A Collection of Problems in Differential Calculus: Problems from Calculus I Final Examinations, 2000-2020 Department of Mathematics, Simon Fraser University

Section 6.1 Limits

1. 1. $20\text{.}$

   2. Does not exist.

   3. $0\text{.}$

   4. $100\text{.}$

   5. Does not exist. Consider the domain of $g(x) =\sqrt{-x^2+20x-100}=\sqrt{-(x-10)^2}\text{.}$

2. $-8\ln 4\text{.}$

3. $0\text{.}$ Note the exponential function in the denominator.

4. $\displaystyle \frac{3}{5}\text{.}$ Divide the numerator and denominator by the highest power.

5. $\displaystyle \frac{5}{2}\text{.}$

6. $3\text{.}$

7. $2\text{.}$

8. $\ds \frac{a}{c}\text{.}$

9. $0\text{.}$ What is the value of $3x+|1-3x|$ if $x\lt \frac{1}{3}\text{?}$

10. $1\text{.}$

11. $1\text{.}$

12. $\displaystyle \frac{3}{\sqrt{2}}\text{.}$

13. $\displaystyle -\frac{2}{3}\text{.}$

14. $\ds -\frac{1}{2}\text{.}$

15. $\infty\text{.}$ Note that $x^2-1=(x-1)(x+1)\text{.}$

16. $-2\text{.}$ Which statement is true for $x\lt 1\text{:}$ $|x-1|=x-1$ or $|x-1|=1-x\text{?}$

17. 1. $1.5\text{.}$

    2. $-1.5\text{.}$

    3. No. The left-hand limit and the right-hand limit are not equal.

18. $1\text{.}$

19. $\ds \frac{4}{5}\text{.}$

20. $\infty\text{.}$

21. Does not exist

22. Does not exist.

23. $0\text{.}$

24. $\displaystyle \frac{1}{8}\text{.}$ Rationalize the numerator.

25. $\displaystyle \frac{1}{12}\text{.}$ Note that $x-8=(\sqrt[3]{x}-2)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)\text{.}$

26. $\displaystyle \ds \frac{1}{16}$

27. $3\text{.}$

28. $a=b=4\text{.}$ Rationalize the numerator. Choose the value of $b$ so that $x$ becomes a factor in the numerator.

29. $\ds e^4\text{.}$

30. $\ds e^{1/6}\text{.}$

31. $\ds \frac{1}{2}\text{.}$

32. $\displaystyle\frac{1}{12}\text{.}$ Note that $x-8=(\sqrt[3]{x}-2)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)\text{.}$

33. $\displaystyle \frac{1}{2}\text{.}$ Rationalize the numerator.

34. $\displaystyle -\frac{3}{2}\text{.}$ Rationalize the numerator. Note that $x\to -\infty$ and use the fact that if $x\lt 0$ then $x=-\sqrt{x^2}\text{.}$

35. $\displaystyle -\frac{1}{2}\text{.}$

36. $\displaystyle \frac{3}{2}\text{.}$

37. Since the denominator approaches $0$ as $x\to -2\text{,}$ the necessary condition for this limit to exist is that the numerator approaches $0$ as $x\to -2\text{.}$ Thus we solve $4b-30+15+b=0$ to obtain $b=3\text{.}$ $\ds \lim _{x\to -2}\frac{3x^2+15x+18}{x^2+x-2}=-1\text{.}$

38. $a=4\text{.}$ Write $\ds f(x)=x+\frac{(a-1)x+5}{x+1}\text{.}$

39. $\displaystyle \lim _{x\to \infty}\frac{\ln x}{x}=0\text{.}$

40. From $\displaystyle \lim _{x\to 4}(x+2)=6$ and $\displaystyle \lim _{x\to 4}(x^2-10)=6\text{,}$ by the Squeeze Theorem, it follows that $\displaystyle \lim _{x\to 4}f(x)=6\text{.}$

41. 1.

42. 1. Use the fact $\ds -x^4\leq x^4\sin\left(\frac{1}{x}\right)\leq x^4\text{,}$ $x\not= 0\text{.}$

    2. From the fact that $\displaystyle \left| \sin (1/x)\right|\leq 1$ for all $x\not= 0$ and the fact that the function $\displaystyle y=e^x$ is increasing conclude that $\displaystyle e^{-1}\leq e^{\sin (1/x)}\leq e$ for all $x\not= 0\text{.}$ Thus $\displaystyle e^{-1} \cdot \sqrt{x} \leq \sqrt{x}e^{\sin (1/x)} \leq e\cdot \sqrt{x}$ for all $x>0\text{.}$ By the Squeeze Theorem, $\displaystyle \lim _{x\to 0^+}\left( \sqrt{x}e^{\sin (1/x)}\right) =0\text{.}$

43. $0\text{.}$ Squeeze Theorem.

44. $0\text{.}$ Squeeze Theorem.

45. $0\text{.}$ Squeeze Theorem.

46. $0\text{.}$ Squeeze Theorem.

47. $0\text{.}$ Squeeze Theorem.

48. $-\infty\text{.}$

49. $0\text{.}$

50. $2\text{.}$

51. Does not exist.

52. $\displaystyle \ds -\frac{2}{7}$

53. $\infty\text{.}$

54. $\displaystyle \frac{76}{45}\text{.}$ This is the case “$0/0$”. Apply L'Hospital's rule.

55. $\ds \frac{a}{b}\text{.}$

56. $\displaystyle \frac{1}{2}\text{.}$ Write $\displaystyle \frac{1}{2}\cdot \left( \frac{\sin x}{x}\right) ^{100}\cdot \frac{2x}{\sin 2x}\text{.}$

57. 7. Write $\displaystyle 7\cdot \left( \frac{x}{\sin x}\right) ^{101}\cdot \frac{\sin 7x}{7x}\text{.}$

58. 7.

59. $\displaystyle \frac{3}{5}\text{.}$ This is the case “$0/0$”. Apply L'Hospital's rule.

60. $\displaystyle \frac{3}{5}\text{.}$

61. $0\text{.}$ Write $\displaystyle x^2 \cdot \frac{x}{\sin x}\cdot \sin \left( \frac{1}{x^2}\right)\text{.}$

62. Does not exist. $\displaystyle \frac{\sin x}{2|x|}\cdot \frac{1}{\sqrt{\frac{\sin 4x}{4x}}}\text{.}$

63. $\displaystyle \frac{1}{2}\text{.}$ Write $\displaystyle \frac{1-\cos x}{x^2}\cdot\frac{x}{\sin x}\text{.}$

64. $-1\text{.}$

65. $-1\text{.}$

66. $1\text{.}$ Substitute $\displaystyle t=\frac{1}{x}\text{.}$

67. $0\text{.}$ This is the case $"\infty - \infty"\text{.}$ Write $\displaystyle \frac{x- \sin x}{x\sin x}$ and apply L'Hospital's rule.

68. $\displaystyle \ds \frac{1}{6}$

69. $0\text{.}$

70. $0\text{.}$ This is the case $``0\cdot \infty ''\text{.}$ Write $\displaystyle \frac{\ln \sin x}{\frac{1}{\sin x}}$ and apply L'Hospital's rule.

71. $-2\text{.}$

72. $\infty\text{.}$

73. $0\text{.}$ This is the case $``\infty /\infty ''\text{.}$ Apply L'Hospital's rule.

74. $0\text{.}$

75. $0\text{.}$

76. $0\text{.}$

77. $1\text{.}$ This is the case $``0/0''\text{.}$ Write $\displaystyle \frac{\ln (1+x)}{x}$ and apply L'Hospital's rule.

78. $\displaystyle \frac{3}{2}\text{.}$ Use properties of logarithms first.

79. $\displaystyle \frac{3}{2}\text{.}$

80. $\ln 2\text{.}$ The denominator approaches 2.

81. $0\text{.}$ This is the case “$0/0$”. Apply L'Hospital's rule.

82. $\displaystyle -\frac{1}{\pi ^2}\text{.}$ Apply L'Hospital's rule twice.

83. $\infty\text{.}$ This is the case "$\infty - \infty$". Write $\displaystyle \frac{\sin x-x^2\cos x}{x^2\sin x}$ and apply L'Hospital's rule.

84. $\ds \frac{1}{2}\text{.}$

85. $\displaystyle e^{\frac{1}{2}}\text{.}$ This is the case "$\displaystyle 1^\infty$". Write $\displaystyle e^{\frac{\ln \cosh x}{x^2}}\text{.}$ Apply L'Hospital's rule and use the fact that the exponential function $f(x)=e^x$ is continuous.

86. $1\text{.}$

87. $\ds e^{-1/2}\text{.}$

88. $1\text{.}$ This is the case "$\displaystyle 0^0$". Write $\displaystyle x^x=e^{x\ln x}=e^{\frac{\ln x}{x^{-1}}}\text{.}$ Apply L'Hospital's rule and use the fact that the exponential function $f(x)=e^x$ is continuous.

89. $1\text{.}$

90. $1\text{.}$

91. $1\text{.}$

92. $e\text{.}$

93. $1\text{.}$ This is the case "$\infty^0$".

94. $1\text{.}$

95. $e^2\text{.}$

96. $\displaystyle e^3\text{.}$

97. $\displaystyle 0\text{.}$

98. $1\text{.}$ Write $\displaystyle e^{x\ln \frac{x}{x+1}}=e^{x\ln x}\cdot e^{-x\ln (x+1)}$ and make your conclusion.

99. $\ds e^{\frac{1}{e}}\text{.}$

100. $1\text{.}$

101. $1\text{.}$ Squeeze Theorem.

102. $\displaystyle e^{-2}\text{.}$ Write $\displaystyle \left( (1-2x)^{-\frac{1}{2x}}\right) ^{-2}\text{.}$

103. $\displaystyle e^{\frac{7}{5}}\text{.}$ Write $\displaystyle \left( (1+7x)^{\frac{1}{7x}}\right) ^{\frac{7}{5}}\text{.}$

104. $\displaystyle e^{\frac{3}{8}}\text{.}$ Write $\displaystyle \left( (1+3x)^{\frac{1}{3x}}\right) ^{\frac{3}{8}}\text{.}$

105. $\displaystyle e^{\frac{3}{2}}\text{.}$ Write $\displaystyle \left( \left( 1+\frac{x}{2}\right) ^{\frac{2}{x}}\right) ^{\frac{3}{2}}\text{.}$

106. $10\text{.}$ Use the fact that $\displaystyle L=\lim _{n\to \infty }x_n$ to conclude $L^2=100\text{.}$ Can $L$ be negative?

107. 1. $\displaystyle \frac{1}{2}\text{.}$ Write $\displaystyle \frac{2\sin ^2 \frac{x}{2}}{x^2}\text{,}$ or use L'Hospital's rule.

     2. $0\text{.}$

     3. Does not exist. Note that the domain of $f(x)=\arcsin x$ is the interval $[-1,1]\text{.}$

108. 1. $\displaystyle \frac{1}{64}\text{.}$

     2. $\displaystyle -\frac{1}{\pi }\text{.}$ Use L'Hospital's rule.

     3. $1\text{.}$ Divide the numerator and denominator by $u\text{.}$

     4. $e^{-2}\text{.}$

     5. $\displaystyle \displaystyle \frac{1}{2}$

     6. $\infty\text{.}$ Think, exponential vs. polynomial.

109. (a) $2016\cdot 2^{2015}\text{;}$ (b) $\displaystyle \frac{1}{2}\text{;}$ (c) $\displaystyle \frac{7}{3}\text{;}$ (d) 1; (e) 0; (f) $\displaystyle \frac{\sin 3 - 3}{27}\text{.}$

110. (a) Does not exist. Consider $\displaystyle \lim _{x\to 0}\frac{xf(x)}{x|x|}\text{;}$ (b) $\displaystyle \frac{1}{3}\text{;}$ (c) $\displaystyle -\frac{1}{4}\text{.}$ Note that $x\lt 0\text{;}$ (d) $0\text{.;}$ (e) $e\text{.}$

111. (a) $\ds \frac{1}{4}\text{;}$ (b) $-1\text{;}$ (c) $0\text{.}$

112. (a) $-\infty\text{;}$ (b) 3; (c) 2; (d) 0.

113. Let $\varepsilon >0$ be given. We need to find $\delta =\delta (\varepsilon )>0$ such that $|x-0|\lt \delta \Rightarrow |x^3-0|\lt \varepsilon\text{,}$ which is the same as $|x|\lt \delta \Rightarrow |x^3|\lt \varepsilon\text{.}$ Clearly, we can take $\delta =\sqrt[3]{\varepsilon }\text{.}$ Indeed, for any $\varepsilon >0$ we have that $|x|\lt \sqrt[3]{\varepsilon } \Rightarrow |x|^3=|x^3|\lt \varepsilon$ and, by definition, $\displaystyle \lim _{x\to 0}x^3=0\text{.}$

114. (c) For any $\varepsilon >0$ there exists $\delta =\delta (\varepsilon )>0$ such that $|x-1|\lt \delta \Rightarrow |2x^2-2|\lt \varepsilon\text{.}$

115. Take, for example, $f(x)=\mbox{sign} (x)\text{,}$ $g(x)=-\mbox{sign} (x)\text{,}$ and $a=0\text{.}$

116. $\displaystyle \ds \frac{2}{3}$

117. $\displaystyle \lim _{h\to 0}\frac{f(x+h)-f(x-h)}{2h}=\lim _{h\to 0}\frac{f'(x+h)+f'(x-h)}{2}$ and, since $f'$ is continuous, $\displaystyle \lim _{h\to 0}f'(x+h)=\lim _{h\to 0}f'(x-h)=f'(x)\text{.}$