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Section 6.1 Limits

These answers correspond to the problems in Section 1.2.

    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 \(|x-1|=1-x\text{?}\)

    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.

    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?

    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 \([-1,1]\text{.}\)

    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, \(\displaystyle \lim _{h\to 0}f'(x+h)=\lim _{h\to 0}f'(x-h)=f'(x)\text{.}\)