) = 13 > Therefore by the Intermediate Value Theorem, there is at least one value c in (1, 2) such that 2c − = Therefore, x = is a vertical asymptote x3 x3 = x −9 ( x + 3)( x − 3) x3 x3 = − ∞ and lim = ∞ x → −3− x − x → −3+ x − lim Therefore, x = − is a vertical asymptote lim x → −3− x3 x3 = − ∞ and lim = ∞ + x −9 x→3 x − Therefore, x = is a vertical asymptote 72 f ( x) = 2t lim Nonremovable discontinuity every months ( x − 2) 5 = ∞ = lim − 2) x → 2+ ( x − 2) lim x → 2− ( x 71 f ( x ) = x → 2+ 66 A = 5000(1.06) 70 f ( x ) = x → −6− 6x 6x = − 36 − x ( x + 6)( x − 6) 6x 6x = ∞ and lim = −∞ 36 − x x → −6+ 36 − x Therefore, x = − is a vertical asymptote 9000 lim x → 6− 6x 6x = ∞ and lim = −∞ 36 − x x → 6+ 36 − x Therefore, x = is a vertical asymptote 4000 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 132 Chapter 2x + 2x + = x − 64 ( x + 8)( x − 8) 73 g ( x) = lim x → −8− Limits and Their Properties 79 2x + 2x + = − ∞ and lim = ∞ x − 64 x → −8+ x − 64 80 Therefore, x = − is a vertical asymptote lim x → 8− lim x +1 1 = lim = + x +1 x → −1 x − x + lim x +1 1 = lim = − − x4 − x → −1 ( x + 1)( x − 1) x → −1+ x → −1− 1⎞ ⎛ 81 lim ⎜ x − ⎟ = −∞ x ⎠ x → 0+ ⎝ 2x + 2x + = − ∞ and lim = ∞ + x − 64 x − 64 x →8 Therefore, x = is a vertical asymptote 82 lim x → 2− sin π x sin π x = for x = n, where n is an integer 74 f ( x) = csc π x = x − = −∞ 83 lim sin x ⎡ ⎛ sin x ⎞⎤ = lim ⎢ ⎜ ⎟⎥ = 5x x → 0+ ⎣ ⎝ x ⎠⎦ 84 lim sec x = ∞ x 85 lim csc x = lim = ∞ x x → 0+ x sin x x → 0+ lim f ( x) = ∞ or −∞ x→n Therefore, the graph has vertical asymptotes at x = n x → 0+ 75 g ( x) = ln ( 25 − x ) = ln ⎡⎣(5 + x)(5 − x)⎤⎦ lim ln ( 25 − x ) = x → 0+ x →5 lim ln ( 25 − x ) = cos x = −∞ x x → 0− x → −5 86 lim Therefore, the graph has holes at x = ± The graph does not have any vertical asymptotes 87 lim ln (sin x) = −∞ x → 0+ 76 f ( x ) = 7e − x lim 7e − x = ∞ 88 lim 12e − x → 0− x → 0− Therefore, x = is a vertical asymptote 89 C = x2 + x + 77 lim = −∞ x −1 x →1− 78 lim x → (1 2)+ (c) C (90) = $720,000 –0.1 –0.01 –0.001 0.001 0.01 0.1 f(x) 2.0271 2.0003 2.0000 2.0000 2.0003 2.0271 lim lim 80,000 p = ∞ − p p →100− 100 tan 2x x x x→0 80,000 p , ≤ p < 100 100 − p (b) C (50) = $80.000 (d) (a) = ∞ (a) C (15) ≈ $14,117.65 x = ∞ 2x − 90 f ( x ) = x tan x = x ⎧ tan x , x ≠ ⎪ (b) Yes, define f ( x ) = ⎨ x ⎪2, x = ⎩ Now f ( x) is continuous at x = © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Problem Solving for Chapter 133 Problem Solving for Chapter x + ( y − 1) + (a) Perimeter ∆PAO = x + ( x − 1) + = x2 + x4 + Perimeter ∆PBO = (x − 1) + y + x2 + y2 + = (x − 1) + x + x2 + x4 + 2 x + ( x − 1) + x2 + x4 + (x x2 + x4 + (b) r ( x ) = (c) x2 + y2 + − 1) + x + x 0.1 0.01 Perimeter ∆PAO 33.02 9.08 3.41 2.10 2.01 Perimeter ∆PBO 33.77 9.60 3.41 2.00 2.00 r ( x) 0.98 0.95 1.05 1.005 lim r ( x) = x → 0+ 1+ 0+1 = =1 1+ 0+1 1 x bh = (1)( x) = 2 1 y x2 Area ∆PBO = bh = (1)( y ) = = 2 2 (a) Area ∆PAO = (b) a( x) = (c) Area ∆PBO x2 = = x Area ∆PAO x x 0.1 0.01 Area ∆PAO 12 20 200 Area ∆PBO 12 200 20,000 a( x) 1 10 100 lim a( x ) = lim x = x → 0+ x → 0+ (a) There are triangles, each with a central angle of 60° = π So, π⎤ 3 ⎡1 ⎤ ⎡1 ≈ 2.598 Area hexagon = 6⎢ bh⎥ = 6⎢ (1) sin ⎥ = 3⎦ ⎣2 ⎦ ⎣2 h = sin θ h = sin 60° 1 60° θ Error = Area (Circle) − Area (Hexagon) = π − 3 ≈ 0.5435 (b) There are n triangles, each with central angle of θ = 2π n So, n sin ( 2π n) 2π ⎤ ⎡1 ⎤ ⎡1 An = n ⎢ bh⎥ = n ⎢ (1) sin ⎥ = n⎦ ⎣2 ⎦ ⎣2 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 134 Chapter (c) Limits and Their Properties n 12 24 48 96 An 2.598 3.106 3.133 3.139 (d) As n gets larger and larger, 2π n approaches Letting x = 2π n, An = sin ( 2π n) sin ( 2π n) = 2n (2π n) π = sin x π x which approaches (1)π = π (a) Slope = 4− = 3−0 (b) Slope = − (a) Slope = − 3 Tangent line: y − = − ( x − 3) 4 25 y = − x + 4 ( (c) Let Q = ( x, y ) = x, 25 − x (b) Slope of tangent line is ( x − 5) 12 169 Tangent line y = x − 12 12 ) ( (c) Q = ( x, y ) = x, − 169 − x 25 − x − ⋅ x −3 (d) lim mx = lim x →3 x →3 = lim x →3 = lim x →3 mx = 25 − x + 25 − x + 25 − x − 16 ( − 3) 25 − x + (3 − x)(3 + x) ( x − 3)( 25 − x + = lim x →3 x →5 ) 12 − x→5 x→5 −6 = = − + 4 25 − x + x→5 = lim This is the slope of the tangent line at P (x ( − 5) 12 + 169 − x x − 25 169 − x 169 − x ) = lim −(3 + x ) 169 − x 12 + ⋅ x −5 12 + 144 − (169 − x ) = lim ) ) − 169 − x + 12 x −5 (d) lim mx = lim (x 12 y + 12 = 25 − x − x −3 mx = 12 (x ( − 5) 12 + (x 12 + x→5 169 − x + 5) 169 − x = ) 10 = 12 + 12 12 This is the same slope as part (b) a + bx − x a + bx − x = ⋅ a + bx + a + bx + = x ( (a + bx) − ) a + bx + b + bx + Letting a = simplifies the numerator So, lim + bx − x Setting b + x→0 3 = lim x→0 x = ( bx + bx + ) = lim x→0 3, you obtain b = So, a = and b = © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Problem Solving for Chapter + x1 − ⋅ x −1 (d) lim f ( x) = lim (a) + x1 ≥ x →1 x1 ≥ −3 x ≥ −27 x →1 = lim x →1 Domain: x ≥ −27, x ≠ or [−27, 1) ∪ (1, ∞) (b) = lim 0.5 x →1 = lim − 30 x →1 12 ( − 1) (c) 13 lim f ( x ) = − −27 − ≈ 0.0714 x → −27 + = −2 = −28 14 lim f ( x) = lim ( a − 2) = a − x → 0− x → 0− (x 23 a2 − = a a = −1, + 2) = + x1 + ) 12 2 −3 −4 (a) f (1) = + −1 = + ( −1) = f ( 0) = f (b) f continuous at 2: g1 lim f ( x ) = 3: g1 , g3 , g −2 x→2 x → 2− x −4 −3 −2 −1 (a) lim f ( x) = 3: g1 , g ( 12 ) = + (−1) = −1 f ( −2.7) = −3 + = −1 (b) lim f ( x) = −1 x →1− lim f ( x) = −1 y x →1+ lim f ( x) = −1 x →1 x −1 ( + 1) (a − 2)( a + 1) = 10 + x 13 y a2 − a − = (c) + x1 + ax tan x ⎛ ⎞ = a⎜ because lim = 1⎟ x→0 x x → 0+ tan x ⎝ ⎠ Thus, ) ( x1 − 1)( x + x1 + 1)( (1 + + 1)( 11 lim f ( x) = lim x → 0+ + x1 + = + x1 + x1 − − 0.1 + ( −27) + x1 + + x1 − (x 135 −1 (c) f is continuous for all real numbers except x = 0, ±1, ± 2, ± 3, … −2 (a) f ( 14 ) = f (3) = = = f (1) = = (b) lim f ( x) = x →1− lim f ( x) = x →1+ lim f ( x) = −∞ x → 0− lim f ( x) = ∞ x → 0+ (c) f is continuous for all real numbers except x = 0, ±1, ± 12 , ± 13 , … © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part ) 136 Chapter Limits and Their Properties v2 = 12 (a) 192,000 + v0 − 48 r 13 (a) 192,000 = v − v0 + 48 r 192,000 r = v − v0 + 48 x 192,000 lim r = v→0 48 − v0 Let v0 = (b) v2 = a x → a− (iii) lim Pa , b ( x) = x → b+ (iv) lim Pa , b ( x) = x → b− (c) Pa , b is continuous for all positive real numbers except x = a, b 1920 lim r = − v0 2.17 v→0 (≈ 1.47 mi/sec) 10,600 r = v − v0 + 6.99 (c) lim r = v→0 Let v0 = x → a+ (ii) lim Pa , b ( x ) = 1920 + v0 − 2.17 r 2.17 mi sec b (b) (i) lim Pa , b ( x) = 48 = mi sec 1920 = v − v0 + 2.17 r 1920 r = v − v0 + 2.17 Let v0 = y 10,600 6.99 − v0 (d) The area under the graph of U, and above the x-axis, is 14 Let a ≠ and let ε > be given There exists δ1 > such that if < x − < δ1 then f ( x) − L < ε Let δ = δ1 a Then for < x − < δ = δ1 a , you have x < 6.99 ≈ 2.64 mi sec Because this is smaller than the escape velocity for Earth, the mass is less δ1 a ax < δ1 f ( ax) − L < ε As a counterexample, let ⎧1, x ≠ a = and f ( x ) = ⎨ ⎩2, x = Then lim f ( x ) = = L, but x→0 lim f ( ax) = lim f (0) = lim = x→0 x→0 x→0 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part ... ft/sec)(15 sec) = 300 ft P Calculus required: Slope of the tangent line at x = is the rate of change, and equals about 0.16 Precalculus: rate of change = slope = 0.08 (a) Precalculus: Area = bh 2... the left side of and approaches from the right side of Continuity and One-Sided Limits sec x − x2 (a) The domain of f is all x ≠ 0, π /2 + nπ 129 f ( x ) = (b) −3 − 3␲ 3␲ −2 x2 and g ( x) = x... by considering values of x close to Answers will vary Sample answer: x = The instantaneous rate of change of an automobile’s position is the velocity of the automobile, and can be determined