Solutions for Section 1.9

3. The limit does not exist since canceling, we have lim𝑥→0

4. Factoring the numerator and simplifying, we get 𝑥²− 3𝑥

5. Factoring the numerator and the denominator, we obtain 𝑡⁴+ 𝑡² 6. Factoring the numerator, we obtain

𝑥³− 3𝑥

7. Factoring the denominator and simplifying, we get 8. Factoring the numerator and simplifying, we get

𝑦²− 5𝑦 + 4

9. Factoring the numerator and denominator and simplifying, we get 𝑥²+ 2𝑥 − 3

10. Factoring the numerator and denominator and simplifying, we get 2𝑡²+ 3𝑡 − 2

11. Factoring the numerator and the denominator and simplifying, we get 𝑥²− 9 12. Factoring the numerator and the denominator, we obtain

2𝑦²+ 𝑦 − 1 13. Expanding the numerator, we get

(3 + ℎ)²− 9 = 9 + 6ℎ + ℎ²− 9 = 6ℎ + ℎ². Next, factoring and simplifying, we get

(3 + ℎ)²− 9

14. Simplifying and factoring the numerator and the denominator, we obtain 15. Factoring the numerator and the denominator, we obtain

5𝑥²− 15 18. Rewriting the numerator as a single quotient, we have

1

19. Rewriting the numerator as a single quotient we get 1

20. There are two ways to do this. One way is to multiply the numerator and denominator by√

21. There are two ways to do this. One way is to multiply the numerator and denominator by√

9 + ℎ + 3 and then simplify:

22. There are two ways to do this. First, factoring the numerator and simplifying, we get 4^𝑥− 1

23. Expanding the numerator, we get

(1 + ℎ)⁴− 1 = 1 + 4ℎ + 6ℎ²+ 4ℎ³+ ℎ⁴− 1.

Next, factoring and simplifying, we get (1 + ℎ)⁴− 1

(b) Rationalizing the denominator, we have lim𝑥→3

27. Dividing both numerator and denominator by 𝑧³, we get

𝑧→∞lim

28. Dividing both numerator and denominator by 𝑥², we get

𝑥→∞lim

29. Dividing both numerator and denominator by 𝑒^𝑡, we get

35. First we use algebra: Multiply by(√

4 + ℎ + 2)

in numerator and denominator, then expand and simplify:

√4 + ℎ − 2 = Now we can find the limit:

limℎ→0

36. First we combine the fractions:

√ 1

in numerator and denominator, then expand and simplify:

(2 −√ Now we find the limit:

limℎ→0

37. Given that −4𝑥 + 6 ≤ 𝑓(𝑥) ≤ 𝑥²− 2𝑥 + 7, we see that

(using the sum property since each limit exists) 4 − 0 ≤ lim

40. Because the denominator equals 0 when 𝑥 = 4, so must the numerator. This means 𝑘²= 16 and the choices for 𝑘 are 4 or

−4.

41. Because the denominator equals 0 when 𝑥 = 1, so must the numerator. So 1 − 𝑘 + 4 = 0. The only possible value of 𝑘 is 5.

42. Because the denominator equals 0 when 𝑥 = −2, so must the numerator. So 4 − 8 + 𝑘 = 0 and the only possible value of 𝑘 is 4.

43. Because the denominator equals 0 when 𝑥 = 5, so must the numerator. So 25 − 5𝑘 + 5 = 0 and the only possible value of 𝑘 is 6.

44. Because the denominator equals 0 when 𝑥 = 0, so must the numerator. So 𝑒^𝑘− 8 = 0 and the only possible value of 𝑘 is ln 8.

45. Because the denominator equals 0 when 𝑥 = 1, so must the numerator. So 𝑘²− 49 = 0 and the choices for 𝑘 are 7 or −7.

46. Division of numerator and denominator by 𝑥²yields 𝑥²+ 3𝑥 + 5

4𝑥 + 1 + 𝑥^𝑘 = 1 + 3∕𝑥 + 5∕𝑥² 4∕𝑥 + 1∕𝑥²+ 𝑥^𝑘−2.

As𝑥 → ∞, the limit of the numerator is 1. The limit of the denominator depends upon 𝑘. If 𝑘 > 2, the denominator approaches ∞ as𝑥 → ∞, so the limit of the quotient is 0. If 𝑘 = 2, the denominator approaches 1 as 𝑥 → ∞, so the limit of the quotient is 1. If𝑘 < 2 the denominator approaches 0⁺as𝑥 → ∞, so the limit of the quotient is ∞. Therefore the values of𝑘 we are looking for are 𝑘 ≥ 2.

47. For the numerator, lim

48. Division of numerator and denominator by 𝑥³yields 𝑥³− 6

𝑥^𝑘+ 3= 1 − 6∕𝑥³ 𝑥^𝑘−3+ 3∕𝑥³.

As𝑥 → ∞, the limit of the numerator is 1. The limit of the denominator depends upon 𝑘. If 𝑘 > 3, the denominator approaches ∞ as𝑥 → ∞, so the limit of the quotient is 0. If 𝑘 = 3, the denominator approaches 1 as 𝑥 → ∞, so the limit of the quotient is 1. If𝑘 < 3 the denominator approaches 0⁺as𝑥 → ∞, so the limit of the quotient is ∞. Therefore the values of𝑘 we are looking for are 𝑘 ≥ 3.

49. We divide both the numerator and denominator by 𝑒^5𝑥, giving

𝑥→∞lim

50. We divide both the numerator and denominator by 3^2𝑥, giving

𝑥→∞lim so the quotient has a limit of ∞.

51. In the denominator, we have lim

𝑥→−∞3^2𝑥+ 4 = 4. In the numerator, if 𝑘 < 0, we have lim

𝑥→−∞3^𝑘𝑥+ 6 = ∞, so the quotient has a limit of ∞. If𝑘 = 0, we have lim

𝑥→−∞3^𝑘𝑥+ 6 = 7, so the quotient has a limit of 7∕4. If 𝑘 > 0, we have lim

𝑥→−∞3^𝑘𝑥+ 6 = 6, so the quotient has a limit of 6∕4.

52. Let 𝑡 =√

54. Let 𝑡 =√

Thus, lim

66. There are many possible correct answers.

67. There are many possible choices for 𝑔(𝑥). One possibility is 𝑔(𝑥) =( 𝑥 − 3

Observe that𝑓(𝑥) = 𝑔(𝑥) wherever 𝑓(𝑥) is defined except for 𝑥 = 3 since 𝑔(3) is undefined. This means the graphs of 𝑓(𝑥) and 𝑔(𝑥) will be the same except at 𝑥 = 3 where the graph of 𝑔(𝑥) has a hole.

Since the graphs of𝑓(𝑥) and 𝑔(𝑥) are identical except at 𝑥 = 3, the hole will have the same coordinates as the function 𝑓(𝑥) at 𝑥 = 3, so the coordinates of the hole will be (3, 10).

68. There are many possible choices for 𝑔(𝑥). One possibility is 𝑔(𝑥) =( 𝑥

Observe that𝑓(𝑥) = 𝑔(𝑥) wherever 𝑓(𝑥) is defined except for 𝑥 = 0 since 𝑔(0) is undefined. This means the graphs of 𝑓(𝑥) and 𝑔(𝑥) will be the same except at 𝑥 = 0 where the graph of 𝑔(𝑥) has a hole.

Since the graphs of𝑓(𝑥) and 𝑔(𝑥) are identical except at 𝑥 = 0, the hole will have the same coordinates as the function 𝑓(𝑥) at 𝑥 = 0, so the coordinates of the hole will be (0, 1).

69. There are many possible choices for 𝑔(𝑥). One possibility is 𝑔(𝑥) =( 𝑥 − 1

Observe that𝑓(𝑥) = 𝑔(𝑥) wherever 𝑓(𝑥) is defined except for 𝑥 = 1 since 𝑔(1) is undefined. This means the graphs of 𝑓(𝑥) and 𝑔(𝑥) will be the same except at 𝑥 = 1 where the graph of 𝑔(𝑥) has a hole.

Since the graphs of𝑓(𝑥) and 𝑔(𝑥) are identical except at 𝑥 = 1, the hole will have the same coordinates as the function 𝑓(𝑥) at 𝑥 = 1, so the coordinates of the hole will be (1, 0).

70. There are many possible choices for 𝑔(𝑥). One possibility is 𝑔(𝑥) =( 𝑥 − 𝜋

Observe that𝑓(𝑥) = 𝑔(𝑥) wherever 𝑓(𝑥) is defined except for 𝑥 = 𝜋 since 𝑔(𝜋) is undefined. This means the graphs of 𝑓(𝑥) and 𝑔(𝑥) will be the same except at 𝑥 = 𝜋 where the graph of 𝑔(𝑥) has a hole.

Since the graphs of𝑓(𝑥) and 𝑔(𝑥) are identical except at 𝑥 = 𝜋, the hole will have the same coordinates as the function 𝑓(𝑥) at 𝑥 = 𝜋, so the coordinates of the hole will be (𝜋, 0).

73. We have In both cases𝐿 = 0. The Squeeze Theorem cannot be applied to other cases since for any other 𝑐,

lim𝑥→𝑐− 1

78. If we factor the numerator of 𝑓 and simplify, we see that

𝑓(𝑥) = 𝑥 − 1, for 𝑥 ≠ −1.

This show that the two functions are identical (that is, have the same values) where they are both defined, the functions have different domains:𝑥 = −1 is in the domain of 𝑔 but not in the domain of 𝑓.

79. The limit lim

𝑥→1𝑓(𝑥)∕𝑔(𝑥) depends on the values of 𝑓(𝑥)∕𝑔(𝑥) as 𝑥 approaches 1 but is not necessarily dependent on the value at𝑥 = 1. For example, we can take 𝑔(𝑥) = 1 and

80. False, since the limit lim

𝑥→0𝑓(𝑥) may not even exist. For example if 𝑏(𝑥) = −1, 𝑎(𝑥) = 1 and 𝑓(𝑥) = sin(1∕𝑥), then the conditions are met but sin(1∕𝑥) has no limit as 𝑥 → 0.

81. True, by the Squeeze Theorem, using the constant function 𝑏(𝑥) = 0.

82. False, this is a confused, and incorrect, version of the Squeeze Theorem. For example if 𝑏(𝑥) = 𝑓(𝑥) = 𝑥, and 𝑎(𝑥) = 1 then the conditions are met but lim

𝑥→0𝑓(𝑥) = 0 and lim

𝑔(𝑥) does not exist, since lim

𝑥→0⁺

86. True, this follows from the multiplicative limit property: if lim

𝑥→0𝑎(𝑥) exists and lim

(b) No for𝛿 = 0.75. The graph of 𝑓(𝑥) on the interval 2.25 < 𝑥 < 3.75 extends outside the shaded region.

(c) Yes for𝛿 = 0.5. The graph of 𝑓(𝑥) on the interval 2.5 < 𝑥 < 3.5 fits inside the shaded region.

(d) Yes for𝛿 = 0.25. The graph of 𝑓(𝑥) on the interval 2.75 < 𝑥 < 3.25 fits inside the shaded region.

(e) Yes for𝛿 = 0.1. The graph of 𝑓(𝑥) on the interval 2.9 < 𝑥 < 3.1 fits inside the shaded region.

2. For each 𝛿 we determine whether the graph of 𝑔(𝑥) on the interval 20 − 𝛿 < 𝑥 < 20 + 𝛿 stays in the shaded region 𝐿 − 𝜖 < 𝑦 < 𝐿 + 𝜖 in the figure. Notice that as 𝛿 gets smaller, the interval gets shorter.

(a) No for𝛿 = 0.5. The graph of 𝑔(𝑥) on the interval 19.5 < 𝑥 < 20.5 extends outside the shaded region.

(b) No for𝛿 = 0.4. The graph of 𝑔𝑥) on the interval 19.6 < 𝑥 < 20.4 extends outside the shaded region.

(c) No for𝛿 = 0.3. The graph of 𝑔(𝑥) on the interval 19.7 < 𝑥 < 20.3 extends outside the shaded region.

(d) Yes for𝛿 = 0.2. The graph of 𝑔(𝑥) on the interval 19.8 < 𝑥 < 20.2 fits inside the shaded region.

(e) Yes for𝛿 = 0.1. The graph of 𝑔(𝑥) on the interval 19.9 < 𝑥 < 20.1 fits inside the shaded region.

3. By tracing on a calculator or solving equations, we find the following values of 𝛿:

For𝜖 = 0.2, 𝛿 ≤ 0.1.

4. By tracing on a calculator or solving equations, we find the following values of 𝛿:

For𝜖 = 0.1, 𝛿 ≤ 0.46.