{"id":3775,"date":"2021-05-12T20:24:07","date_gmt":"2021-05-12T20:24:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-section-2-5-the-precise-definition-of-a-limit\/"},"modified":"2021-05-12T20:24:07","modified_gmt":"2021-05-12T20:24:07","slug":"review-for-section-2-5-the-precise-definition-of-a-limit","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-section-2-5-the-precise-definition-of-a-limit\/","title":{"raw":"Skills Review for The Precise Definition of a Limit","rendered":"Skills Review for The Precise Definition of a Limit"},"content":{"raw":"\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n \t<li style=\"orphans: 1;\">Solve inequalities containing absolute values<\/li>\n \t<li>Express solutions to inequalities containing absolute values<\/li>\n<\/ul>\n<\/div>\nThe Precise Definition of a Limit section covers the precise (formal) definition of a limit. Using and understanding the formal definition of a limit requires us to have some background knowledge about how to solve absolute value inequalities. What you need to know about absolute value inequalities is reviewed here.\n<h2>Solve Absolute Value Inequalities<\/h2>\nAs we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at [latex]\\left(-x,0\\right)[\/latex] has an absolute value of [latex]x[\/latex] as it is <em>x <\/em>units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero.\n\nAn <strong>absolute value inequality<\/strong> is an equation of the form\n<div style=\"text-align: center;\">[latex]|A|&lt;B,|A|\\le B,|A|&gt;B,\\text{or }|A|\\ge B[\/latex],<\/div>\nwhere <em>A<\/em>, and sometimes <em>B<\/em>, represents an algebraic expression dependent on a variable <em>x. <\/em>Solving the inequality means finding the set of all [latex]x[\/latex] <em>-<\/em>values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.\n\nThere are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.\n\nSuppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of <em>x-<\/em>values such that the distance between [latex]x[\/latex] and 600 is less than 200. We represent the distance between [latex]x[\/latex] and 600 as [latex]|x - 600|[\/latex], and therefore, [latex]|x - 600|\\le 200[\/latex] or\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}-200\\le x - 600\\le 200\\\\ -200+600\\le x - 600+600\\le 200+600\\\\ 400\\le x\\le 800\\end{array}[\/latex]<\/div>\nThis means our returns would be between $400 and $800.\n\nTo solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.\n<div class=\"textbox\">\n<h3>A General Note: Absolute Value Inequalities<\/h3>\nFor an algebraic expression <em>X&nbsp;<\/em>and [latex]k&gt;0[\/latex], an <strong>absolute value inequality<\/strong> is an inequality of the form:\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}|X|&lt; k\\text{ which is equivalent to }-k&lt; X&lt; k\\end{array}[\/latex]. Note that absolute value inequalities can include any type of inequality sign, although the focus here is absolute value inequalities with less than or less than or equal to signs.<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Absolute Value Inequality<\/h3>\nSolve [latex]|x - 1|\\le 3[\/latex].\n\n[reveal-answer q=\"4865\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"4865\"]\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}|x - 1|\\le 3\\hfill \\\\ \\hfill \\\\ -3\\le x - 1\\le 3\\hfill \\\\ \\hfill \\\\ -2\\le x\\le 4\\hfill \\\\ \\hfill \\\\ \\left[-2,4\\right]\\hfill \\end{array}[\/latex]<\/p>\n\n<div>[\/hidden-answer]<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nSolve [latex]2|k - 4|\\le 6[\/latex].\n\n[reveal-answer q=\"96760\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"96760\"]\n\nIn interval notation, this would be [latex][1,7][\/latex]. Be sure you isolate the absolute value first by dividing both sides of the inequality by 2.\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n[ohm_question]89935[\/ohm_question]\n\n<\/div>\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li style=\"orphans: 1;\">Solve inequalities containing absolute values<\/li>\n<li>Express solutions to inequalities containing absolute values<\/li>\n<\/ul>\n<\/div>\n<p>The Precise Definition of a Limit section covers the precise (formal) definition of a limit. Using and understanding the formal definition of a limit requires us to have some background knowledge about how to solve absolute value inequalities. What you need to know about absolute value inequalities is reviewed here.<\/p>\n<h2>Solve Absolute Value Inequalities<\/h2>\n<p>As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at [latex]\\left(-x,0\\right)[\/latex] has an absolute value of [latex]x[\/latex] as it is <em>x <\/em>units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero.<\/p>\n<p>An <strong>absolute value inequality<\/strong> is an equation of the form<\/p>\n<div style=\"text-align: center;\">[latex]|A|<B,|A|\\le B,|A|>B,\\text{or }|A|\\ge B[\/latex],<\/div>\n<p>where <em>A<\/em>, and sometimes <em>B<\/em>, represents an algebraic expression dependent on a variable <em>x. <\/em>Solving the inequality means finding the set of all [latex]x[\/latex] <em>&#8211;<\/em>values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.<\/p>\n<p>There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.<\/p>\n<p>Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of <em>x-<\/em>values such that the distance between [latex]x[\/latex] and 600 is less than 200. We represent the distance between [latex]x[\/latex] and 600 as [latex]|x - 600|[\/latex], and therefore, [latex]|x - 600|\\le 200[\/latex] or<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}-200\\le x - 600\\le 200\\\\ -200+600\\le x - 600+600\\le 200+600\\\\ 400\\le x\\le 800\\end{array}[\/latex]<\/div>\n<p>This means our returns would be between $400 and $800.<\/p>\n<p>To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Absolute Value Inequalities<\/h3>\n<p>For an algebraic expression <em>X&nbsp;<\/em>and [latex]k>0[\/latex], an <strong>absolute value inequality<\/strong> is an inequality of the form:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}|X|< k\\text{ which is equivalent to }-k< X< k\\end{array}[\/latex]. Note that absolute value inequalities can include any type of inequality sign, although the focus here is absolute value inequalities with less than or less than or equal to signs.<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Absolute Value Inequality<\/h3>\n<p>Solve [latex]|x - 1|\\le 3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q4865\">Show Solution<\/span><\/p>\n<div id=\"q4865\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}|x - 1|\\le 3\\hfill \\\\ \\hfill \\\\ -3\\le x - 1\\le 3\\hfill \\\\ \\hfill \\\\ -2\\le x\\le 4\\hfill \\\\ \\hfill \\\\ \\left[-2,4\\right]\\hfill \\end{array}[\/latex]<\/p>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]2|k - 4|\\le 6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q96760\">Show Solution<\/span><\/p>\n<div id=\"q96760\" class=\"hidden-answer\" style=\"display: none\">\n<p>In interval notation, this would be [latex][1,7][\/latex]. Be sure you isolate the absolute value first by dividing both sides of the inequality by 2.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm89935\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=89935&theme=oea&iframe_resize_id=ohm89935&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\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-3775\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Modification and Revision. <strong>Provided by<\/strong>: Lumen Learning. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/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":17533,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"College Algebra Corequisite\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/precalculus\/\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Modification and Revision\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"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-3775","chapter","type-chapter","status-publish","hentry"],"part":3770,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3775","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":0,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3775\/revisions"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/3770"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3775\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=3775"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=3775"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=3775"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=3775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}