Calculus Volume 1

# Chapter 2

## Checkpoint

2.1

2.25

2.2

12.006001

2.3

17 unit2

2.4

$lim x → 1 1 x − 1 x − 1 = −1 lim x → 1 1 x − 1 x − 1 = −1$

2.5

$lim x → 2 h ( x ) = −1 . lim x → 2 h ( x ) = −1 .$

2.6

$limx→2|x2−4|x−2limx→2|x2−4|x−2$ does not exist.

2.7

a. $limx→2−|x2−4|x−2=−4;limx→2−|x2−4|x−2=−4;$ b. $limx→2+|x2−4|x−2=4limx→2+|x2−4|x−2=4$

2.8

a. $limx→0−1x2=+∞;limx→0−1x2=+∞;$ b. $limx→0+1x2=+∞;limx→0+1x2=+∞;$ c. $limx→01x2=+∞limx→01x2=+∞$

2.9

a. $limx→2−1(x−2)3=−∞;limx→2−1(x−2)3=−∞;$ b. $limx→2+1(x−2)3=+∞;limx→2+1(x−2)3=+∞;$ c. $limx→21(x−2)3limx→21(x−2)3$ DNE. The line $x=2x=2$ is the vertical asymptote of $f(x)=1/(x−2)3.f(x)=1/(x−2)3.$

2.10

Does not exist.

2.11

$11 10 11 10$

2.12

−13;

2.13

$1 3 1 3$

2.14

$1 4 1 4$

2.15

−1;

2.16

$1 4 1 4$

2.17

$limx→−1−f(x)=−1limx→−1−f(x)=−1$

2.18

+∞

2.19

0

2.20

0

2.21

f is not continuous at 1 because $f(1)=2≠3=limx→1f(x).f(1)=2≠3=limx→1f(x).$

2.22

$f(x)f(x)$ is continuous at every real number.

2.23

Discontinuous at 1; removable

2.24

$[ −3 , + ∞ ) [ −3 , + ∞ )$

2.25

0

2.26

$f(0)=1>0,f(1)=−2<0;f(x)f(0)=1>0,f(1)=−2<0;f(x)$ is continuous over $[0,1].[0,1].$ It must have a zero on this interval.

2.27

Let $ε>0;ε>0;$ choose $δ=ε3;δ=ε3;$ assume $0<|x−2|<δ.0<|x−2|<δ.$

Thus, $|(3x−2)−4|=|3x−6|=|3|·|x−2|<3·δ=3·(ε/3)=ε.|(3x−2)−4|=|3x−6|=|3|·|x−2|<3·δ=3·(ε/3)=ε.$

Therefore, $limx→23x−2=4.limx→23x−2=4.$

2.28

Choose $δ=min{9−(3−ε)2,(3+ε)2−9}.δ=min{9−(3−ε)2,(3+ε)2−9}.$

2.29

$| x 2 − 1 | = | x − 1 | · | x + 1 | < ε / 3 · 3 = ε | x 2 − 1 | = | x − 1 | · | x + 1 | < ε / 3 · 3 = ε$

2.30

$δ = ε 2 δ = ε 2$

## Section 2.1 Exercises

1.

a. 2.2100000; b. 2.0201000; c. 2.0020010; d. 2.0002000; e. (1.1000000, 2.2100000); f. (1.0100000, 2.0201000); g. (1.0010000, 2.0020010); h. (1.0001000, 2.0002000); i. 2.1000000; j. 2.0100000; k. 2.0010000; l. 2.0001000

3.

$y = 2 x y = 2 x$

5.

3

7.

a. 2.0248457; b. 2.0024984; c. 2.0002500; d. 2.0000250; e. (4.1000000,2.0248457); f. (4.0100000,2.0024984); g. (4.0010000,2.0002500); h. (4.00010000,2.0000250); i. 0.24845673; j. 0.24984395; k. 0.24998438; l. 0.24999844

9.

$y = x 4 + 1 y = x 4 + 1$

11.

π

13.

a. −0.95238095; b. −0.99009901; c. −0.99502488; d. −0.99900100; e. (−1;.0500000,−0;.95238095); f. (−1;.0100000,−0;.9909901); g. (−1;.0050000,−0;.99502488); h. (1.0010000,−0;.99900100); i. −0.95238095; j. −0.99009901; k. −0.99502488; l. −0.99900100

15.

$y = − x − 2 y = − x − 2$

17.

−49 m/sec (velocity of the ball is 49 m/sec downward)

19.

5.2 m/sec

21.

−9.8 m/sec

23.

6 m/sec

25.

Under, 1 unit2; over: 4 unit2. The exact area of the two triangles is $12(1)(1)+12(2)(2)=2.5units2.12(1)(1)+12(2)(2)=2.5units2.$

27.

Under, 0.96 unit2; over, 1.92 unit2. The exact area of the semicircle with radius 1 is $π(1)22=π2π(1)22=π2$ unit2.

29.

Approximately 1.3333333 unit2

## Section 2.2 Exercises

31.

$limx→1f(x)limx→1f(x)$ does not exist because $limx→1−f(x)=−2≠limx→1+f(x)=2.limx→1−f(x)=−2≠limx→1+f(x)=2.$

33.

$lim x → 0 ( 1 + x ) 1 / x = 2.7183 lim x → 0 ( 1 + x ) 1 / x = 2.7183$

35.

a. 1.98669331; b. 1.99986667; c. 1.99999867; d. 1.99999999; e. 1.98669331; f. 1.99986667; g. 1.99999867; h. 1.99999999; $limx→0sin2xx=2limx→0sin2xx=2$

37.

$lim x → 0 sin a x x = a lim x → 0 sin a x x = a$

39.

a. −0.80000000; b. −0.98000000; c. −0.99800000; d. −0.99980000; e. −1.2000000; f. −1.0200000; g. −1.0020000; h. −1.0002000; $limx→1(1−2x)=−1limx→1(1−2x)=−1$

41.

a. −37.931934; b. −3377.9264; c. −333,777.93; d. −33,337,778; e. −29.032258; f. −3289.0365; g. −332,889.04; h. −33,328,889 $limx→0z−1z2(z+3)=−∞limx→0z−1z2(z+3)=−∞$

43.

a. 0.13495277; b. 0.12594300; c. 0.12509381; d. 0.12500938; e. 0.11614402; f. 0.12406794; g. 0.12490631; h. 0.12499063; $∴limx→21−2xx2−4=0.1250=18∴limx→21−2xx2−4=0.1250=18$

45.

a. 10.00000; b. 100.00000; c. 1000.0000; d. 10,000.000; Guess: $limα→0+1αcos(πα)=∞,limα→0+1αcos(πα)=∞,$ actual: DNE

47.

False; $limx→−2+f(x)=+∞limx→−2+f(x)=+∞$

49.

False; $limx→6f(x)limx→6f(x)$ DNE since $limx→6−f(x)=2limx→6−f(x)=2$ and $limx→6+f(x)=5.limx→6+f(x)=5.$

51.

2

53.

1

55.

1

57.

DNE

59.

0

61.

DNE

63.

2

65.

3

67.

DNE

69.

0

71.

−2

73.

DNE

75.

0

77.

79.

81.

a. $ρ2ρ2$ b. $ρ1ρ1$ c. DNE unless $ρ1=ρ2.ρ1=ρ2.$ As you approach $xSFxSF$ from the left, you are in the high-density area of the shock. When you approach from the right, you have not experienced the “shock” yet and are at a lower density.

## Section 2.3 Exercises

83.

Use constant multiple law and difference law: $limx→0(4x2−2x+3)=4limx→0x2−2limx→0x+limx→03=3limx→0(4x2−2x+3)=4limx→0x2−2limx→0x+limx→03=3$

85.

Use root law: $limx→−2x2−6x+3=limx→−2(x2−6x+3)=19limx→−2x2−6x+3=limx→−2(x2−6x+3)=19$

87.

49

89.

1

91.

$− 5 7 − 5 7$

93.

$limx→4x2−16x−4=16−164−4=00;limx→4x2−16x−4=16−164−4=00;$ then, $limx→4x2−16x−4=limx→4(x+4)(x−4)x−4=8limx→4x2−16x−4=limx→4(x+4)(x−4)x−4=8$

95.

$limx→63x−182x−12=18−1812−12=00;limx→63x−182x−12=18−1812−12=00;$ then, $limx→63x−182x−12=limx→63(x−6)2(x−6)=32limx→63x−182x−12=limx→63(x−6)2(x−6)=32$

97.

$limx→9t−9t−3=9−93−3=00;limx→9t−9t−3=9−93−3=00;$ then, $limt→9t−9t−3=limt→9t−9t−3t+3t+3=limt→9(t+3)=6limt→9t−9t−3=limt→9t−9t−3t+3t+3=limt→9(t+3)=6$

99.

$limθ→πsinθtanθ=sinπtanπ=00;limθ→πsinθtanθ=sinπtanπ=00;$ then, $limθ→πsinθtanθ=limθ→πsinθsinθcosθ=limθ→πcosθ=−1limθ→πsinθtanθ=limθ→πsinθsinθcosθ=limθ→πcosθ=−1$

101.

$limx→1/22x2+3x−22x−1=12+32−21−1=00;limx→1/22x2+3x−22x−1=12+32−21−1=00;$ then, $limx→1/22x2+3x−22x−1=limx→1/2(2x−1)(x+2)2x−1=52limx→1/22x2+3x−22x−1=limx→1/2(2x−1)(x+2)2x−1=52$

103.

−∞

105.

−∞

107.

$lim x → 6 2 f ( x ) g ( x ) = 2 lim x → 6 f ( x ) lim x → 6 g ( x ) = 72 lim x → 6 2 f ( x ) g ( x ) = 2 lim x → 6 f ( x ) lim x → 6 g ( x ) = 72$

109.

$lim x → 6 ( f ( x ) + 1 3 g ( x ) ) = lim x → 6 f ( x ) + 1 3 lim x → 6 g ( x ) = 7 lim x → 6 ( f ( x ) + 1 3 g ( x ) ) = lim x → 6 f ( x ) + 1 3 lim x → 6 g ( x ) = 7$

111.

$lim x → 6 g ( x ) − f ( x ) = lim x → 6 g ( x ) − lim x → 6 f ( x ) = 5 lim x → 6 g ( x ) − f ( x ) = lim x → 6 g ( x ) − lim x → 6 f ( x ) = 5$

113.

$lim x → 6 [ ( x + 1 ) f ( x ) ] = ( lim x → 6 ( x + 1 ) ) ( lim x → 6 f ( x ) ) = 28 lim x → 6 [ ( x + 1 ) f ( x ) ] = ( lim x → 6 ( x + 1 ) ) ( lim x → 6 f ( x ) ) = 28$

115.

a. 9; b. 7

117.

a. 1; b. 1

119.

$lim x → −3 − ( f ( x ) − 3 g ( x ) ) = lim x → −3 − f ( x ) − 3 lim x → −3 − g ( x ) = 0 + 6 = 6 lim x → −3 − ( f ( x ) − 3 g ( x ) ) = lim x → −3 − f ( x ) − 3 lim x → −3 − g ( x ) = 0 + 6 = 6$

121.

$lim x → −5 2 + g ( x ) f ( x ) = 2 + ( lim x → −5 g ( x ) ) lim x → −5 f ( x ) = 2 + 0 2 = 1 lim x → −5 2 + g ( x ) f ( x ) = 2 + ( lim x → −5 g ( x ) ) lim x → −5 f ( x ) = 2 + 0 2 = 1$

123.

$lim x → 1 f ( x ) − g ( x ) 3 = lim x → 1 f ( x ) − lim x → 1 g ( x ) 3 = 2 + 5 3 = 7 3 lim x → 1 f ( x ) − g ( x ) 3 = lim x → 1 f ( x ) − lim x → 1 g ( x ) 3 = 2 + 5 3 = 7 3$

125.

$lim x → −9 ( x f ( x ) + 2 g ( x ) ) = ( lim x → −9 x ) ( lim x → −9 f ( x ) ) + 2 lim x → −9 ( g ( x ) ) = ( −9 ) ( 6 ) + 2 ( 4 ) = −46 lim x → −9 ( x f ( x ) + 2 g ( x ) ) = ( lim x → −9 x ) ( lim x → −9 f ( x ) ) + 2 lim x → −9 ( g ( x ) ) = ( −9 ) ( 6 ) + 2 ( 4 ) = −46$

127.

The limit is zero.

129.

a.

b. ∞. The magnitude of the electric field as you approach the particle q becomes infinite. It does not make physical sense to evaluate negative distance.

## Section 2.4 Exercises

131.

The function is defined for all x in the interval $(0,∞).(0,∞).$

133.

Removable discontinuity at $x=0;x=0;$ infinite discontinuity at $x=1x=1$

135.

Infinite discontinuity at $x=ln2x=ln2$

137.

Infinite discontinuities at $x=(2k+1)π4,x=(2k+1)π4,$ for $k=0,±1,±2,±3,…k=0,±1,±2,±3,…$

139.

No. It is a removable discontinuity.

141.

Yes. It is continuous.

143.

Yes. It is continuous.

145.

$k = −5 k = −5$

147.

$k = −1 k = −1$

149.

$k = 16 3 k = 16 3$

151.

Since both s and $y=ty=t$ are continuous everywhere, then $h(t)=s(t)−th(t)=s(t)−t$ is continuous everywhere and, in particular, it is continuous over the closed interval $[2,5].[2,5].$ Also, $h(2)=3>0h(2)=3>0$ and $h(5)=−3<0.h(5)=−3<0.$ Therefore, by the IVT, there is a value $x=cx=c$ such that $h(c)=0.h(c)=0.$

153.

The function $f(x)=2x−x3f(x)=2x−x3$ is continuous over the interval $[1.25,1.375][1.25,1.375]$ and has opposite signs at the endpoints.

155.

a.

b. It is not possible to redefine $f(1)f(1)$ since the discontinuity is a jump discontinuity.

157.

Answers may vary; see the following example:

159.

Answers may vary; see the following example:

161.

False. It is continuous over $(−∞,0)∪(0,∞).(−∞,0)∪(0,∞).$

163.

False. Consider $f(x)={xifx≠04ifx=0.f(x)={xifx≠04ifx=0.$

165.

False. IVT only says that there is at least one solution; it does not guarantee that there is exactly one. Consider $f(x)=cos(x)f(x)=cos(x)$ on $[−π,2π].[−π,2π].$

167.

False. The IVT does not work in reverse! Consider $(x−1)2(x−1)2$ over the interval $[−2,2].[−2,2].$

169.

$R = 0.0001519 m R = 0.0001519 m$

171.

$D = 345,826 km D = 345,826 km$

173.

For all values of $a,f(a)a,f(a)$ is defined, $limθ→af(θ)limθ→af(θ)$ exists, and $limθ→af(θ)=f(a).limθ→af(θ)=f(a).$ Therefore, $f(θ)f(θ)$ is continuous everywhere.

175.

Nowhere

## Section 2.5 Exercises

177.

For every $ε>0,ε>0,$ there exists a $δ>0,δ>0,$ so that if $0<|t−b|<δ,0<|t−b|<δ,$ then $|g(t)−M|<ε|g(t)−M|<ε$

179.

For every $ε>0,ε>0,$ there exists a $δ>0,δ>0,$ so that if $0<|x−a|<δ,0<|x−a|<δ,$ then $|φ(x)−A|<ε|φ(x)−A|<ε$

181.

$δ ≤ 0.25 δ ≤ 0.25$

183.

$δ ≤ 2 δ ≤ 2$

185.

$δ ≤ 1 δ ≤ 1$

187.

$δ < 0.3900 δ < 0.3900$

189.

Let $δ=ε.δ=ε.$ If $0<|x−3|<ε,0<|x−3|<ε,$ then $|x+3−6|=|x−3|<ε.|x+3−6|=|x−3|<ε.$

191.

Let $δ=ε4.δ=ε4.$ If $0<|x|<ε4,0<|x|<ε4,$ then $|x4|=x4<ε.|x4|=x4<ε.$

193.

Let $δ=ε2.δ=ε2.$ If $5−ε2 then $|5−x|=5−x<ε.|5−x|=5−x<ε.$

195.

Let $δ=ε/5.δ=ε/5.$ If $1−ε/5 then $|f(x)−3|=5x−5<ε.|f(x)−3|=5x−5<ε.$

197.

Let $δ=3M.δ=3M.$ If $0<|x+1|<3M,0<|x+1|<3M,$ then $f(x)=3(x+1)2>M.f(x)=3(x+1)2>M.$

199.

0.328 cm, $ε=8,δ=0.33,a=12,L=144ε=8,δ=0.33,a=12,L=144$

201.

203.

0

205.

$lim x → a f x + lim x → a g x = L + M lim x → a f x + lim x → a g x = L + M$

207.

## Review Exercises

209.

False

211.

False. A removable discontinuity is possible.

213.

5

215.

$8 / 7 8 / 7$

217.

DNE

219.

$2 / 3 2 / 3$

221.

−4;

223.

Since $−1≤cos(2πx)≤1,−1≤cos(2πx)≤1,$ then $−x2≤x2cos(2πx)≤x2.−x2≤x2cos(2πx)≤x2.$ Since $limx→0x2=0=limx→0−x2,limx→0x2=0=limx→0−x2,$ it follows that $limx→0x2cos(2πx)=0.limx→0x2cos(2πx)=0.$

225.

$[ 2 , ∞ ] [ 2 , ∞ ]$

227.

$c = −1 c = −1$

229.

$δ = ε 3 δ = ε 3$

231.

$0 m / sec 0 m / sec$

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