{"id":1451,"date":"2023-06-05T14:51:52","date_gmt":"2023-06-05T14:51:52","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/chapter\/solutions-for-finding-limits-numerical-and-graphical-approaches\/"},"modified":"2023-06-05T14:51:52","modified_gmt":"2023-06-05T14:51:52","slug":"solutions-for-finding-limits-numerical-and-graphical-approaches","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/chapter\/solutions-for-finding-limits-numerical-and-graphical-approaches\/","title":{"raw":"Solutions 70: Finding Limits: Numerical and Graphical Approaches","rendered":"Solutions 70: Finding Limits: Numerical and Graphical Approaches"},"content":{"raw":"\n<h2>Solutions to Odd-Numbered Exercises<\/h2>\n1.&nbsp;The value of the function, the output, at [latex]x=a[\/latex] is [latex]f\\left(a\\right)[\/latex]. When the [latex]\\underset{x\\to a}{\\mathrm{lim}}f\\left(x\\right)[\/latex] is taken, the values of [latex]x[\/latex] get infinitely close to [latex]a[\/latex] but never equal [latex]a[\/latex]. As the values of [latex]x[\/latex] approach [latex]a[\/latex] from the left and right, the limit is the value that the function is approaching.\n\n3.&nbsp;\u20134\n\n5.&nbsp;\u20134\n\n7.&nbsp;2\n\n9.&nbsp;does not exist\n\n11.&nbsp;4\n\n13.&nbsp;does not exist\n\n15.&nbsp;Answers will vary.\n\n17.&nbsp;Answers will vary.\n\n19.&nbsp;Answers will vary.\n\n21.&nbsp;Answers will vary.\n\n23.&nbsp;7.38906\n\n25.&nbsp;54.59815\n\n27.&nbsp;[latex]{e}^{6}\\approx 403.428794[\/latex], [latex]{e}^{7}\\approx 1096.633158[\/latex], [latex]{e}^{n}[\/latex]\n\n29.&nbsp;[latex]\\underset{x\\to -2}{\\mathrm{lim}}f\\left(x\\right)=1[\/latex]\n\n31.&nbsp;[latex]\\underset{x\\to 3}{\\mathrm{lim}}\\left(\\frac{{x}^{2}-x - 6}{{x}^{2}-9}\\right)=\\frac{5}{6}\\approx 0.83[\/latex]\n\n33.&nbsp;[latex]\\underset{x\\to 1}{\\mathrm{lim}}\\left(\\frac{{x}^{2}-1}{{x}^{2}-3x+2}\\right)=-2.00[\/latex]\n\n35.&nbsp;[latex]\\underset{x\\to 1}{\\mathrm{lim}}\\left(\\frac{10 - 10{x}^{2}}{{x}^{2}-3x+2}\\right)=20.00[\/latex]\n\n37.&nbsp;[latex]\\underset{x\\to \\frac{-1}{2}}{\\mathrm{lim}}\\left(\\frac{x}{4{x}^{2}+4x+1}\\right)[\/latex] does not exist. Function values decrease without bound as [latex]x[\/latex] approaches \u20130.5 from either left or right.\n\n39.&nbsp;[latex]\\underset{x\\to 0}{\\mathrm{lim}}\\frac{7\\tan x}{3x}=\\frac{7}{3}[\/latex]\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27185252\/CNX_Precalc_Figure_12_01_2022.jpg\" alt=\"Table shows as the function approaches 0, the value is 7 over 3 but the function is undefined at 0.\">\n\n41.&nbsp;[latex]\\underset{x\\to 0}{\\mathrm{lim}}\\frac{2\\sin x}{4\\tan x}=\\frac{1}{2}[\/latex]\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27185255\/CNX_Precalc_Figure_12_01_2042.jpg\" alt=\"Table shows as the function approaches 0, the value is 1 over 2, but the function is undefined at 0.\">\n\n43.&nbsp;[latex]\\underset{x\\to 0}{\\mathrm{lim}}{e}^{{e}^{-\\text{ }\\frac{1}{{x}^{2}}}}=1.0[\/latex]\n\n45.&nbsp;[latex]\\underset{x\\to -{1}^{-}}{\\mathrm{lim}}\\frac{|x+1|}{x+1}=\\frac{-\\left(x+1\\right)}{\\left(x+1\\right)}=-1[\/latex] and [latex]\\underset{x\\to -{1}^{+}}{\\mathrm{lim}}\\frac{|x+1|}{x+1}=\\frac{\\left(x+1\\right)}{\\left(x+1\\right)}=1[\/latex]; since the right-hand limit does not equal the left-hand limit, [latex]\\underset{x\\to -1}{\\mathrm{lim}}\\frac{|x+1|}{x+1}[\/latex] does not exist.\n\n47.&nbsp;[latex]\\underset{x\\to -1}{\\mathrm{lim}}\\frac{1}{{\\left(x+1\\right)}^{2}}[\/latex] does not exist. The function increases without bound as [latex]x[\/latex] approaches [latex]-1[\/latex] from either side.\n\n49.&nbsp;[latex]\\underset{x\\to 0}{\\mathrm{lim}}\\frac{5}{1-{e}^{\\frac{2}{x}}}[\/latex] does not exist. Function values approach 5 from the left and approach 0 from the right.\n\n51.&nbsp;Through examination of the postulates and an understanding of relativistic physics, as [latex]v\\to c[\/latex], [latex]m\\to \\infty [\/latex]. Take this one step further to the solution,\n<p style=\"text-align: center;\">[latex]\\underset{v\\to {c}^{-}}{\\mathrm{lim}}m=\\underset{v\\to {c}^{-}}{\\mathrm{lim}}\\frac{{m}_{o}}{\\sqrt{1-\\left({v}^{2}\/{c}^{2}\\right)}}=\\infty [\/latex]<\/p>\n","rendered":"<h2>Solutions to Odd-Numbered Exercises<\/h2>\n<p>1.&nbsp;The value of the function, the output, at [latex]x=a[\/latex] is [latex]f\\left(a\\right)[\/latex]. When the [latex]\\underset{x\\to a}{\\mathrm{lim}}f\\left(x\\right)[\/latex] is taken, the values of [latex]x[\/latex] get infinitely close to [latex]a[\/latex] but never equal [latex]a[\/latex]. As the values of [latex]x[\/latex] approach [latex]a[\/latex] from the left and right, the limit is the value that the function is approaching.<\/p>\n<p>3.&nbsp;\u20134<\/p>\n<p>5.&nbsp;\u20134<\/p>\n<p>7.&nbsp;2<\/p>\n<p>9.&nbsp;does not exist<\/p>\n<p>11.&nbsp;4<\/p>\n<p>13.&nbsp;does not exist<\/p>\n<p>15.&nbsp;Answers will vary.<\/p>\n<p>17.&nbsp;Answers will vary.<\/p>\n<p>19.&nbsp;Answers will vary.<\/p>\n<p>21.&nbsp;Answers will vary.<\/p>\n<p>23.&nbsp;7.38906<\/p>\n<p>25.&nbsp;54.59815<\/p>\n<p>27.&nbsp;[latex]{e}^{6}\\approx 403.428794[\/latex], [latex]{e}^{7}\\approx 1096.633158[\/latex], [latex]{e}^{n}[\/latex]<\/p>\n<p>29.&nbsp;[latex]\\underset{x\\to -2}{\\mathrm{lim}}f\\left(x\\right)=1[\/latex]<\/p>\n<p>31.&nbsp;[latex]\\underset{x\\to 3}{\\mathrm{lim}}\\left(\\frac{{x}^{2}-x - 6}{{x}^{2}-9}\\right)=\\frac{5}{6}\\approx 0.83[\/latex]<\/p>\n<p>33.&nbsp;[latex]\\underset{x\\to 1}{\\mathrm{lim}}\\left(\\frac{{x}^{2}-1}{{x}^{2}-3x+2}\\right)=-2.00[\/latex]<\/p>\n<p>35.&nbsp;[latex]\\underset{x\\to 1}{\\mathrm{lim}}\\left(\\frac{10 - 10{x}^{2}}{{x}^{2}-3x+2}\\right)=20.00[\/latex]<\/p>\n<p>37.&nbsp;[latex]\\underset{x\\to \\frac{-1}{2}}{\\mathrm{lim}}\\left(\\frac{x}{4{x}^{2}+4x+1}\\right)[\/latex] does not exist. Function values decrease without bound as [latex]x[\/latex] approaches \u20130.5 from either left or right.<\/p>\n<p>39.&nbsp;[latex]\\underset{x\\to 0}{\\mathrm{lim}}\\frac{7\\tan x}{3x}=\\frac{7}{3}[\/latex]<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27185252\/CNX_Precalc_Figure_12_01_2022.jpg\" alt=\"Table shows as the function approaches 0, the value is 7 over 3 but the function is undefined at 0.\" \/><\/p>\n<p>41.&nbsp;[latex]\\underset{x\\to 0}{\\mathrm{lim}}\\frac{2\\sin x}{4\\tan x}=\\frac{1}{2}[\/latex]<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27185255\/CNX_Precalc_Figure_12_01_2042.jpg\" alt=\"Table shows as the function approaches 0, the value is 1 over 2, but the function is undefined at 0.\" \/><\/p>\n<p>43.&nbsp;[latex]\\underset{x\\to 0}{\\mathrm{lim}}{e}^{{e}^{-\\text{ }\\frac{1}{{x}^{2}}}}=1.0[\/latex]<\/p>\n<p>45.&nbsp;[latex]\\underset{x\\to -{1}^{-}}{\\mathrm{lim}}\\frac{|x+1|}{x+1}=\\frac{-\\left(x+1\\right)}{\\left(x+1\\right)}=-1[\/latex] and [latex]\\underset{x\\to -{1}^{+}}{\\mathrm{lim}}\\frac{|x+1|}{x+1}=\\frac{\\left(x+1\\right)}{\\left(x+1\\right)}=1[\/latex]; since the right-hand limit does not equal the left-hand limit, [latex]\\underset{x\\to -1}{\\mathrm{lim}}\\frac{|x+1|}{x+1}[\/latex] does not exist.<\/p>\n<p>47.&nbsp;[latex]\\underset{x\\to -1}{\\mathrm{lim}}\\frac{1}{{\\left(x+1\\right)}^{2}}[\/latex] does not exist. The function increases without bound as [latex]x[\/latex] approaches [latex]-1[\/latex] from either side.<\/p>\n<p>49.&nbsp;[latex]\\underset{x\\to 0}{\\mathrm{lim}}\\frac{5}{1-{e}^{\\frac{2}{x}}}[\/latex] does not exist. Function values approach 5 from the left and approach 0 from the right.<\/p>\n<p>51.&nbsp;Through examination of the postulates and an understanding of relativistic physics, as [latex]v\\to c[\/latex], [latex]m\\to \\infty[\/latex]. Take this one step further to the solution,<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{v\\to {c}^{-}}{\\mathrm{lim}}m=\\underset{v\\to {c}^{-}}{\\mathrm{lim}}\\frac{{m}_{o}}{\\sqrt{1-\\left({v}^{2}\/{c}^{2}\\right)}}=\\infty[\/latex]<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1451\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":503070,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1451","chapter","type-chapter","status-publish","hentry"],"part":1445,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapters\/1451","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/users\/503070"}],"version-history":[{"count":0,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapters\/1451\/revisions"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/parts\/1445"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapters\/1451\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/media?parent=1451"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapter-type?post=1451"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/contributor?post=1451"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/license?post=1451"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}