CHAPTER 2
Limits and Continuity SECTION 2.1 1.
The function f (x) is graphed. lim f (x) = x→0−
A. 0
B. 4
C. 2
D. undefined
2.
Answer true or false. For the function graphed lim f (x) is undefined. x→2
x2 − 9 x2 − 9 by evaluating f (x) = at x = 4, 3.5, 3.1, 3.01, 3.001, 2, 2.5, x→3 x − 3 x−3 2.9, 2.99, and 2.999.
3. Approximate the lim
B. −9
A. 6
D. −6
C. 0
4. Answer true or false. If lim f (x) = 6 and lim f (x) = 6, then lim f (x) = 0. x→0+
5. Approximate the A. 1
lim
x→−6−
x→0−
x x by evaluating f (x) = at appropriate values of x. x+6 x+6
B. 5
7. Approximate the limit by evaluating f (x) = A. 1
C. ∞
B. 0
6. Approximate the limit by evaluating f (x) = A. 1
x→0
D. −∞
5x 5x at appropriate values of x. lim− = sin x x→0 sin x 1 C. D. ∞ 5 sin x sin x at appropriate values of x. lim = x→0 x x
B. −1
C. 0 √
8. Approximate the limit by evaluating f (x) = √ x+1−1 lim = − x x→0 1 A. B. 0 2
x+1−1 at appropriate values of x. x
C. ∞
1
D. ∞
D. −∞
2
True/False and Multiple Choice Questions
9. Use a graphing utility to approximate the y-coordinates of any horizontal asymptote of 6x − 8 . y = f (x) = x+2 A. 6
B. 1
D. −4
C. None exist.
10. Use a graphing utility to approximate the y-coordinate of any horizontal asymptote of sin x y = f (x) = . x A. 0
C. −1 and 1
B. 1
D. −1
11. Use a graphing utility to approximate the y-coordinate of any horizontal asymptote of x3 + 5 y = f (x) = . x−3 A. 0
B. None exist.
D. −1 and 1
C. 1
12. Answer true or false. A graphing utility can be used to show f (x) =
5 1+ x
x has a horizontal
asymptote. 13. Answer true or false. A graphing utility can be used to show f (x) =
1 10 + 2x
2x has a horizontal
asymptote. 4 − 1 4−x 14. Answer true or false. lim is equivalent to lim− x . 3 x→−∞ 3 + x x→0 +1 x
15. Answer true or false. f (x) =
x3 has no horizontal asymptote. −2
x5
16. Approximate the horizontal asymptote of f (x) = A. It does not exist. 17.
5 − 2x . 3+x
B. y = 0
C. y = −2
D. y = 2
B. 0
C. It does not exist.
D. −4
sin(8x) = x→0 sin(2x) lim
A. 4
Section 2.2
3
SECTION 2.2 1. Given that lim f (x) = 3 and lim g(x) = 5, find, if it exists, lim [2f (x) − 3g(x)]2 . x→a
x→a
A. −81 C. 9 2.
x→a
B. 81 D. It does not exist.
lim 5 =
x→3
A. 3
B. 5
C. 8
D. 15
C. 12
D. 1
3. Answer true or false. lim 9x = 18 x→2
4.
x2 − 36 = x→−6 x + 6 lim
A. −∞ 5.
B. −12
10 = x+5
lim
x→−5
B. −∞ D. It does not exist.
A. +∞ C. 0 6. Find lim+ f (x) where f (x) = x→0
7.
8.
A. 0
B. 3
C. −3
D. It does not exist.
lim
x→5+
x = x−5
A. −∞
B. ∞
C. 0
D. It does not exist.
lim
x→4−
3 = x−4
A. ∞ C. It does not exist. 9.
|3x| . x
lim
x→6
B. 0 D. −∞
3x − 2 = x−6
A. ∞ C. It does not exist.
B. 0 D. −∞
4
True/False and Multiple Choice Questions
10.
Use the graph of f (x) above. lim f (x) = x→−1
B. −1 D. 0
A. 1 C. It does not exist. 11.
Use the graph of f (x) above. lim f (x) = x→1
B. −1 D. 1
A. 0 C. It does not exist. 12. Let f (x) =
x3 , x − 2,
x≤2 . lim f (x) = x > 2 x→2+
A. 8 C. 0
B. 4 D. It does not exist.
13. Let g(x) =
x3 − 3, 5
x ,
x≤1 . lim g(x) = x > 1 x→1
A. 5 C. 3 14.
B. 1 D. It does not exist.
16 − x = lim √ x−4
x→16
A. 8
B. 0
C. −8
D. It does not exist. √
15. Answer true or false. lim
x→0+
x2 + 25 − 5 1 = . x 4
Section 2.3
5
SECTION 2.3 1. Answer true or false.
6x2 − 6x + 5 =0 x→+∞ 3x2 + 2x + 9 lim
5x3 + 4x2 − 3x + 8 = −5 x→−∞ x3 + 2 √ x2 + 9 =2 3. Answer true or false. lim x→+∞ x − 7 √ 9x2 + 5 = −3 4. Answer true or false. lim x→−∞ x+3
2. Answer true or false.
5. Answer true or false. 6.
7.
lim
x→+∞
lim
x→−∞
A. 0
B. 3
C. 1
D. It does not exist.
4x3 − 2 = x→−∞ x3 lim
B. −∞
2x2 = x→−1 x8 − 2x2 − x B. −∞ D. It does not exist.
x+5 = lim √ x−3
x→9
B. −∞ D. It does not exist.
A. +∞ C. 84
10.
lim
x→−∞
20x10 − 2x5 + 2 = 5x10 + x5 − 3
A. +∞ C. 2 11.
C. ∞
lim
A. +∞ C. 0 9.
1 =0 3x − 2
4x − 3 = x4 − 3
A. 4 8.
lim
B. −∞ D. It does not exist.
lim (x6 − 400x5 − x4 + x)
x→+∞
A. +∞ C. −500
B. −∞ D. It does not exist.
D. −4
6
True/False and Multiple Choice Questions
√ 3 12.
x→+∞
A. 13.
x3 = 3x
lim
1 3
√ 3
lim
x→+∞
B. +∞
C. −∞
D. 0
B. +∞
C. −∞
D. 0
27x3 = 3x
A. 1
√ 14. Answer true or false.
15.
lim
x→+∞
x2 − 7 + 2 does not exist. x
18x3 + 5 = x→−∞ 6x3 − 2 lim
A. −3
B. 3
C. 0
D. 18
Section 2.4
7
SECTION 2.4 1. Find a least number δ such that |f (x) − L| < if 0 < |x − a| < δ. lim 10x = 50; = 0.1 x→5
A. 0.1
B. 0.01
C. 0.5
D. 0.025
2. Find a least number δ such that |f (x) − L| < if 0 < |x − a| < δ. lim 3x − 5 = 1; = 0.1 x→2
A. 0.033
B. 0.33
C. 3.0
D. 0.3
3. Answer true or false. It can be shown that if |f (x) − L| < when 0 < |x − a| < δ, |x2 − 9| < if |x − 3| < δ for arbitrarily small positive . x2 − 25 = −10; = 0.001 x→−5 x + 5
4. Find a least number δ such that |f (x) − L| < if 0 < |x − a| < δ. lim A. 0.001
B. 0.000001
C. 0.005
D. 0.025
5. Find a least positive number N such that |f (x) − L| < if x > N . A. N = 100
B. N = 1,000
lim
x→+∞
C. N = 10
D. N = 10,000
6. Find a greatest negative number N such that |f (x) − L| < if x < N . A. N = −100,000
100 = 0; = 0.1 x
B. N = −10,000
C. N = −100
7. Answer true or false. It is possible to prove that lim
1 = 0. x3 + 9
8. Answer true or false. It is possible to prove that lim
1 = 0. 4x + 16
9. Answer true or false. It is possible to prove that lim
3x = 0. 5x + 2
x→+∞
x→−∞
x→+∞
10. Answer true or false. It is possible to prove that lim
x→3
x2
10 = 0; = 0.1 x→−∞ x lim
D. N = −10
1 = +∞. −9
11. To prove that lim (x − 2) = 3 a reasonable relationship between δ and would be x→5
A. δ = 5
B. δ =
C. δ =
√
12. Answer true or false. To use a δ- approach to show that lim
x→0+
would be to change the limit to lim x2 = 0.
D. δ =
1 .
1 = +∞, a reasonable first step x2
x→+∞
−1 = −∞. x→4 |x − 4|
13. Answer true or false. It is possible to show that lim 14. To prove that lim f (x) = 9 where f (x) = x→3
3x,
x<3
x + 6, x ≥ 3
a reasonable relationship between δ
and would be A. δ = 3
B. δ =
C. δ = + 3
D. δ = 2 + 3.
8
True/False and Multiple Choice Questions
15. Answer true or false. It is possible to show that lim
x→0
16. If lim
x→4
√
x = 0. 5
x + 3 = 5 and ε = 0.01, find a smallest positive number δ such that |f (x) − 5| < 0.01 if
0 < |x − 4| < δ. A. 0.01
B. 0.5
C. 0.1
D. 0.04
17. If lim 10x = 40 and ε = 0.01, find a smallest positive number δ such that |f (x) − 40| < 0.01 if x→4
0 < |x − 4| < δ. A. 0.01
B. 0.1
C. 0.001
D. 0.0001
Section 2.5
9
SECTION 2.5 1.
On the interval of [−10, 10], where is f not continuous? A. −2, 2
C. −2
B. 2
D. nowhere
2.
On the interval of [−10, 10], where is f not continuous? A. 3
B. 0, 3
C. 0
D. nowhere
3. Answer true or false. f (x) = x7 − 2x5 + 3 has no point of discontinuity. 4. Answer true or false. f (x) = |x2 − 4| has points of discontinuity at x = −2 and x = 2. 5. Find the x-coordinates for all points of discontinuity for f (x) = A. 5, 7
B. −7
D. −5, −7
C. 7
6. Find the x-coordinates for all points of discontinuity for f (x) = A. 0
x−5 . x2 − 12x + 35
9x2 + 36 . |3x + 6|
C. −2
B. 2
D. −2, 2
7. Find the x-coordinates for all points of discontinuity for f (x) = A. 1
B.
√ 3
2
C. 1,
√ 3
x3 + 2, x ≤ 1 . −5, x>1
2
D. None exists.
8. Find the value of k, if possible, that will make the function continuous. x + 2k, x≤1 f (x) = kx2 + x + 1, x > 1 A. 1
B. −1
C. 2
D. None exists.
x+5 has a removable discontinuity at x = 1. x−1 x≤2 x3 , is continuous everywhere. 10. Answer true or false. The function f (x) = 2 x + 4, x > 2 9. Answer true or false. The function f (x) =
10
True/False and Multiple Choice Questions
11. Answer true or false. If f and g are each continuous at c, f /g may be discontinuous at c. 12. Answer true or false. The Intermediate-Value Theorem can be used to approximate the locations −3x3 + 2x + 1 . of all discontinuities for f (x) = x 13. Answer true or false. f (x) = x2 − 3x + 1 = 0 has at least one solution on the interval [−1, 0]. 14. Answer true or false. f (x) = x4 − 2x2 + 3 = 0 has at least one solution on the interval [0, 1]. 15. Use the fact that
√ 4
A. 1.65
8 is a solution of x4 − 8 = 0 to approximate B. 1.66
C. 1.68
√ 4
8 with an error of at most 0.005. D. 1.69
2x − 5, if x ≤ 2 . Find the value for the constant k that will make the function conkx + 3, if x > 2 tinuous everywhere.
16. f (x) =
A. −2
B. 0
C. 2
D. −1
if x ≤ 0 . Find the value for the constant k that will make the function con3x + k, if x > 0 tinuous everywhere.
17. f (x) =
A. 2
x2 − 2,
B. −2
C. 0
D. 1
Section 2.6
11
SECTION 2.6 1. Answer true or false. f (x) = tan(x2 − 3) has no point of discontinuity. 2. A point of discontinuity of f (x) = π 2
1 is at |0.5 − sin x|
π 3 5 4 3. Find the limit. lim sin = cos x→+∞ x x A.
A. 0 4. Find the limit. lim
x→0−
π 4
D.
π . 6
B.
C.
B. 1
C. −1
D. +∞
C. 1
D. −∞
sin3 x = x3
A. +∞
B. 0 sin(7x) = x→0 sin(9x)
5. Find the limit. lim A. +∞
B. 0
6. Find the limit. lim
x→0
A. 1
1 − cos x = 6 1 B. 6
7 9
D. 1
C. 3
D. 0
C. −∞
D. 0
C. +∞
D. −∞
C.
sin2 x = x→0 tan2 x
7. Find the limit. lim A. +∞
B. 1
8. Find the limit. lim
x→0
sin x = sin(−x)
A. −1
B. 1
9. Find the limit. lim− tan x→0
1 = x B. −1 D. does not exist
A. 1 C. −∞ x2 = x→0 sin x
10. Find the limit. lim A. 0
B. 1
C. −1
D. +∞
11. Answer true or false. The value of k that makes f continuous for f (x)
sin x , x≤0 is 0. x cos x + k, x > 0
12
True/False and Multiple Choice Questions
sin x sin2 x = 1 and that lim = 1 guarantees that x→0 x x→0 x2
12. Answer true or false. The fact that lim sin2 x = 1 by the Squeeze Theorem. x→0 x lim
13. Answer true or false. The Squeeze Theorem can be used to show lim x + 1 = 1 utilizing lim x = 0 x→0
and lim 1 = 1.
x→0
x→0
14. Answer true or false. The Intermediate-Value Theorem can be used to show that the equation y 5 = cos x has at least one solution on the interval [−5π/6, 5π/6]. 15.
lim
x→0
x sin x +2 3x 3 sin x
A. 1
= B. 2
16. Find all points of discontinuity, if any, for f (x) = A. x = 2 C. x = 0 17.
lim
x→0
C.
1 2
1 . 5 − 2 sin x B. x = π D. None exist.
sin 4x = x
A. 0 C. 4
B. 1 D. It does not exist.
D. 0
Chapter 2 Test
13
CHAPTER 2 TEST 1.
The function f is graphed. lim f (x) = x→−2
B. −2
A. 2
C. 0
D. undefined
x2 − 49 x2 − 49 by evaluating f (x) = at x = −6, −6.5, −6.9, −6.99, −6.999, x→−7 x + 7 x+7 −7, −7.5, −7.1, −7.01, and −7.001.
2. Approximate lim
B. −7
A. 7
D. −14
C. 14
3. Use a graphing utility to approximate the y-coordinate of the horizontal asymptote of 10x + 3 . y = f (x) = 2x − 5 3 3 A. 5 B. C. − D. −5 5 5 3x 3 has a 4. Answer true or false. A graphing utility can be used to show that f (x) = 12 + 3x horizontal asymptote. 5. Answer true or false.
5 x4 is equivalent to lim . x→−∞ x4 x→0− 5 lim
6. Given that lim f (x) = 5 and lim g(x) = −5, find lim [6f 2 (x) − g(x)]. x→a
A. 0 7.
x→a
B. 150
C. 155
D. 145
B. −1
C. 7
D. Does not exist
B. 0
C. +∞
D. Does not exist
C. 0
D. Does not exist
lim 7 =
x→6
A. 1 8.
x→a
x = x→3 x − 3 lim
A. 1 √ 9. Let f (x) = A. 1
x, x ≤ 1 . lim f (x) = √ 3 x, x > 1 x→1 B. −1
10. Find a least number δ such that |f (x) − L| < if 0 < |x − a| < δ. lim 2x = 20; < 0.01 x→10
A. 0.01
B. 0.005
C. 0.05
D. 0.0025
14
True/False and Multiple Choice Questions
x2 − 81 = −18; < 0.001 x→−9 x + 9
11. Find a least number δ such that |f (x) − L| < if 0 < |x − a| < δ. lim A. 0.001
B. 0.000001
C. 0.006
12. Answer true or false. It is possible to prove that lim
x→−∞
D. 0.03
1 = 0. x7 + 2
13. To prove lim (7x + 2) = 16, a reasonable relationship between δ and would be x→2
A. δ = 7
B. δ = 7
D. δ = − 7.
C. δ =
14. Answer true or false. It is possible to show that lim (x2 + 2) = 2. x→+∞
15. Find the x-coordinate of each point of discontinuity of f (x) = B. −3, 8
A. 3
16. Answer true or false. f (x) =
x−3 . x2 + 5x − 24
C. −8, 3
D. −3, −8
1 has a removable discontinuity at x = 2. x2 − 4
17. Answer true or false. f (x) = x2 − 3 = 0 has at least one solution on the interval [1, 4]. 18. Find lim
x→0
sin(2x) . sin(−5x) 2 5
C. −
B. −1
C. 1
B. −
A. 0
5 2
D. not defined
sin3 x . x→0 tan2 x
19. Find lim A. 0
20. Answer true or false. lim
x→0
1 − cos x =0 sin x
D. undefined
SOLUTIONS
SECTION 2.1 1. B 13. F
2. F 3. A 4. T 5. C 6. B 14. T 15. F 16. C 17. A
7. A
8. A
9. A
10. A
11. B
12. T
SECTION 2.2 1. C 13. D
2. B 3. T 4. B 14. C 15. F
5. D
6. B
7. B
8. D
9. C
10. C
11. D
12. C
5. T
6. A
7. A
8. A
9. C
10. C
11. A
12. A
2. A 3. T 4. A 5. B 6. C 14. B 15. T 16. D 17. C
7. T
8. T
9. F
10. F
11. B
12. T
7. A
8. A
9. F
10. T
11. T
12. T
7. A
8. A
9. D
10. A
11. F
12. F
2. D 3. A 4. F 5. F 6. C 7. C 14. F 15. C 16. F 17. T 18. B
8. D 19. A
9. A 10. B 20. T
11. A
12. T
SECTION 2.3 1. F 13. A
2. F 3. F 4. T 14. F 15. B
SECTION 2.4 1. B 13. T
SECTION 2.5 1. A 13. T
2. A 3. T 4. F 5. A 6. C 14. F 15. C 16. A 17. B
SECTION 2.6 1. F 13. F
2. D 3. A 4. C 5. C 6. B 14. F 15. A 16. D 17. C
CHAPTER 2 TEST 1. D 13. A
15