{"id":151,"date":"2026-01-12T15:55:24","date_gmt":"2026-01-12T15:55:24","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/?post_type=chapter&p=151"},"modified":"2026-01-12T15:55:24","modified_gmt":"2026-01-12T15:55:24","slug":"limits-and-continuity-get-stronger-answer-key","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/limits-and-continuity-get-stronger-answer-key\/","title":{"raw":"Limits and Continuity: Get Stronger Answer Key","rendered":"Limits and Continuity: Get Stronger Answer Key"},"content":{"raw":"

The Limit Laws<\/h2>\r\n

In the following exercises (1-2), use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).<\/strong><\/p>\r\n\r\n

    \r\n \t
  1. Use constant multiple law and difference law: [latex]\\underset{x\\to 0}{\\lim}(4x^2-2x+3)=4\\underset{x\\to 0}{\\lim}x^2-2\\underset{x\\to 0}{\\lim}x+\\underset{x\\to 0}{\\lim}3=3[\/latex]<\/li>\r\n \t
  2. Use root law: [latex]\\underset{x\\to -2}{\\lim}\\sqrt{x^2-6x+3}=\\sqrt{\\underset{x\\to -2}{\\lim}(x^2-6x+3)}=\\sqrt{19}[\/latex]<\/li>\r\n \t
  3. [latex]49[\/latex]<\/li>\r\n \t
  4. [latex]1[\/latex]<\/li>\r\n \t
  5. [latex]-\\frac{5}{7}[\/latex]<\/li>\r\n \t
  6. [latex]\\underset{x\\to 4}{\\lim}\\frac{x^2-16}{x-4}=\\frac{16-16}{4-4}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{x\\to 4}{\\lim}\\frac{x^2-16}{x-4}=\\underset{x\\to 4}{\\lim}\\frac{(x+4)(x-4)}{x-4}=8[\/latex]<\/li>\r\n \t
  7. [latex]\\underset{x\\to 6}{\\lim}\\frac{3x-18}{2x-12}=\\frac{18-18}{12-12}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{x\\to 6}{\\lim}\\frac{3x-18}{2x-12}=\\underset{x\\to 6}{\\lim}\\frac{3(x-6)}{2(x-6)}=\\frac{3}{2}[\/latex]<\/li>\r\n \t
  8. [latex]\\underset{t \\to 9}{\\lim}\\frac{t-9}{\\sqrt{t}-3}=\\frac{9-9}{3-3}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{t\\to 9}{\\lim}\\frac{t-9}{\\sqrt{t}-3}=\\underset{t\\to 9}{\\lim}\\frac{t-9}{\\sqrt{t}-3}\\frac{\\sqrt{t}+3}{\\sqrt{t}+3}=\\underset{t\\to 9}{\\lim}(\\sqrt{t}+3)=6[\/latex]<\/li>\r\n \t
  9. [latex]\\underset{\\theta \\to \\pi}{\\lim}\\frac{\\sin \\theta}{\\tan \\theta}=\\frac{\\sin \\pi}{\\tan \\pi}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{\\theta \\to \\pi}{\\lim}\\frac{\\sin \\theta}{\\tan \\theta}=\\underset{\\theta \\to \\pi}{\\lim}\\frac{\\sin \\theta}{\\frac{\\sin \\theta}{\\cos \\theta}}=\\underset{\\theta \\to \\pi}{\\lim}\\cos \\theta =-1[\/latex].<\/li>\r\n \t
  10. [latex]\\underset{x\\to 1\/2}{\\lim}\\frac{2x^2+3x-2}{2x-1}=\\frac{\\frac{1}{2}+\\frac{3}{2}-2}{1-1}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{x\\to 1\/2}{\\lim}\\frac{2x^2+3x-2}{2x-1}=\\underset{x\\to 1\/2}{\\lim}\\frac{(2x-1)(x+2)}{2x-1}=\\frac{5}{2}[\/latex]<\/li>\r\n \t
  11. [latex]\\underset{x\\to 6}{\\lim}2f(x)g(x)=2\\underset{x\\to 6}{\\lim}f(x)\\underset{x\\to 6}{\\lim}g(x)=72[\/latex]<\/li>\r\n \t
  12. [latex]\\underset{x\\to 6}{\\lim}(f(x)+\\frac{1}{3}g(x))=\\underset{x\\to 6}{\\lim}f(x)+\\frac{1}{3}\\underset{x\\to 6}{\\lim}g(x)=7[\/latex]<\/li>\r\n \t
  13. [latex]\\underset{x\\to 6}{\\lim}\\sqrt{g(x)-f(x)}=\\sqrt{\\underset{x\\to 6}{\\lim}g(x)-\\underset{x\\to 6}{\\lim}f(x)}=\\sqrt{5}[\/latex]<\/li>\r\n \t
  14. [latex]\\underset{x\\to 6}{\\lim}[(x+1)\\cdot f(x)]=(\\underset{x\\to 6}{\\lim}(x+1))(\\underset{x\\to 6}{\\lim}f(x))=28[\/latex].<\/li>\r\n \t
  15. \"The\r\n
      \r\n \t
    1. [latex]9[\/latex]<\/li>\r\n \t
    2. [latex]7[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
    3. \"The\r\n
        \r\n \t
      1. [latex]1[\/latex]<\/li>\r\n \t
      2. [latex]1[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
      3. [latex]\\underset{x\\to -3^-}{\\lim}(f(x)-3g(x))=\\underset{x\\to -3^-}{\\lim}f(x)-3\\underset{x\\to -3^-}{\\lim}g(x)=0+6=6[\/latex]<\/li>\r\n \t
      4. [latex]\\underset{x\\to -5}{\\lim}\\frac{2+g(x)}{f(x)}=\\frac{2+(\\underset{x\\to -5}{\\lim}g(x))}{\\underset{x\\to -5}{\\lim}f(x)}=\\frac{2+0}{2}=1[\/latex]<\/li>\r\n \t
      5. [latex]\\underset{x\\to 1}{\\lim}\\sqrt[3]{f(x)-g(x)}=\\sqrt[3]{\\underset{x\\to 1}{\\lim}f(x)-\\underset{x\\to 1}{\\lim}g(x)}=\\sqrt[3]{2+5}=\\sqrt[3]{7}[\/latex]<\/li>\r\n \t
      6. [latex]\\underset{x\\to -9}{\\lim}(x\\cdot f(x)+2g(x))=(\\underset{x\\to -9}{\\lim}x)(\\underset{x\\to -9}{\\lim}f(x))+2\\underset{x\\to -9}{\\lim}(g(x))=(-9)(6)+2(4)=-46[\/latex]<\/li>\r\n<\/ol>\r\n

        Continuity<\/h2>\r\n

        For the following exercises (1-4), determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.<\/strong><\/p>\r\n\r\n

          \r\n \t
        1. The function is defined for all [latex]x[\/latex] in the interval [latex](0,\\infty)[\/latex].<\/li>\r\n \t
        2. Removable discontinuity at [latex]x=0[\/latex]; infinite discontinuity at [latex]x=1[\/latex]<\/li>\r\n \t
        3. [latex]f(x)=\\dfrac{5}{e^x-2}[\/latex] Infinite discontinuity at [latex]x=\\ln 2[\/latex]<\/li>\r\n \t
        4. [latex]H(x)= \\tan 2x[\/latex] Infinite discontinuities at [latex]x=\\frac{(2k+1)\\pi}{4}[\/latex], for [latex]k=0, \\, \\pm 1, \\, \\pm 2, \\, \\pm 3, \\cdots[\/latex]<\/li>\r\n \t
        5. No. It is a removable discontinuity.<\/li>\r\n \t
        6. Yes. It is continuous.<\/li>\r\n \t
        7. Yes. It is continuous.<\/li>\r\n \t
        8. [latex]k=-5[\/latex]<\/li>\r\n \t
        9. [latex]k=-1[\/latex]<\/li>\r\n \t
        10. [latex]k=\\frac{16}{3}[\/latex]<\/li>\r\n \t
        11. \"AIt is not possible to redefine [latex]f(1)[\/latex] since the discontinuity is a jump discontinuity.<\/li>\r\n \t
        12. Answers may vary; see the following example:\r\n\r\n\"A<\/span><\/li>\r\n \t
        13. \r\n

          Answers may vary; see the following example:<\/p>\r\n\"The<\/span>\/li><\/li>\r\n \t

        14. False. It is continuous over [latex](\u2212\\infty,0) \\cup (0,\\infty)[\/latex].<\/li>\r\n \t
        15. False. Consider [latex]f(x)=\\begin{cases} x & \\text{ if } \\, x \\ne 0 \\\\ 4 & \\text{ if } \\, x = 0 \\end{cases}[\/latex]<\/li>\r\n \t
        16. False. Consider [latex]f(x)= \\cos (x)[\/latex] on [latex][-\\pi, 2\\pi][\/latex].<\/li>\r\n \t
        17. False. The IVT does not<\/em> work in reverse! Consider [latex](x-1)^2[\/latex] over the interval [latex][-2,2][\/latex].<\/li>\r\n \t
        18. For all values of [latex]a, \\, f(a)[\/latex] is defined, [latex]\\underset{\\theta \\to a}{\\lim}f(\\theta)[\/latex] exists, and [latex]\\underset{\\theta \\to a}{\\lim}f(\\theta)=f(a)[\/latex]. Therefore, [latex]f(\\theta)[\/latex] is continuous everywhere.<\/li>\r\n \t
        19. Nowhere<\/li>\r\n<\/ol>\r\n

          The Precise Definition of a Limit<\/h2>\r\n

          In the following exercises (1-2), write the appropriate [latex]\\varepsilon[\/latex]-[latex]\\delta[\/latex] definition for each of the given statements.<\/strong><\/p>\r\n\r\n

            \r\n \t
          1. For every [latex]\\varepsilon > 0[\/latex], there exists a [latex]\\delta > 0[\/latex] so that if [latex]0 < |t-b| < \\delta[\/latex], then [latex]|g(t)-M| < \\varepsilon[\/latex]<\/li>\r\n \t
          2. For every [latex]\\varepsilon > 0[\/latex], there exists a [latex]\\delta > 0[\/latex] so that if [latex]0 < |x-a| < \\delta[\/latex], then [latex]|\\phi(x)-A| < \\varepsilon[\/latex]<\/li>\r\n \t
          3. [latex]\\delta \\le 0.25[\/latex]<\/li>\r\n \t
          4. [latex]\\delta \\le 2[\/latex]<\/li>\r\n \t
          5. [latex]\\delta \\le 1[\/latex]<\/li>\r\n \t
          6. [latex]\\delta < 0.3900[\/latex]<\/li>\r\n \t
          7. Let [latex]\\delta =\\varepsilon[\/latex]. If [latex]0 < |x-3| < \\varepsilon[\/latex], then [latex]|x+3-6|=|x-3| < \\varepsilon[\/latex].<\/li>\r\n \t
          8. Let [latex]\\delta =\\sqrt[4]{\\varepsilon}[\/latex]. If [latex]0 < |x| < \\sqrt[4]{\\varepsilon}[\/latex], then [latex]|x^4|=x^4 < \\varepsilon[\/latex].<\/li>\r\n \t
          9. Let [latex]\\delta =\\varepsilon^2[\/latex]. If [latex]5-\\varepsilon^2 < x < 5[\/latex], then [latex]|\\sqrt{5-x}|=\\sqrt{5-x} < \\varepsilon[\/latex].<\/li>\r\n \t
          10. Let [latex]\\delta =\\varepsilon\/5[\/latex]. If [latex]1-\\varepsilon\/5 < x < 1[\/latex], then [latex]|f(x)-3|=5x-5 < \\varepsilon[\/latex].<\/li>\r\n \t
          11. Let [latex]\\delta =\\sqrt{\\frac{3}{N}}[\/latex]. If [latex]0 < |x+1| < \\sqrt{\\frac{3}{N}}[\/latex], then [latex]f(x)=\\frac{3}{(x+1)^2} > N[\/latex].<\/li>\r\n \t
          12. [latex]f(x)-g(x)=f(x)+(-1)g(x)[\/latex]<\/li>\r\n \t
          13. Answers may vary.<\/li>\r\n<\/ol>","rendered":"

            The Limit Laws<\/h2>\n

            In the following exercises (1-2), use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).<\/strong><\/p>\n

              \n
            1. Use constant multiple law and difference law: [latex]\\underset{x\\to 0}{\\lim}(4x^2-2x+3)=4\\underset{x\\to 0}{\\lim}x^2-2\\underset{x\\to 0}{\\lim}x+\\underset{x\\to 0}{\\lim}3=3[\/latex]<\/li>\n
            2. Use root law: [latex]\\underset{x\\to -2}{\\lim}\\sqrt{x^2-6x+3}=\\sqrt{\\underset{x\\to -2}{\\lim}(x^2-6x+3)}=\\sqrt{19}[\/latex]<\/li>\n
            3. [latex]49[\/latex]<\/li>\n
            4. [latex]1[\/latex]<\/li>\n
            5. [latex]-\\frac{5}{7}[\/latex]<\/li>\n
            6. [latex]\\underset{x\\to 4}{\\lim}\\frac{x^2-16}{x-4}=\\frac{16-16}{4-4}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{x\\to 4}{\\lim}\\frac{x^2-16}{x-4}=\\underset{x\\to 4}{\\lim}\\frac{(x+4)(x-4)}{x-4}=8[\/latex]<\/li>\n
            7. [latex]\\underset{x\\to 6}{\\lim}\\frac{3x-18}{2x-12}=\\frac{18-18}{12-12}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{x\\to 6}{\\lim}\\frac{3x-18}{2x-12}=\\underset{x\\to 6}{\\lim}\\frac{3(x-6)}{2(x-6)}=\\frac{3}{2}[\/latex]<\/li>\n
            8. [latex]\\underset{t \\to 9}{\\lim}\\frac{t-9}{\\sqrt{t}-3}=\\frac{9-9}{3-3}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{t\\to 9}{\\lim}\\frac{t-9}{\\sqrt{t}-3}=\\underset{t\\to 9}{\\lim}\\frac{t-9}{\\sqrt{t}-3}\\frac{\\sqrt{t}+3}{\\sqrt{t}+3}=\\underset{t\\to 9}{\\lim}(\\sqrt{t}+3)=6[\/latex]<\/li>\n
            9. [latex]\\underset{\\theta \\to \\pi}{\\lim}\\frac{\\sin \\theta}{\\tan \\theta}=\\frac{\\sin \\pi}{\\tan \\pi}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{\\theta \\to \\pi}{\\lim}\\frac{\\sin \\theta}{\\tan \\theta}=\\underset{\\theta \\to \\pi}{\\lim}\\frac{\\sin \\theta}{\\frac{\\sin \\theta}{\\cos \\theta}}=\\underset{\\theta \\to \\pi}{\\lim}\\cos \\theta =-1[\/latex].<\/li>\n
            10. [latex]\\underset{x\\to 1\/2}{\\lim}\\frac{2x^2+3x-2}{2x-1}=\\frac{\\frac{1}{2}+\\frac{3}{2}-2}{1-1}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{x\\to 1\/2}{\\lim}\\frac{2x^2+3x-2}{2x-1}=\\underset{x\\to 1\/2}{\\lim}\\frac{(2x-1)(x+2)}{2x-1}=\\frac{5}{2}[\/latex]<\/li>\n
            11. [latex]\\underset{x\\to 6}{\\lim}2f(x)g(x)=2\\underset{x\\to 6}{\\lim}f(x)\\underset{x\\to 6}{\\lim}g(x)=72[\/latex]<\/li>\n
            12. [latex]\\underset{x\\to 6}{\\lim}(f(x)+\\frac{1}{3}g(x))=\\underset{x\\to 6}{\\lim}f(x)+\\frac{1}{3}\\underset{x\\to 6}{\\lim}g(x)=7[\/latex]<\/li>\n
            13. [latex]\\underset{x\\to 6}{\\lim}\\sqrt{g(x)-f(x)}=\\sqrt{\\underset{x\\to 6}{\\lim}g(x)-\\underset{x\\to 6}{\\lim}f(x)}=\\sqrt{5}[\/latex]<\/li>\n
            14. [latex]\\underset{x\\to 6}{\\lim}[(x+1)\\cdot f(x)]=(\\underset{x\\to 6}{\\lim}(x+1))(\\underset{x\\to 6}{\\lim}f(x))=28[\/latex].<\/li>\n
            15. \"The\n
                \n
              1. [latex]9[\/latex]<\/li>\n
              2. [latex]7[\/latex]<\/li>\n<\/ol>\n<\/li>\n
              3. \"The\n
                  \n
                1. [latex]1[\/latex]<\/li>\n
                2. [latex]1[\/latex]<\/li>\n<\/ol>\n<\/li>\n
                3. [latex]\\underset{x\\to -3^-}{\\lim}(f(x)-3g(x))=\\underset{x\\to -3^-}{\\lim}f(x)-3\\underset{x\\to -3^-}{\\lim}g(x)=0+6=6[\/latex]<\/li>\n
                4. [latex]\\underset{x\\to -5}{\\lim}\\frac{2+g(x)}{f(x)}=\\frac{2+(\\underset{x\\to -5}{\\lim}g(x))}{\\underset{x\\to -5}{\\lim}f(x)}=\\frac{2+0}{2}=1[\/latex]<\/li>\n
                5. [latex]\\underset{x\\to 1}{\\lim}\\sqrt[3]{f(x)-g(x)}=\\sqrt[3]{\\underset{x\\to 1}{\\lim}f(x)-\\underset{x\\to 1}{\\lim}g(x)}=\\sqrt[3]{2+5}=\\sqrt[3]{7}[\/latex]<\/li>\n
                6. [latex]\\underset{x\\to -9}{\\lim}(x\\cdot f(x)+2g(x))=(\\underset{x\\to -9}{\\lim}x)(\\underset{x\\to -9}{\\lim}f(x))+2\\underset{x\\to -9}{\\lim}(g(x))=(-9)(6)+2(4)=-46[\/latex]<\/li>\n<\/ol>\n

                  Continuity<\/h2>\n

                  For the following exercises (1-4), determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.<\/strong><\/p>\n

                    \n
                  1. The function is defined for all [latex]x[\/latex] in the interval [latex](0,\\infty)[\/latex].<\/li>\n
                  2. Removable discontinuity at [latex]x=0[\/latex]; infinite discontinuity at [latex]x=1[\/latex]<\/li>\n
                  3. [latex]f(x)=\\dfrac{5}{e^x-2}[\/latex] Infinite discontinuity at [latex]x=\\ln 2[\/latex]<\/li>\n
                  4. [latex]H(x)= \\tan 2x[\/latex] Infinite discontinuities at [latex]x=\\frac{(2k+1)\\pi}{4}[\/latex], for [latex]k=0, \\, \\pm 1, \\, \\pm 2, \\, \\pm 3, \\cdots[\/latex]<\/li>\n
                  5. No. It is a removable discontinuity.<\/li>\n
                  6. Yes. It is continuous.<\/li>\n
                  7. Yes. It is continuous.<\/li>\n
                  8. [latex]k=-5[\/latex]<\/li>\n
                  9. [latex]k=-1[\/latex]<\/li>\n
                  10. [latex]k=\\frac{16}{3}[\/latex]<\/li>\n
                  11. \"AIt is not possible to redefine [latex]f(1)[\/latex] since the discontinuity is a jump discontinuity.<\/li>\n
                  12. Answers may vary; see the following example:\n

                    \"A<\/span><\/li>\n

                  13. \n

                    Answers may vary; see the following example:<\/p>\n

                    \"The<\/span>\/li><\/li>\n

                  14. False. It is continuous over [latex](\u2212\\infty,0) \\cup (0,\\infty)[\/latex].<\/li>\n
                  15. False. Consider [latex]f(x)=\\begin{cases} x & \\text{ if } \\, x \\ne 0 \\\\ 4 & \\text{ if } \\, x = 0 \\end{cases}[\/latex]<\/li>\n
                  16. False. Consider [latex]f(x)= \\cos (x)[\/latex] on [latex][-\\pi, 2\\pi][\/latex].<\/li>\n
                  17. False. The IVT does not<\/em> work in reverse! Consider [latex](x-1)^2[\/latex] over the interval [latex][-2,2][\/latex].<\/li>\n
                  18. For all values of [latex]a, \\, f(a)[\/latex] is defined, [latex]\\underset{\\theta \\to a}{\\lim}f(\\theta)[\/latex] exists, and [latex]\\underset{\\theta \\to a}{\\lim}f(\\theta)=f(a)[\/latex]. Therefore, [latex]f(\\theta)[\/latex] is continuous everywhere.<\/li>\n
                  19. Nowhere<\/li>\n<\/ol>\n

                    The Precise Definition of a Limit<\/h2>\n

                    In the following exercises (1-2), write the appropriate [latex]\\varepsilon[\/latex]–[latex]\\delta[\/latex] definition for each of the given statements.<\/strong><\/p>\n

                      \n
                    1. For every [latex]\\varepsilon > 0[\/latex], there exists a [latex]\\delta > 0[\/latex] so that if [latex]0 < |t-b| < \\delta[\/latex], then [latex]|g(t)-M| < \\varepsilon[\/latex]<\/li>\n
                    2. For every [latex]\\varepsilon > 0[\/latex], there exists a [latex]\\delta > 0[\/latex] so that if [latex]0 < |x-a| < \\delta[\/latex], then [latex]|\\phi(x)-A| < \\varepsilon[\/latex]<\/li>\n
                    3. [latex]\\delta \\le 0.25[\/latex]<\/li>\n
                    4. [latex]\\delta \\le 2[\/latex]<\/li>\n
                    5. [latex]\\delta \\le 1[\/latex]<\/li>\n
                    6. [latex]\\delta < 0.3900[\/latex]<\/li>\n
                    7. Let [latex]\\delta =\\varepsilon[\/latex]. If [latex]0 < |x-3| < \\varepsilon[\/latex], then [latex]|x+3-6|=|x-3| < \\varepsilon[\/latex].<\/li>\n
                    8. Let [latex]\\delta =\\sqrt[4]{\\varepsilon}[\/latex]. If [latex]0 < |x| < \\sqrt[4]{\\varepsilon}[\/latex], then [latex]|x^4|=x^4 < \\varepsilon[\/latex].<\/li>\n
                    9. Let [latex]\\delta =\\varepsilon^2[\/latex]. If [latex]5-\\varepsilon^2 < x < 5[\/latex], then [latex]|\\sqrt{5-x}|=\\sqrt{5-x} < \\varepsilon[\/latex].<\/li>\n
                    10. Let [latex]\\delta =\\varepsilon\/5[\/latex]. If [latex]1-\\varepsilon\/5 < x < 1[\/latex], then [latex]|f(x)-3|=5x-5 < \\varepsilon[\/latex].<\/li>\n
                    11. Let [latex]\\delta =\\sqrt{\\frac{3}{N}}[\/latex]. If [latex]0 < |x+1| < \\sqrt{\\frac{3}{N}}[\/latex], then [latex]f(x)=\\frac{3}{(x+1)^2} > N[\/latex].<\/li>\n
                    12. [latex]f(x)-g(x)=f(x)+(-1)g(x)[\/latex]<\/li>\n
                    13. Answers may vary.<\/li>\n<\/ol>\n","protected":false},"author":15,"menu_order":4,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":108,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/151"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":1,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/151\/revisions"}],"predecessor-version":[{"id":164,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/151\/revisions\/164"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/parts\/108"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/151\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/media?parent=151"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapter-type?post=151"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/contributor?post=151"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/license?post=151"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}