{"id":2487,"date":"2018-02-01T15:33:28","date_gmt":"2018-02-01T15:33:28","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/?post_type=chapter&p=2487"},"modified":"2018-02-01T15:33:28","modified_gmt":"2018-02-01T15:33:28","slug":"chapter-2-review-exercises","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/chapter\/chapter-2-review-exercises\/","title":{"raw":"Chapter 2 Review Exercises","rendered":"Chapter 2 Review Exercises"},"content":{"raw":"

Chapter Review Exercises<\/h1>\r\n

True or False<\/em>. In the following exercises, justify your answer with a proof or a counterexample.<\/p>\r\n\r\n

\r\n
\r\n

A function has to be continuous at [latex]x=a[\/latex] if the [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] exists.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n

You can use the quotient rule to evaluate [latex]\\underset{x\\to 0}{\\text{lim}}\\frac{ \\sin x}{x}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n

[reveal-answer q=\"fs-id1170571653983\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571653983\"]\r\n

False<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n

If there is a vertical asymptote at [latex]x=a[\/latex] for the function [latex]f(x),[\/latex] then [latex]f[\/latex] is undefined at the point [latex]x=a.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n

If [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] does not exist, then [latex]f[\/latex] is undefined at the point [latex]x=a.[\/latex]<\/p>\r\n\r\n<\/div>\r\n

[reveal-answer q=\"fs-id1170571597426\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571597426\"]\r\n

False. A removable discontinuity is possible.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n

Using the graph, find each limit or explain why the limit does not exist.<\/p>\r\n\r\n

    \r\n \t
  1. [latex]\\underset{x\\to -1}{\\text{lim}}f(x)[\/latex]<\/li>\r\n \t
  2. [latex]\\underset{x\\to 1}{\\text{lim}}f(x)[\/latex]<\/li>\r\n \t
  3. [latex]\\underset{x\\to {0}^{+}}{\\text{lim}}f(x)[\/latex]<\/li>\r\n \t
  4. [latex]\\underset{x\\to 2}{\\text{lim}}f(x)[\/latex]<\/li>\r\n<\/ol>\r\n\"A<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n

    In the following exercises, evaluate the limit algebraically or explain why the limit does not exist.<\/p>\r\n\r\n

    \r\n
    \r\n

    [latex]\\underset{x\\to 2}{\\text{lim}}\\frac{2{x}^{2}-3x-2}{x-2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

    [reveal-answer q=\"fs-id1170572330176\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572330176\"]\r\n

    5<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    [latex]\\underset{x\\to 0}{\\text{lim}}3{x}^{2}-2x+4[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    [latex]\\underset{x\\to 3}{\\text{lim}}\\frac{{x}^{3}-2{x}^{2}-1}{3x-2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

    [reveal-answer q=\"fs-id1170571733970\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571733970\"]\r\n

    [latex]8\\text{\/}7[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    [latex]\\underset{x\\to \\pi \\text{\/}2}{\\text{lim}}\\frac{ \\cot x}{ \\cos x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    [latex]\\underset{x\\to -5}{\\text{lim}}\\frac{{x}^{2}+25}{x+5}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

    [reveal-answer q=\"fs-id1170571636177\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571636177\"]\r\n

    DNE<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    [latex]\\underset{x\\to 2}{\\text{lim}}\\frac{3{x}^{2}-2x-8}{{x}^{2}-4}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    [latex]\\underset{x\\to 1}{\\text{lim}}\\frac{{x}^{2}-1}{{x}^{3}-1}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

    [reveal-answer q=\"fs-id1170572510140\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572510140\"]\r\n

    [latex]2\\text{\/}3[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    [latex]\\underset{x\\to 1}{\\text{lim}}\\frac{{x}^{2}-1}{\\sqrt{x}-1}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    [latex]\\underset{x\\to 4}{\\text{lim}}\\frac{4-x}{\\sqrt{x}-2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

    [reveal-answer q=\"fs-id1170572419312\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572419312\"]\r\n

    \u22124;<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    [latex]\\underset{x\\to 4}{\\text{lim}}\\frac{1}{\\sqrt{x}-2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    In the following exercises, use the squeeze theorem to prove the limit.<\/p>\r\n\r\n

    \r\n
    \r\n

    [latex]\\underset{x\\to 0}{\\text{lim}}{x}^{2} \\cos (2\\pi x)=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n

    [reveal-answer q=\"fs-id1170572333059\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572333059\"]\r\n

    Since [latex]-1\\le \\cos (2\\pi x)\\le 1,[\/latex] then [latex]-{x}^{2}\\le {x}^{2} \\cos (2\\pi x)\\le {x}^{2}.[\/latex] Since [latex]\\underset{x\\to 0}{\\text{lim}}{x}^{2}=0=\\underset{x\\to 0}{\\text{lim}}-{x}^{2},[\/latex] it follows that [latex]\\underset{x\\to 0}{\\text{lim}}{x}^{2} \\cos (2\\pi x)=0.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    [latex]\\underset{x\\to 0}{\\text{lim}}{x}^{3} \\sin (\\frac{\\pi }{x})=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    Determine the domain such that the function [latex]f(x)=\\sqrt{x-2}+x{e}^{x}[\/latex] is continuous over its domain.<\/p>\r\n\r\n<\/div>\r\n

    [reveal-answer q=\"fs-id1170571697178\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571697178\"]\r\n

    [latex]\\left[2,\\infty \\right][\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    In the following exercises, determine the value of [latex]c[\/latex] such that the function remains continuous. Draw your resulting function to ensure it is continuous.<\/p>\r\n\r\n

    \r\n
    \r\n

    [latex]f(x)=\\bigg\\{\\begin{array}{l}{x}^{2}+1,x>c\\\\ 2x,x\\le c\\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    [latex]f(x)=\\bigg\\{\\begin{array}{l}\\sqrt{x+1},x>\\text{\u2212}1\\\\ {x}^{2}+c,x\\le -1\\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

    [reveal-answer q=\"fs-id1170572609485\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572609485\"]\r\n

    [latex]c=-1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    In the following exercises, use the precise definition of limit to prove the limit.<\/p>\r\n\r\n

    \r\n
    \r\n

    [latex]\\underset{x\\to 1}{\\text{lim}}(8x+16)=24[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    [latex]\\underset{x\\to 0}{\\text{lim}}{x}^{3}=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n

    [reveal-answer q=\"fs-id1170571710699\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571710699\"]\r\n

    [latex]\\delta =\\sqrt[3]{\\epsilon }[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    A ball is thrown into the air and the vertical position is given by [latex]x(t)=-4.9{t}^{2}+25t+5.[\/latex] Use the Intermediate Value Theorem to show that the ball must land on the ground sometime between 5 sec and 6 sec after the throw.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    A particle moving along a line has a displacement according to the function [latex]x(t)={t}^{2}-2t+4,[\/latex] where [latex]x[\/latex] is measured in meters and [latex]t[\/latex] is measured in seconds. Find the average velocity over the time period [latex]t=\\left[0,2\\right].[\/latex]<\/p>\r\n\r\n<\/div>\r\n

    [reveal-answer q=\"fs-id1170572386121\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572386121\"]\r\n

    [latex]0\\text{m}\\text{\/} \\sec [\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    From the previous exercises, estimate the instantaneous velocity at [latex]t=2[\/latex] by checking the average velocity within [latex]t=0.01 \\sec \\text{.}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>","rendered":"

    Chapter Review Exercises<\/h1>\n

    True or False<\/em>. In the following exercises, justify your answer with a proof or a counterexample.<\/p>\n

    \n
    \n

    A function has to be continuous at [latex]x=a[\/latex] if the [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] exists.<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    You can use the quotient rule to evaluate [latex]\\underset{x\\to 0}{\\text{lim}}\\frac{ \\sin x}{x}.[\/latex]<\/p>\n<\/div>\n

    \n
    Show Solution<\/span><\/p>\n
    \n

    False<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    \n
    \n

    If there is a vertical asymptote at [latex]x=a[\/latex] for the function [latex]f(x),[\/latex] then [latex]f[\/latex] is undefined at the point [latex]x=a.[\/latex]<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    If [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] does not exist, then [latex]f[\/latex] is undefined at the point [latex]x=a.[\/latex]<\/p>\n<\/div>\n

    \n
    Show Solution<\/span><\/p>\n
    \n

    False. A removable discontinuity is possible.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    \n
    \n

    Using the graph, find each limit or explain why the limit does not exist.<\/p>\n

      \n
    1. [latex]\\underset{x\\to -1}{\\text{lim}}f(x)[\/latex]<\/li>\n
    2. [latex]\\underset{x\\to 1}{\\text{lim}}f(x)[\/latex]<\/li>\n
    3. [latex]\\underset{x\\to {0}^{+}}{\\text{lim}}f(x)[\/latex]<\/li>\n
    4. [latex]\\underset{x\\to 2}{\\text{lim}}f(x)[\/latex]<\/li>\n<\/ol>\n

      \"A<\/span><\/p>\n<\/div>\n<\/div>\n

      In the following exercises, evaluate the limit algebraically or explain why the limit does not exist.<\/p>\n

      \n
      \n

      [latex]\\underset{x\\to 2}{\\text{lim}}\\frac{2{x}^{2}-3x-2}{x-2}[\/latex]<\/p>\n<\/div>\n

      \n
      Show Solution<\/span><\/p>\n
      \n

      5<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      \n
      \n

      [latex]\\underset{x\\to 0}{\\text{lim}}3{x}^{2}-2x+4[\/latex]<\/p>\n<\/div>\n<\/div>\n

      \n
      \n

      [latex]\\underset{x\\to 3}{\\text{lim}}\\frac{{x}^{3}-2{x}^{2}-1}{3x-2}[\/latex]<\/p>\n<\/div>\n

      \n
      Show Solution<\/span><\/p>\n
      \n

      [latex]8\\text{\/}7[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      \n
      \n

      [latex]\\underset{x\\to \\pi \\text{\/}2}{\\text{lim}}\\frac{ \\cot x}{ \\cos x}[\/latex]<\/p>\n<\/div>\n<\/div>\n

      \n
      \n

      [latex]\\underset{x\\to -5}{\\text{lim}}\\frac{{x}^{2}+25}{x+5}[\/latex]<\/p>\n<\/div>\n

      \n
      Show Solution<\/span><\/p>\n
      \n

      DNE<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      \n
      \n

      [latex]\\underset{x\\to 2}{\\text{lim}}\\frac{3{x}^{2}-2x-8}{{x}^{2}-4}[\/latex]<\/p>\n<\/div>\n<\/div>\n

      \n
      \n

      [latex]\\underset{x\\to 1}{\\text{lim}}\\frac{{x}^{2}-1}{{x}^{3}-1}[\/latex]<\/p>\n<\/div>\n

      \n
      Show Solution<\/span><\/p>\n
      \n

      [latex]2\\text{\/}3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      \n
      \n

      [latex]\\underset{x\\to 1}{\\text{lim}}\\frac{{x}^{2}-1}{\\sqrt{x}-1}[\/latex]<\/p>\n<\/div>\n<\/div>\n

      \n
      \n

      [latex]\\underset{x\\to 4}{\\text{lim}}\\frac{4-x}{\\sqrt{x}-2}[\/latex]<\/p>\n<\/div>\n

      \n
      Show Solution<\/span><\/p>\n
      \n

      \u22124;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      \n
      \n

      [latex]\\underset{x\\to 4}{\\text{lim}}\\frac{1}{\\sqrt{x}-2}[\/latex]<\/p>\n<\/div>\n<\/div>\n

      In the following exercises, use the squeeze theorem to prove the limit.<\/p>\n

      \n
      \n

      [latex]\\underset{x\\to 0}{\\text{lim}}{x}^{2} \\cos (2\\pi x)=0[\/latex]<\/p>\n<\/div>\n

      \n
      Show Solution<\/span><\/p>\n
      \n

      Since [latex]-1\\le \\cos (2\\pi x)\\le 1,[\/latex] then [latex]-{x}^{2}\\le {x}^{2} \\cos (2\\pi x)\\le {x}^{2}.[\/latex] Since [latex]\\underset{x\\to 0}{\\text{lim}}{x}^{2}=0=\\underset{x\\to 0}{\\text{lim}}-{x}^{2},[\/latex] it follows that [latex]\\underset{x\\to 0}{\\text{lim}}{x}^{2} \\cos (2\\pi x)=0.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      \n
      \n

      [latex]\\underset{x\\to 0}{\\text{lim}}{x}^{3} \\sin (\\frac{\\pi }{x})=0[\/latex]<\/p>\n<\/div>\n<\/div>\n

      \n
      \n

      Determine the domain such that the function [latex]f(x)=\\sqrt{x-2}+x{e}^{x}[\/latex] is continuous over its domain.<\/p>\n<\/div>\n

      \n
      Show Solution<\/span><\/p>\n
      \n

      [latex]\\left[2,\\infty \\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      In the following exercises, determine the value of [latex]c[\/latex] such that the function remains continuous. Draw your resulting function to ensure it is continuous.<\/p>\n

      \n
      \n

      [latex]f(x)=\\bigg\\{\\begin{array}{l}{x}^{2}+1,x>c\\\\ 2x,x\\le c\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n

      \n
      \n

      [latex]f(x)=\\bigg\\{\\begin{array}{l}\\sqrt{x+1},x>\\text{\u2212}1\\\\ {x}^{2}+c,x\\le -1\\end{array}[\/latex]<\/p>\n<\/div>\n

      \n
      Show Solution<\/span><\/p>\n
      \n

      [latex]c=-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      In the following exercises, use the precise definition of limit to prove the limit.<\/p>\n

      \n
      \n

      [latex]\\underset{x\\to 1}{\\text{lim}}(8x+16)=24[\/latex]<\/p>\n<\/div>\n<\/div>\n

      \n
      \n

      [latex]\\underset{x\\to 0}{\\text{lim}}{x}^{3}=0[\/latex]<\/p>\n<\/div>\n

      \n
      Show Solution<\/span><\/p>\n
      \n

      [latex]\\delta =\\sqrt[3]{\\epsilon }[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      \n
      \n

      A ball is thrown into the air and the vertical position is given by [latex]x(t)=-4.9{t}^{2}+25t+5.[\/latex] Use the Intermediate Value Theorem to show that the ball must land on the ground sometime between 5 sec and 6 sec after the throw.<\/p>\n<\/div>\n<\/div>\n

      \n
      \n

      A particle moving along a line has a displacement according to the function [latex]x(t)={t}^{2}-2t+4,[\/latex] where [latex]x[\/latex] is measured in meters and [latex]t[\/latex] is measured in seconds. Find the average velocity over the time period [latex]t=\\left[0,2\\right].[\/latex]<\/p>\n<\/div>\n

      \n
      Show Solution<\/span><\/p>\n
      \n

      [latex]0\\text{m}\\text{\/} \\sec[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      \n
      \n

      From the previous exercises, estimate the instantaneous velocity at [latex]t=2[\/latex] by checking the average velocity within [latex]t=0.01 \\sec \\text{.}[\/latex]<\/p>\n<\/div>\n<\/div>\n","protected":false},"author":44985,"menu_order":7,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2487","chapter","type-chapter","status-publish","hentry"],"part":1589,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters\/2487","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/users\/44985"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters\/2487\/revisions"}],"predecessor-version":[{"id":2488,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters\/2487\/revisions\/2488"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/parts\/1589"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters\/2487\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/media?parent=2487"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapter-type?post=2487"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/contributor?post=2487"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/license?post=2487"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}