Solutions manual for university calculus early transcendentals 3rd edition by hass by DuBri111

Solutions Manual for University Calculus Early Transcendentals 3rd Edition by Hass Full download: http://downloadlink.org/p/solutions-manual-for-university-calculus-earlytranscendentals-3rd-edition-by-hass/ Test Bank for University Calculus Early Transcendentals 3rd Edition by Hass IBSN 9780321999573 Full download: http://downloadlink.org/p/test-bank-for-university-calculus-earlytranscendentals-3rd-edition-by-hass-ibsn-9780321999573/

CHAPTER 2 2.1

LIMITS AND CONTINUITY

RATES OF CHANGE AND TANGENTS TO CURVES

Δf f (3)− f (2) = 19 1. (a) Δx = 3−2 = 28−9 1

2. (a)

Δg Δx

= 3. (a) Δh Δt

g (3) − g (1) 3 −1

Δf f (1) − f (−1) =1 (b) Δx = 1−(−1) = 2−0 2

= 3 − 2−1 = 2 (

)

(b)

( 3π4 )−h( π4 ) = −1−1 = − 4 3π − π 4 4

− g ( −2) 8 − 8 = g (4) = 6 =0 4 − (−2)

h( π ) − h( π ) = 2π π 6 = 0−π 3 = −3 3 (b) Δh Δt π −

π 2

Δg Δx

Δg g (π )− g (0) (2−1)−(2+1) = = − π2 4. (a) Δt = π −0 π −0

Δg

(b) Δt =

g (π )− g (−π ) π −(−π )

= (2−1)−(2−1) =0 2π

R (2)− R (0) 1 3−1 = 2−0 = 8+1− = 2 =1 5. ΔR Δθ 2 P (2)− P (1) (8−16+10)−(1−4+5) = 2−2 = 0 = 2−1 = 6. ΔP Δθ 1

7. (a) (b)

Δy Δx

2 2 2 2 2 −5) = ((2+h ) −5)−( = 4+4h +hh −5+1 = 4h +h = 4 + h. As h → 0, 4 + h → 4 at P(2, −1) the slope is 4. h h y − (−1) = 4( x − 2) y + 1 = 4 x − 8 y = 4x − 9

Δy Δx

2 2 2 −3 = −4h−h 2 = −4 − h. As h → 0, −4 − h → −4 at P(2, 3) the slope = (7−(2+h )h )−(7−2 ) = 7−4−4h−h h h is −4. (b) y − 3 = (−4)( x − 2) y − 3 = −4 x + 8  y = −4 x + 11

8. (a)

Δy

((2+ h)2 − 2(2+ h)−3)−(22 −2(2)−3)

4+ 4h + h 2 − 4− 2h −3−(−3)

= 9. (a) Δx = h h P(2, − 3) the slope is 2. (b) y − (−3) = 2(x − 2) y + 3 = 2x − 4 y = 2 x − 7.

= 2h+ h = 2 + h. As h → 0, 2 + h → 2 at h

2 1+ 2h+ h −4−4h−(−3) Δy ((1+ h) −4(1+ h))−(1 − 4(1)) = = h −2h = h − 2. As h → 0, h − 2 → −2 at P(1, − 3) the 10. (a) Δx = h h h slope is −2. (b) y − (−3) = (−2)( x −1) y + 3 = −2x + 2 y = −2 x −1. 2

Δy

(2+ h)3 − 23

8+12h+ 4h 2 + h3 −8

12h +4h 2 +h3

= 12 + 4h + h . As h → 0, 12 + 4h + h → 12, at P(2, 8) 11. (a) Δx = = = h h h the slope is 12. (b) y − 8 = 12( x − 2) y − 8 = 12x − 24 y = 12x −16. 2 3 2 3 Δy 2−(1+ h) −(2−1 ) = 2−1−3h−3h − h −1 = −3h −3h −h = −3 − 3h − h2 . As h → 0, −3 −3h − h 2 → −3, at 12. (a) Δx = h h h P(1, 1) the slope is −3. (b) y − 1 = (−3)( x − 1) y −1 = −3x + 3 y = −3x + 4. 3

Chapter 2 Limits and Continuity

2 1+3h+3h + h −12−12h−(−11) Δy (1+ h) −12(1+ h)−(1 −12(1)) + h3 = −9 + 3h + h 2 . = = −9h+3h 13. (a) Δx = h h h As h → 0, −9 + 3h + h 2 → −9 at P(1, −11) the slope is −9. (b) y − (−11) = (−9)( x −1) y + 11 = −9x + 9 y = −9 x − 2. 3

2 3 2 2 3 Δy (2+ h) −3(2+ h) + 4−(2 −3(2) + 4) = 8+12h +6h +h −12−12h −3h +4−0 = 3h +h = 3h + h2 . 14. (a) Δx = h h h As h → 0, 3h + h 2 → 0 at P(2, 0) the slope is 0. (b) y − 0 = 0(x − 2) y = 0. 3

15. (a)

Q Q1 (10, 225) Q2 (14, 375) Q3 (16.5, 475) Q4 (18, 550)

Δp

Slope of PQ = Δt 650−225 = 42.5 m/sec 20−10 650−375 20−14 650−475 20−16.5 650−550 20−18

= 45.83 m/sec = 50.00 m/sec = 50.00 m/sec

(b) At t = 20, the sportscar was traveling approximately 50 m/sec or 180 km/h.

16. (a)

Δp

Q Q1 (5, 20) Q2 (7, 39) Q3 (8.5, 58) Q4 (9.5, 72)

Slope of PQ = Δt 80− 20 = 12 m/sec 10−5

80−39 10−7 = 13.7 m/sec 80−58 = 14.7 m/sec 10−8.5 80−72 = 16 m/sec 10−9.5

(b) Approximately 16 m/sec p 17. (a) Profit (1000s)

200 160 120 80 40 0

2010 2011 2012 2013 2014 Ye ar

Δp

174−62 = 112 = 56 thousand dollars per year (b) Δt = 2014−2012 2 Δp

62−27 = 35 thousand dollars per year. (c) The average rate of change from 2011 to 2012 is Δt = 2012−2011 Δp

111−62 = 49 thousand dollars per year. The average rate of change from 2012 to 2013 is Δt = 2013−2012 So, the rate at which profits were changing in 2012 is approximately 12(35 + 49) = 42 thousand dollars per year.

18. (a) F ( x) = (x + 2)/(x − 2) x 1.2 1.1 F ( x) −4.0 −3.4 ΔF Δx ΔF Δx ΔF Δx

= = =

−4.0−(−3) = −5.0; 1.2−1 −3.04−( −3) 1.01−1 = −4.04; −3.0004−(−3) 1.0001−1 = −4.0004;

1.01 −3.04

1.001 −3.004 ΔF Δx ΔF Δx

= =

1.0001 −3.0004

−3.4 −( −3) = −4.4; 1.1−1 −3.004−(−3) 1.001−1 = −4.004;

1 −3

Section 2.1 (b) The rate of change of F ( x) at x = 1 is −4. Δg

g (2)− g (1) 2−1 = 2−1 ≈ 0.414213 2−1 g (1+ h)− g (1) = 1+hh −1 (1+h)−1

19. (a) Δx = Δg Δx

(b) g ( x) =

g (1.5)− g (1) 1.5−1

−1 ≈ 0.449489 = 1.5 0.5

1+ h

(

Δg Δx

Rates of Change and Tangents to Curves

1+ h

)

1+ h −1 /h

1.1

1.01

1.001

1.0001

1.00001

1.000001

1.04880

1.004987

1.0004998

1.0000499

1.000005

1.0000005

0.4880

0.4987

0.4998

0.499

0.5

−1

= 3 1 2 = 16 = − 16 1− 1 2− T 1 ,T = = TT −22 = 2TT −22T = 2T2−T = −2T2−T = − 2T / 2 ii) (T −2) (2−T ) 2.1 2.01 2.001 (b) T f (T ) 0.476190 0.497512 0.499750 ( f (T ) − f (2))/(T − 2) −0.2381 −0.2488 −0.2500 (c) The table indicates the rate of change is −0.25 at t = 2. 1 (d) lim −2T = − 41

20. (a) i)

f (3)− f (2) 3−2 f(T)− f (2) T −2

T →2

2.0001 0.4999750 −0.2500

2.00001 0.499997 −0.2500

2.000001 0.499999 −0.2500

( )

NOTE: Answers will vary in Exercises 21 and 22. 30− 20 = 10 mph 21. (a) [0, 1]: Δs = 15−0 = 15 mph; [1, 2.5]: Δs = 20−15 = 103 mph; [2.5, 3.5]: Δs = 3.5−2.5 Δt 1−0 Δt 2.5−1 Δt (b) At P 21 , 7.5 : Since the portion of the graph from t = 0 to t = 1 is nearly linear, the instantaneous rate of

(

)

change will be almost the same as the average rate of change, thus the instantaneous speed at t = 12 is = 15 mi/hr. At P(2, 20): Since the portion of the graph from t = 2 to t = 2.5 is nearly linear, the 20 = 0 mi/hr. instantaneous rate of change will be nearly the same as the average rate of change, thus v = 20− 2.5−2 For values of t less than 2, we have

15−7.5 1−0.5

Q Q1 (1, 15) Q2 (1.5, 19) Q3 (1.9, 19.9)

Slope of PQ = Δs Δt 15− 20 = 5 mi/hr 1−2 19−20 = 2 mi/hr 1.5−2 19.9−20 = 1 mi/hr 1.9−2

Thus, it appears that the instantaneous speed at t = 2 is 0 mi/hr. At P(3, 22) : Slope of PQ = Δs Δt = 13 mi/hr

Q2 (3.5, 30)

35− 22 4−3 30−22 3.5−3

Q Q1 (2, 20)

= 16 mi/hr

Q2 (2.5, 20)

Q3 (3.1, 23)

23−22 3.1−3

= 10 mi/hr

Q3 (2.9, 21.6)

Q Q1 (4, 35)

Thus, it appears that the instantaneous speed at t = 3 is about 7 mi/hr.

Slope of PQ = Δs 20− 22 = 2 2−3 20− 22 = 4 2.5−3 21.6−22 2.9−3

Δt

mi/hr mi/hr

= 4 mi/hr

Chapter 2 Limits and Continuity (c) It appears that the curve is increasing the fastest at t = 3.5. Thus for P(3.5, 30) Slope of PQ = Δs Slope of PQ = Δs Q Q Δt Q1 (4, 35) Q2 (3.75, 34) Q3 (3.6, 32)

35−30 = 10 mi/hr 4−3.5 34−30 = 16 mi/hr 3.75−3.5 32−30 = 20 mi/hr 3.6−3.5

Δt 22−30 = 16 mi/hr 3−3.5 25−30 = 20 mi/hr 3.25−3.5 28−30 = 20 mi/hr 3.4−3.5

Q1 (3, 22) Q2 (3.25, 25) Q3 (3.4, 28)

Thus, it appears that the instantaneous speed at t = 3.5 is about 20 mi/hr. 22. (a) [0, 3]: ΔA = 10−15 ≈ −1.67 day ; [0, 5]: ΔA = 3.9−15 ≈ −2.2 day ; [7, 10]: ΔA = 0−1.4 ≈ −0.5 day Δt 3−0 Δt 5−0 Δt 10−7 (b) At P(1, 14): Slope of PQ = ΔA Slope of PQ = ΔA Q Q Δt Δt 12.2−14 = −1.8 gal/day 15−14 Q1 (2, 12.2) Q1 (0, 15) = −1 gal/day gal

Q2 (1.5, 13.2) Q3 (1.1, 13.85)

gal

2−1 13.2−14

= −1.6 gal/day

1.5−1 13.85−14 1.1−1

= −1.5 gal/day

0−1 14.6−14

Q2 (0.5, 14.6)

= −1.2 gal/day

0.5−1 14.86−14 0.9−1

Q3 (0.9, 14.86)

= −1.4 gal/day

Thus, it appears that the instantaneous rate of consumption at t = 1 is about −1.45 gal/day. At P(4, 6): Slope of PQ = ΔA Q Slope of PQ = ΔA Q Δt Δt 10−6 = −4 gal/day Q (3, 10) 3.9−6 1 Q1 (5, 3.9) = −2.1 gal/day 3−4 5−4 7.8−6 = −3.6 gal/day Q (3.5, 7.8) 4.8−6 2 Q (4.5, 4.8) = −2.4 gal/day 3.5−4 2

Q3 (4.1, 5.7)

4.5−4 5.7 −6 4.1−4

= −3 gal/day

Q3 (3.9, 6.3)

6.3−6 3.9−4

= −3 gal/day

Thus, it appears that the instantaneous rate of consumption at t = 1 is −3 gal/day. At P(8, 1): Slope of PQ = ΔA Q Δt Slope of PQ = ΔA Q Δt 1.4−1 = −0.6 gal/day Q (7, 1.4) 1 0.5−1 = −0.5 gal/day Q1 (9, 0.5) 7−8 9−8 1.3−1 = −0.6 gal/day Q (7.5, 1.3) 2 0.7−1 Q2 (8.5, 0.7) 7.5−8 = −0.6 gal/day 8.5−8 1.04−1 = −0.6 gal/day Q3 (7.9, 1.04) 0.95−1 = −0.5 gal/day 7.9−8 Q3 (8.1, 0.95) 8.1−8

Thus, it appears that the instantaneous rate of consumption at t = 1 is −0.55 gal/day. (c) It appears that the curve (the consumption) is decreasing the fastest at t = 3.5. Thus for P(3.5, 7.8) Slope of PQ = Δs Q Slope of PQ = ΔA Δt Δt Q 11.2−7.8 = − 3.4 gal/day Q (2.5, 11.2) 4.8−7.8 1 Q1 (4.5, 4.8) = −3 gal/day 2.5−3.5 4.5−3.5 10−7.8 = −4.4 gal/day Q (3, 10) 6−7.8 2 Q (4, 6) = −3.6 gal/day 3−3.5 2

Q3 (3.6, 7.4)

4−3.5 7.4−7.8 3.6−3.5

= −4 gal/day

Q3 (3.4, 8.2)

8.2−7.8 3.4−3.5

= −4 gal/day

Thus, it appears that the rate of consumption at t = 3.5 is about −4 gal/day. 2.2

LIMIT OF A FUNCTION AND LIMIT LAWS

1. (a) Does not exist. As x approaches 1 from the right, g ( x) approaches 0. As x approaches 1 from the left, g ( x) approaches 1. There is no single number L that all the values g ( x) get arbitrarily close to as x → 1. (b) 1 (c) 0 (d) 0.5 2. (a) 0 (b) −1

Section 2.2

Limit of a Function and Limit Laws

(c) Does not exist. As t approaches 0 from the left, f (t ) approaches −1. As t approaches 0 from the right, f (t ) approaches 1. There is no single number L that f (t ) gets arbitrarily close to as t → 0. (d) −1 3. (a) True (d) False (g) True

(b) True (e) False

4. (a) False (d) True

(b) False (e) True

5. lim | xx| does not exist because | xx| = xx = 1 if x > 0 and | xx| = −xx = −1 if x < 0. As x approaches 0 from the left, | xx| x→0 approaches −1. As x approaches 0 from the right, | xx | approaches 1. There is no single number L that all the function values get arbitrarily close to as x → 0.

1 become increasingly large and negative. As x approaches 1 6. As x approaches 1 from the left, the values of x−1 from the right, the values become increasingly large and positive. There is no number L that all the function values get arbitrarily close to as x → 1, so lim x1−1 does not exist. x→1

7. Nothing can be said about f ( x) because the existence of a limit as x → x0 does not depend on how the function is defined at x0 . In order for a limit to exist, f ( x) must be arbitrarily close to a single real number L when x is close enough to x0 . That is, the existence of a limit depends on the values of f ( x) for x near x0 , not on the definition of f ( x) at x0 itself. 8. Nothing can be said. In order for lim f ( x) to exist, f ( x) must close to a single value for x near 0 regardless of x→0

the value f (0) itself. 9. No, the definition does not require that f be defined at x = 1 in order for a limiting value to exist there. If f (1) is defined, it can be any real number, so we can conclude nothing about f (1) from lim f ( x) = 5. x→1

10. No, because the existence of a limit depends on the values of f ( x) when x is near 1, not on f (1) itself. If lim f ( x) exists, its value may be some number other than f (1) = 5. We can conclude nothing about lim f ( x), x→1

x→1

whether it exists or what its value is if it does exist, from knowing the value of f (1) alone. 11.

lim ( x 2 − 13) = (−3)2 − 13 = 9 − 13 = −4

x→−3

12. lim (− x 2 + 5x − 2) = −(2) 2 + 5(2) − 2 = −4 + 10 − 2 = 4 x→2

13. lim 8(t − 5)(t − 7) = 8(6 − 5)(6 − 7) = −8 t →6

14.

15.

lim ( x3 − 2x 2 + 4x + 8) = (−2)3 − 2(−2)2 + 4(−2) + 8 = −8 − 8 − 8 + 8 = −16

x→−2

lim 2 x +53 x→2 11− x

2(2)+5 11−(2) 3

= 93 = 3

48 16.

17.

Chapter 2 Limits and Continuity

(

( ) )( 2 ( 23 ) − 1) = (8 − 2) ( ( 43 ) − 1) = (6) ( 13) = 2

lim (8 − 3s)(2s − 1) = 8 − 5 23

t→2/3

( )( ( ) )

lim 4 x(3x + 4)2 = 4 − 12 3 − 12 + 4 x→ −1/2

(

= (−2) − 32 + 4

)

( )

2 = (−2) 5 = − 25 2 2

18.

19. 20.

y +2

lim 2 = 2 2+ 2 y →2 y +5 y +6 (2) +5(2)+6

4 = 4+10+6

= 204

= 51

(

lim (5 − y) 4/3 = [5 − (−3)]4/3 = (8)4/3 = (8)1/3

y →−3

lim

z→4

21. lim h→0

22. lim h→0

)4 = 24 = 16

z 2 − 10 = 4 2 − 10 = 16 − 10 = 6 3 3h+1+1

3 3(0)+1+1

5h + 4 − 2 = lim h h→0

3 1+1

= 23

5h +4 −2 5h +4 +2 ⋅ h 5h+4 +2

= lim

(5h+ 4)− 4

h→0 h

(

5h+4 +2

)

= lim

h→0 h

(

5h 5h+4 +2

)

= lim

x −5 23. lim 2x −5 = lim ( x+5)( = lim 1 = 1 = 1 x−5) x→5 x+5 5+5 10 x→5 x −25 x→5

24.

lim 2 x +3 x→−3 x +4 x +3

x +3 = lim ( x +3)( = lim 1 = 1 = − 21 x+1) x→−3 x +1 −3+1 x→−3

25.

2 −10 = lim ( x +5)( x − 2) = lim ( x − 2) = −5 − 2 = −7 lim x +3x x +5 x+5 x→−5 x→−5 x→−5

2 x −10 = lim ( x −5)( x − 2) = lim ( x − 5) = 2 − 5 = −3 26. lim x −7 x −2 x−2 x→2 x→2 x→2 2 (t + 2)(t −1) 2 =3 27. lim t 2+t −2 = lim (t −1)(t +1) = lim tt++12 = 1+ 1+1 2 t →1 t −1 t →1 t →1

28.

2 (t +2)(t +1) 1 lim t 2+3t + 2 = lim (t −2)(t +1) = lim tt +−22 = −1+2 −1−2= − 3 t −t −2 t →−1 t →−1 t →−1

29.

−2 1 x − 4 = lim −2( x + 2) = lim −2 lim −2 = 4 =−2 2 3 2 2 x→−2 x +2 x x→−2 x ( x +2) x→−2 x 5 y 3 +8 y 2 4 2 y →0 3 y −16 y

30. lim

y 2 (5 y +8) 2 2 y→0 y (3 y −16)

= lim

(

5 y +8 2 y→0 3 y −16

= lim

8 =−1 = −16 2

)

31.

−1 x lim xx−1−1 = lim x−1 = lim 1− x ⋅ 1 = lim − 1 = −1 x x→1 x→1 x→1 x x −1 x→1

32.

lim

1− x

x→0

1 1 x −1+ x +1

= lim x→0

( x +1) + ( x −1) ( x −1)( x +1)

(

)

x 2 = lim ( x −1)( 2x+1) ⋅ 1 = lim( x −1)( x+1) x x→0 x→0

= 2 = −2

−1

h→0

5 5h+4 +2

5 4 +2

= 45

Section 2.2

Limit of a Function and Limit Laws

(u +1)(u +1)(u −1) (u +1)(u +1) (1+1)(1+1) 33. lim u 3 −1 = lim = lim = 1+1+1 = 43 2 2 u→1 u −1 u→1 (u +u +1)(u −1) u →1 u +u +1 2

3 (v −2)(v 2 +2v + 4) 34. lim 4v −8 = lim = lim 2

v→2 v −16

35.

lim x −3 x→9 x −9

v 2 + 2v + 4 2 (v+2)(v +4) v→2

v→2 (v−2)(v+2)(v +4) x −3 x→9 ( x −3)( x +3)

= lim

1 x +3

x→9

= 4+4+4 = 12 =3 (4)(8) 32 8 =61

1 9 +3

(

)

2 36. lim 4 x − x = lim x (4− x ) = lim x( 2+ x )(2− x ) = lim x 2 + x = 4(2 + 2) = 16

x→4 2− x

37. lim x→1

38.

lim x→−1

x −1 x+3 −2

x→4 2− x

= lim x→1

x 2 +8 −3 x +1

( x −1)

(

x→−1

x→2

= lim

(

x→−2

x+2 x +5 −3 2

= lim x→−2

= lim

)(

(

( x + 2)

x +8 +3 −2 = 3+3

)(

x +12 +4

x +5 +3 2

x→−3

)

(

x 2 +8+3

)

9 +3= −4

− 23

( x +3) 2+ x −5 2

42. lim

4− x

x→4 5− x +9 2

= lim

(

)

(

)

(4− x ) 5+ x 2 +9

)

(

x 2 +5 +3

( x 2 +5)−9

x→−2

) = lim

)(

= lim x→4

(

(4− x) 5+ x 2 +9 (4− x)(4+ x)

(

x→−3 ( x +3) 2+ x −5

)

= lim

x→4

43. lim (2 sin x −1) = 2 sin 0 −1 = 0 −1 = −1

(

(4− x) 5+ x 2 +9 25−( x +9) 2

x→4

) = lim 5+

(

x 2 +8 +3

)

x +9 4+ x 2

)

= lim

) = lim

)

( x−2)( x+2)

(

x 2 +12 +4

(

x 2 +5 +3

x→2 ( x−2)

( x + 2)

)

( x+2)( x−2)

9− x 2

= lim

(

x→−3 ( x+3) 2+ x 2 −5

)

= 23

) = lim (4− x)(5+ x→4

x 2 +9

16− x

)

= 5+ 8 25 = 54 2

44.

x→0

45. lim sec x = lim cos1 x = 1 = 1 = 1 cos 0 1 x→0 x→0

4 +2=4

( x +1)( x−1)

x→−1 ( x+1)

x→−2

4−( x 2 −5)

3− x

x→4 5− x +9 5+ x +9 2

= lim

x +12 +4 2

= lim = lim = 6 x→−3 ( x +3) 2+ x 2 −5 x→−3 2+ x 2 −5 2+ 4

(

)

x+3+2 =

(

( x + 2)

= lim

)

(3− x )(3+ x )

(

( x 2 +12)−16

= lim

x +5 +3

)(

(

= 21

x −5 2+ x −5 2

= lim

( x 2 +8)−9

lim

x→2 ( x−2)

x +5 +3= x −2

)

x→1

x→−1 ( x+1)

4 16 +4

)(

x +3 +2

= − 31

x +5 −3

( 2−

)

(

( x+3)−4

x 2 +12 + 4

( x −1)

x→1

(

)

x→4

= lim

x +8 +3

x→−2

41.

(

x +2 x 2 +12 +4

= lim

x 2 −5 lim 2− x+3 x→−3

( x −2)

x→2

lim

x +8 −3

x 2 +12 − 4

x→2

= lim

40.

)(

)

x+3 +2

x −1 x 2 +8 +3

x→−1

x 2 +12 −4 x−2

x +3 + 2

( x+1)

= lim

39. lim

(

x+3 −2

= lim

2− x

x→4

lim sin 2 x = lim sin x  = (sin 0)2 = 02 = 0

x→0

46.

x→0

sin x = sin 0 = 0 = 0 lim tan x = lim cos x cos 0 1 x→0

x→0

Chapter 2 Limits and Continuity

x +sin x = 1+0+sin 0 = 1+0+0 = 1 47. lim 1+3cos 3cos 0 3 3 x x→0

48. lim ( x 2 −1)(2 − cos x) = (0 2 −1)(2 − cos 0) = (−1)(2 − 1) = (−1)(1) = −1 x→0

49.

lim

x→−π

x + 4 cos( x + π ) = lim

x + 4 ⋅ lim cos( x + π ) = −π + 4 ⋅ cos 0 = 4 − π ⋅ 1= 4 − π

x→−π

50. lim 7 + sec2 x =

x→−π

lim (7 + sec 2 x) = 7 + lim sec 2 x = 7 + sec 2 0 = 7 + (1)2 = 2 2

x→0

51. (a) quotient rule (c) sum and constant multiple rules

(b) difference and power rules

52. (a) quotient rule (c) difference and constant multiple rules

(b)

power and product rules

53. (a) lim f ( x) g ( x) = lim f (x) lim g ( x) = (5)(−2) = −10 x→c

x→c

(b) lim 2 f ( x) g ( x) = 2 lim f ( x) lim g ( x) = 2(5)(−2) = −20 x→c

x→c

f (x)

x→c lim f ( x )

(d) lim f ( x)− g ( x) x→c

x→c

lim f ( x)− lim g ( x)

x →c

5 5−(−2)

x→c

54. (a) lim [g (x) + 3] = lim g ( x) + lim 3 = −3 + 3 = 0 x→4

x→4

(b) lim xf ( x) = lim x ⋅ lim f ( x) = (4)(0) = 0 x→4

x→4

= [−3]2 = 9

x→4 lim g( x)

(d) lim f ( x)−1 = limx→4 = −3 = 3 f ( x)− lim 1 0−1 x→4 g(x)

x→ 4

55. (a) lim [ f ( x) + g(x)] = lim f ( x) + lim g ( x) = 7 + (−3) = 4 x→b

x→b

(b) lim f (x) ⋅ g(x) = lim x→b

x→b

f ( x) lim g ( x) = (7)(−3) = −21

x→b

7 =−7 (d) lim f ( x)/g ( x) = lim f ( x)/ lim g ( x) = −3 3 x→b x→b x→b

56. (a) (b) (c)

lim [ p( x) + r(x) + s(x)] =

x→−2

lim p( x) + lim r ( x) + lim s( x) = 4 + 0 + (−3) = 1

x→−2

lim p( x) ⋅ r( x) ⋅ s( x) = lim

x→−2

p(x) lim

x→−2

r ( x) lim

x→−2

lim [−4 p( x) + 5r ( x)]/s( x) = −4 lim x→−2

x→−2

s( x) = (4)(0)(−3) = 0

x→−2

p( x) + 5 lim r( x) 

x→−2

2 (1+h) 2 −12 = lim 1+2h+ h −1 = lim h(2+h) = lim (2 + h) = 2 57. lim h h

h→0

lim s( x) = [−4(4) + 5(0)]/ − 3 = 16 3

x→−2

Section 2.2

Limit of a Function and Limit Laws

58. lim (−2+h) −(−2) = lim 4−4h+ h −4 = lim h(h−4) = lim (h − 4) = −4 h h h 2

h→0

59. lim [3( 2+ h)−4]−[3(2)−4] = lim 3h =3 h h→0 h h→0

( 60. lim

h→0

62. lim h→0

63. lim

−2

−2 + h

h→0

61. lim

)−(

7+ h − 7 h

−2

) = lim

−1

h→0 −2h

= lim

(

7+ h − 7 h

h→0

3(0 + h) +1 − 3(0)+1 h

−2−(−2+h) −h = lim −2h(−2+h) = lim h(4−2h) = − 14 h→0 h→0

−2+ h

(

= lim

)(

7+ h + 7

7+h + 7

)

(

)(

3h+1 −1 h

h→0

(

)

= lim

h→0 h

)

3h +1 +1

)

3h+1+1

5 − 2 x 2 = 5 − 2(0) 2 = 5 and lim

x→0

(

(7+ h)−7 7+h + 7

= lim

h→0 h

)

= lim

h→0 h

(3h +1)−1

(

)

3h+1+1

(

= lim

h 7+h + 7

h→0 h

(

)

= lim

1 7+h + 7

h→0

3h 3h+1+1

)

= lim h→0

= 1

3 3h+1+1

2 7

= 32

5 − x 2 = 5 − (0) 2 = 5; by the sandwich theorem, lim f ( x) = 5 x→0

64. lim (2 − x 2 ) = 2 − 0 = 2 and lim 2 cos x = 2(1) = 2; by the sandwich theorem, lim g ( x) = 2 x→0

x→0

(

x→0

)

2 x sin x = 1 65. (a) lim 1 − x6 = 1 − 60 = 1 and lim 1 = 1; by the sandwich theorem, lim 2−2 cos x x→0 x→0 x→0 (b) For x =/ 0, y = (x sin x)/(2 − 2 cos x) lies between the other two graphs in the figure, and the graphs converge as x → 0.

(

)

x2 x2 x 1 66. (a) lim 12 − 24 = lim 21 − lim 24 = 21 − 0 = 21 and lim 21 = 12 ; by the sandwich theorem, lim 1−cos = 2. x2 x→0 x→0 x→0 x→0 x→0 (b) For all x =/ 0, the graph of f ( x) = (1 − cos x)/x 2 lies between the line y = 12 and the parabola

y = 12 − x 2 / 24, and the graphs converge as x → 0.

67. (a)

f ( x) = ( x 2 − 9)/( x + 3) x

−3.1

−3.01

−3.001

−3.0001

−3.00001

−3.000001

f ( x)

−6.1

−6.01

−6.001

−6.0001

−6.00001

−6.000001

−2.9

−2.99

−2.999

−2.9999

−2.99999

−2.999999

−5.999

−5.9999

−5.99999

−5.999999

−5.9 −5.99 lim f ( x) = −6. The estimate is f ( x)

x→−3

Chapter 2 Limits and Continuity (b)

(c)

2 −9 = ( x+3)( x −3) = x − 3 if x = −3, and lim ( x − 3) = −3 − 3 = −6. f ( x) = xx +3 / x +3 x→−3

(

68. (a) g ( x) = (x 2 − 2)/ x − 2 x g ( x)

1.4 2.81421

)

1.41 2.82421

1.414 2.82821

1.4142 2.828413

1.41421 2.828423

1.414213 2.828426

(b)

( x + 2 )( x − 2 )

x 2 −2

x− 2

( x− 2 )

= x + 2 if x =/ 2, and lim

69. (a) G( x) = ( x + 6)/( x 2 + 4 x − 12) x −5.9 −5.99 G( x) −.126582 −.1251564 x G( x)

−6.1 −.123456

−6.01 −.124843

x→ 2

(x + 2) =

2 + 2 = 2 2.

−5.999 −.1250156

−5.9999 −.1250015

−5.99999 −.1250001

−5.999999 −.1250000

−6.001 −.124984

−6.0001 −.124998

−6.00001 −.124999

−6.000001 −.124999

(b)

x+6

( x +4 x −12) 2

x+6 ( x+6)( x−2)

= 1 if x =/ −6, and lim

1 6 x −2

x−2 x→−

1 −6−2

= − 1 = −0.125. 8

Section 2.2

Limit of a Function and Limit Laws

70. (a) h( x) = ( x − 2 x − 3)/( x − 4 x + 3) 2

x h( x)

2.9 2.052631

2.99 2.005025

2.999 2.000500

2.9999 2.000050

2.99999 2.000005

2.999999 2.0000005

x 3.1 h( x) 1.952380

3.01 1.995024

3.001 1.999500

3.0001 1.999950

3.00001 1.999995

3.000001 1.999999

(b)

2 ( x −3)( x +1) x +1 if x = 3, and li (c) h( x) = x2 −2 x −3 = ( x −3)( x−1) = x−1 m x +1 = 3+1 = 24 = 2. / x −4 x+3 x→ 3 x −1 3−1

71. (a) f ( x) = ( x 2 − 1)/(|x | −1) x f ( x)

−1.1 2.1

−1.01 2.01

−1.001 2.001

−1.0001 2.0001

−1.00001 2.00001

−1.000001 2.000001

x f ( x)

−.9 1.9

−.99 1.99

−.999 1.999

−.9999 1.9999

−.99999 1.99999

−.999999 1.999999

f ( x) =

( x +1)( x −1) x−1 2 x= −1  x −1 ( x +1)( x −1) −( x +1)

(b)

(c)

= x + 1, x ≥ 0 and x =/ 1 = 1 − x, x < 0 and x =/ −1

, and lim (1 − x) = 1 − (−1) = 2. x→−1

72. (a) F ( x) = ( x 2 + 3x + 2)/(2− |x |) x F ( x)

−2.1 −1.1

−2.01 −1.01

−2.001 −1.001

−2.0001 −1.0001

−2.00001 −1.00001

−2.000001 −1.000001

x F ( x)

−1.9 −.9

−1.99 −.99

−1.999 −.999

−1.9999 −.9999

−1.99999 −.99999

−1.999999 −.999999

Chapter 2 Limits and Continuity (b)

x 2 +3 x + 2 2− x

73. (a) g (θ ) = (sinθ )/θ g (θ )

x≥ 0 ( x + 2)(2−xx+1) , = , and lim ( x + 1) = −2 + 1 = −1. ( x + 2)( x +1) x→−2 = x + 1, x < 0 and x = −2 / 2+ x

.1 .998334

−.1 g (θ ) .998334 lim g(θ ) = 1

.01 .999983

.001 .999999

.0001 .999999

.00001 .999999

.000001 .999999

−.01 .999983

−.001 .999999

−.0001 .999999

−.00001 .999999

−.000001 .999999

θ →0

(b)

74. (a) G(t ) = (1 − cos t )/t 2 t

.1 G(t ) .499583

.01 .499995

.001 .499999

.0001 .5

−.1 G(t ) .499583 lim G(t ) = 0.5

−.01 .499995

−.001 .499999

−.0001 −.00001 −.000001 .5 .5 .5

t →0

(b)

.00001 .5

.000001 .5

Section 2.2 f ( x) = x x .9 f(x) .348678

Limit of a Function and Limit Laws

1/(1− x)

75. (a)

.99 .366032

.999 .367695

.9999 .367861

.99999 .367877

.999999 .367879

x 1.1 1.01 f(x) .385543 .369711 lim f ( x) ≈ 0.36788

1.001 .368063

1.0001 .367897

1.00001 .367881

1.000001 .367878

x→1

(b)

Graph is NOT TO SCALE. Also, the intersection of the axes is not the origin: the axes intersect at the point (1, 2.71820). f ( x) = (3 x −1)/x x .1 .01 f(x) 1.161231 1.104669

76. (a)

x −.1 −.01 f(x) 1.040415 1.092599 lim f ( x) ≈ 1.0986

.001 1.099215

.0001 1.098672

−.001 1.098009

−.0001 1.098551

.00001 1.098618

.000001 1.098612

−.00001 1.098606

−.000001 1.098611

x→1

(b)

4 2 4 2 2 2 77. lim f ( x) exists at those points c where lim x = lim x . Thus, c = c c (1 − c ) = 0 c = 0, 1, or −1. x→c

x→c

Moreover, lim f (x) = lim x 2 = 0 and lim f ( x) = lim f ( x) = 1. x→0

x→−1

x→0

x→1

78. Nothing can be concluded about the values of f , g, and h at x = 2. Yes, f (2) could be 0. Since the conditions of the sandwich theorem are satisfied, lim f (x) = −5 =/ 0. x→2

79. 1

lim

f (x )−5

= x→4 x −2

80. (a) 1 = lim

lim f ( x)− lim 5

x →4

lim f ( x)−5

= lim x− lim 2

x →4

f (x ) 2 = x→−2 x

4−2

x →4

lim f ( x )

x→−2

lim x

x →−2

lim

x→ 4

x→4

lim f ( x)

x →−2

f (x) − 5 = 2(1) lim

lim x→−2

x→4

f (x) = 4.

f ( x) = 2 + 5 = 7.

Chapter 2 Limits  and Continuity  f (x )

(b) 1 = lim

x→−2 x

= lim

x→−2

f (x ) x

lim x→−2

1 x



= lim x→−2

(

f (x) x 

( −21) 

f (x ) x→−2 x

lim

= −2.

)

81. (a) 0 = 3⋅ 0 = lim f (x )−5 lim ( x − 2) = lim f (x )−5 ( x − 2) = lim [ f ( x) − 5] x−2 x→2 x−2 x→2 x→2  x→2 = lim f ( x) − 5 lim f ( x) = 5. x→2

x→2 f (x)−5 lim ( x − 2) x→2 x−2 x→2

(b) 0 = 4 ⋅ 0 = lim

lim

x→2

f ( x) = 5 as in part (a).

2  f (x )  f (x ) 2 2 lim x = lim lim x = lim  2 ⋅ x = lim f ( x). 2 2 x x→0 x x→0  x→0 x x→0 x→0 x→0 That is, lim f ( x) = 0. x→0    f (x) f (x ) f (x) (b) 0 = 1⋅ 0 = lim lim x = lim  f (x2) ⋅ x = lim . That is, lim = 0. 2 x x→0 x x→0 x x→0  x→0 x x→0

82. (a) 0 = 1⋅ 0 = lim

f (x )

83. (a) lim x sin 1 = 0 x x→0

(b) −1 ≤ sin 1x ≤ 1 for x = 0: x > 0 − x ≤ x sin 1 ≤x/ x lim x sin 1 = 0 xby the sandwich theorem; x→0 x < 0 − x ≥ x sin 1 ≥xx lim x sin 1 = 0 xby the sandwich theorem. x→0

( )

84. (a) lim x 2 cos 13 = 0 x

→0

(b) −1 ≤ cos

( ) ≤ 1 for x =/ 0 − x 3

1 x

≤ x cos 2

lim x 2 = 0. x→0

85-90. Example CAS commands: Maple: f := x -> (x^4 − 16)/(x − 2); x0 := 2; plot( f (x), x = x0-1..x0+1, color = black, title = "Section 2.2, #85(a )" ); limit( f (x), x = x0 );

( )≤ x x3 1

lim x 2

cos 2

x→0

( ) = 0 by the sandwich theorem since 3

1 x

Section 2.3

The Precise Definition of a Limit

In Exercise 87, note that the standard cube root, x^(1/3), is not defined for x<0 in many CASs. This can be overcome in Maple by entering the function as f := x -> (surd(x +1, 3) − 1)/x. Mathematica: (assigned function and values for x0 and h may vary) Clear[f , x] f [x _]:=(x 3 − x 2 − 5x − 3)/(x + 1)2 x0= −1; h = 0.1; Plot[f [x],{x, x0 − h, x0 + h}] Limit[f [x], x → x0] 2.3 1.

THE PRECISE DEFINITION OF A LIMIT Step 1: x − 5 < δ −δ < x − 5 < δ −δ + 5 < x < δ + 5 Step 2: δ + 5 = 7 δ = 2,or −δ + 5 = 1 δ = 4. The value of δ which assures x − 5 < δ 1 < x < 7 is the smaller value, δ = 2.

2. Step 1: Step 2:

x − 2 < δ −δ < x − 2 < δ −δ + 2 < x < δ + 2 −δ + 2 = 1 δ = 1, or δ + 2 = 7 δ = 5. The value of δ which assures x − 2 < δ 1 < x < 7 is the smaller value, δ = 1.

Step 1: Step 2:

x − (−3) < δ −δ < x + 3 < δ −δ − 3 < x < δ − 3 −δ − 3 = − 7 δ = 1 , or δ − 3 = − 1 δ = 5 .

The value of δ which assures x − (−3) < δ − 7 <2 x < − 1 is the smaller value, δ = 1 . 2 2 4. Step 1: Step 2:

( )

x − − 23 < δ −δ < x + 3 <2 δ −δ − 3 < x2< δ − 3 −δ − 32 = − 27 δ = 2, or δ − 3 =2 − 1 δ2 = 1.

( )

The value of δ which assures x − − 32 < δ − 7 <2 x < − 1 is the smaller value, δ = 1. 2

5. Step 1: Step 2:

x − 21 < δ −δ < x − 1 <2 δ −δ + 1 < x2< δ + 1 −δ + 12 = 49 δ = 1 ,18or δ + 1 =2 4 δ7 = 1 . 14

The value of δ which assures x − 12 < δ 4 <9 x < 4 is7 the smaller value, δ = 1 .18

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