Section 5.2 Practice

Questions

  1. Show that each sequence \((a_n)\) is increasing:

    1. \(a_n=\tfrac{n}{n+2}\)
    2. \(a_n=\tfrac{3n-1}{2n+1}\)
    3. \(a_n=\tfrac{n^2}{n+2}\)

  2. Show that each sequence \((a_n)\) is decreasing:

    1. \(a_n=\tfrac{n+3}{n+1}\)
    2. \(a_n=\tfrac{n+1}{5n+1}\)
    3. \(a_n=\tfrac{n}{n^2+1}\)

  3. Identify a sequence \((a_n)\) with the following properties:

    1. \((a_n)\) is strictly increasing and converges to \(2.\)
    2. \((a_n)\) is strictly decreasing and converges to \(-4.\)
    3. \((a_n)\) is strictly decreasing and converges to \(1.\)
    4. \((a_n)\) is strictly increasing and converges to \(\frac{2}{3}.\)

  4. Calculate the following limits. Carefully justify your reasoning.

    1. \(\displaystyle\lim_{n\to\infty}\sqrt{4+\frac{1}{n}}\)
    2. \(\displaystyle\lim_{n\to\infty}\sqrt{25+\frac{1}{n}}\)
    3. \(\displaystyle\lim_{n\to\infty}\sqrt[3]{8+\frac{1}{n}}\)

  5. Take \((a_n)\), \((b_n)\), and \((c_n)\) to be sequences with \[ \lim_{n\to\infty}a_n = 1, \quad \lim_{n\to\infty}b_n = -1, \quad {\rm and} \quad \lim_{n\to\infty}c_n = 11.\] Use the limit laws to compute the following:

    1. \(\displaystyle \lim_{n\to\infty}(a_nb_n+3c_n)\)
    2. \(\displaystyle \lim_{n\to\infty}\left(b_n^2-\frac{b_n-2}{c_n}\right)\)
    3. \(\displaystyle \lim_{n\to\infty}\left(\tfrac{a_n+b_n}{c_n+(b_n)^2}\right)\)

  6. For each sequence \((a_n)\), use the limit laws to determine \(\displaystyle\lim_{n\to\infty}a_n\):

    1. \(a_n=5\left(3+\frac{7}{n}\right)\)
    2. \(a_n=12\)
    3. \(a_n=\left(\frac{1}{2}\right)^n+5\)
    4. \(a_n=3^{-n}+2\)

  7. Use the limit laws to determine the following limits:

    1. \(\displaystyle \lim_{n\to\infty}\tfrac{-4n+4}{5n+1}\)
    2. \(\displaystyle \lim_{n\to\infty}\tfrac{4n^3+5n^2+2n-1}{8n^3-n}\)
    3. \(\displaystyle \lim_{n\to\infty}\tfrac{n^3}{n^6-2n^2-3n-4}\)

  8. Calculate the following limits. Carefully justify your reasoning.

    1. \(\displaystyle\lim_{n\to \infty} n\left(\sqrt{4+\frac{1}{n}} - 2 \right)\)
    2. \(\displaystyle\lim_{n\to \infty} n\left(\sqrt{25+\frac{1}{n}} - 5 \right)\)
    3. \(\displaystyle\lim_{n\to\infty}\left(2^n+3^n\right)^\frac{1}{n}\)

  9. Take \((a_n)\) to be a strictly decreasing sequence that is convergent to \(-10\) and take \(f\) to be the function that is given by \[ f(x)= \begin{cases} 7x^2-6&\text{if }x <-10\\ -9&\text{if }x=-10\\ \frac{5x}{10x^2-10}&\text{if }x>-10. \end{cases} \] Determine the quantity \(\lim\limits_{n\to\infty}f(a_n)\).

  10. Take \((a_n)\) to be a strictly decreasing sequence that converges to \(0\), \((b_n)\) to be a strictly increasing sequence that converges to \(8\), and \(f\) to be the function given by \[ f(x)= \begin{cases} 5x+1&\text{if }-1\leq x <0\\ x^2-1&\text{if }0 < x \leq 8\\ \sqrt{x}&\text{if }8<x\leq 10 \end{cases} \] Determine the limit of the sequences \((f(a_n))\), and \((f(b_n))\).

  11. For each sequence \((a_n)\) and real number \(L\), use the Archimedean property to show that for any positive real number \(\varepsilon\), there is a natural number \(N\) so that if \(n\) is greater than or equal to \(N\), then \(|a_n-L|<\varepsilon\).

    1. \(a_n= \frac{1}{n+2}\), \(L=0\)
    2. \(a_n= \frac{4}{n^2+2n}\), \(L=0\)
    3. \(a_n=\frac{2n+1}{4n+1}\), \(L=\tfrac{1}{2}\)
    4. \(a_n=\sqrt{81+\frac{1}{n^2}}\), \(L=9\).

  12. Identify which of the following limits are of indeterminate form and the form.

    1. \(\lim_\limits{n\to\infty}\frac{n+1}{n+2}\)
    2. \(\lim_\limits{n\to\infty}\frac{1}{n+2}\)
    3. \(\lim_\limits{n\to\infty}\frac{\arctan(n)}{n+2}\)
    4. \(\lim_\limits{n\to\infty}\mathrm{e}^n\)
    5. \(\lim_\limits{n\to\infty}\left(\mathrm{e}^n-\ln(n)\right)\)
    6. \(\lim_\limits{n\to\infty}\left(\mathrm{e}^{-n}-\ln(n)\right)\)
    7. \(\lim_\limits{n\to\infty}\left(\mathrm{e}^{-n}\ln(n)\right)\)
    8. \(\lim_\limits{n\to\infty}\frac{\arctan\left(\frac{1}{n}\right)}{\sin\left(\frac{1}{n}\right)}\)

Answers

  1. The following are brief key points you need to show in your argument for each part. You will need to fill in the details in your responses.

    1. Show \(a_{n+1}-a_n>0\) by calculating \(a_{n+1}-a_{n}\). The difference is \(a_{n+1}-a_{n}=\frac{2}{(n+2)(n+3)}.\)
    2. Show \(a_{n+1}-a_n>0\) by calculating \(a_{n+1}-a_{n}\). The difference is \(a_{n+1}-a_{n}=\frac{5}{(2n+1)(2n+3)}.\)
    3. Show \(a_{n+1}-a_n>0\) by calculating \(a_{n+1}-a_{n}\). The difference is \(a_{n+1}-a_{n}=\frac{n^2+5n+2}{(n+2)(n+3)}.\)

  2. The following are brief key points you need to show in your argument for each part. You will need to fill in the details in your responses.

    1. Show \(a_{n}-a_{n+1}>0\) by calculating \(a_{n}-a_{n+1}\). The difference is \(a_{n}-a_{n+1}=\frac{2}{(n+1)(n+2)}.\)
    2. Show \(a_{n}-a_{n+1}>0\) by calculating \(a_{n}-a_{n+1}\). The difference is \(a_{n}-a_{n+1}=\frac{4}{(5n+1)(5n+6)}.\)
    3. Show \(a_{n}-a_{n+1}>0\) by calculating \(a_{n}-a_{n+1}\). The difference is \(a_{n}-a_{n+1}=\frac{n^2+n-1}{(n^2+1)(n^2+2n+2)}.\)

  3. Answers may vary,

    1. \(a_n=2-\frac{1}{n}\)
    2. \(a_n=-4+\frac{1}{n}\)
    3. \(a_n=1+\frac{1}{n}\)
    4. \(a_n=\frac{2}{3}-\frac{1}{n}\)

  4. The outline is given below for each problem

    1. For each \(n\) there is an \(\varepsilon_n>0\) so that \[\sqrt{4+\frac{1}{n}}=2+\varepsilon_n.\] Square both sides to get \[4+\frac{1}{n}=4+4\varepsilon_n+\varepsilon_n^2.\] Because \[4+4\varepsilon_n+\varepsilon_n^2\geq 4+4\varepsilon_n,\] then \[4+\frac{1}{n}\geq 4+4\varepsilon_n.\] Solve for \(\varepsilon_n\) to get \[\frac{1}{4n}\geq \varepsilon_n.\] Because \[0<\varepsilon_n\leq \frac{1}{4n}\] and \(a_n=0\) and \(c_n=\frac{1}{4n}\) are null sequences, the conclusion of the null sequence theorem is that \(\varepsilon_n\) is a null sequence. Hence \[\displaystyle\lim_{n\to\infty}\sqrt{4+\frac{1}{n}}=\lim_{n\to\infty}(2+\varepsilon_n)=2+0=2.\]
    2. For each \(n\) there is an \(\varepsilon_n>0\) so that \[\sqrt{25+\frac{1}{n}}=5+\varepsilon_n.\] Square both sides to get \[25+\frac{1}{n}=25+10\varepsilon_n+\varepsilon_n^2.\] Because \[25+10\varepsilon_n+\varepsilon_n^2\geq 25+10\varepsilon_n,\] then \[25+\frac{1}{n}\geq 25+10\varepsilon_n.\] Solve for \(\varepsilon_n\) to get \[\frac{1}{10n}\geq \varepsilon_n.\] Because \[0<\varepsilon_n\leq \frac{1}{10n}\] and \(a_n=0\) and \(c_n=\frac{1}{10n}\) are null sequences, the conclusion of the null sequence theorem is that \(\varepsilon_n\) is a null sequence. Hence \[\displaystyle\lim_{n\to\infty}\sqrt{25+\frac{1}{n}}=\lim_{n\to\infty}(5+\varepsilon_n)=5+0=5.\]
    3. For each \(n\) there is an \(\varepsilon_n>0\) so that \[\sqrt[3]{8+\frac{1}{n}}=2+\varepsilon_n.\] Cube both sides to get \[8+\frac{1}{n}=8+12\varepsilon_n+6\varepsilon_n^2+\varepsilon_n^3.\] Because \[8+12\varepsilon_n+6\varepsilon_n^2+\varepsilon_n^3\geq 8+12\varepsilon_n,\] then \[8+\frac{1}{n}\geq 8+12\varepsilon_n.\] Solve for \(\varepsilon_n\) to get \[\frac{1}{12n}\geq \varepsilon_n.\] Because \[0<\varepsilon_n\leq \frac{1}{12n}\] and \(a_n=0\) and \(c_n=\frac{1}{12n}\) are null sequences, the conclusion of the null sequence theorem is that \(\varepsilon_n\) is a null sequence. Hence \[\displaystyle\lim_{n\to\infty}\sqrt[3]{8+\frac{1}{n}}=\lim_{n\to\infty}(2+\varepsilon_n)=2+0=2.\]
    1. \(32\)
    2. \(\frac{14}{11}\)
    3. \(0\)
    1. \(15\)
    2. \(12\)
    3. \(5\)
    4. \(2\)
    1. \(-\tfrac{4}{5}\)
    2. \(\tfrac{1}{2}\)
    3. \(0\)
    1. \(\frac{1}{4}\)
    2. \(\frac{1}{10}\)
    3. \(3\)

  9. \(-\frac{5}{99}\)

  10. \(\lim\limits_{n\to\infty}f(a_n)=-1\) and \(\lim\limits_{n\to\infty}f(b_n)=63\)

  11. The following are brief key points you need to show in your argument for each part. You will need to fill in the details in your responses.

    1. Show that for each positive real number \(\varepsilon\), there is a natural number \(N\) so that \(\tfrac{1}{\varepsilon}-2< N\). Show that this implies for all natural numbers \(n\) greater than or equal to \(N\), \(\frac{1}{3n+2}<\varepsilon\).
    2. Show that for each positive real number \(\varepsilon\), there is a natural number \(N\) so that \(\tfrac{4}{\varepsilon}< N\). Show that this implies for all natural numbers \(n\) greater than or equal to \(N\), \(\frac{4}{n^2+2n}<\varepsilon\).
    3. Show that for each positive real number \(\varepsilon\), there is a natural number \(N\) so that \(\tfrac{1}{8\varepsilon}-\frac{1}{4}< N\). Show that this implies for all natural numbers \(n\) greater than or equal to \(N\), \(\left|\frac{2n+1}{4n+1}-\tfrac{1}{2}\right|<\varepsilon\).
    4. Show that for each positive real number \(\varepsilon\), there is a natural number \(N\) so that \(\tfrac{1}{9\varepsilon}< N\). Show that this implies for all natural numbers \(n\) greater than or equal to \(N\), \(\left|\sqrt{81+\tfrac{1}{n^2}}-9\right|<\varepsilon\).
    1. \(\frac{\infty}{\infty}\) indeterminate form
    2. not indeterminate form
    3. not indeterminate form
    4. not indeterminate form
    5. \(\infty-\infty\) indeterminate form
    6. not indeterminate form
    7. \(0\cdot (-\infty)\) indeterminate form
    8. \(\frac{0}{0}\) indeterminate form

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