{"id":3308,"date":"2025-08-16T01:54:56","date_gmt":"2025-08-16T01:54:56","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=3308"},"modified":"2026-03-25T05:18:35","modified_gmt":"2026-03-25T05:18:35","slug":"properties-of-limits-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/properties-of-limits-apply-it\/","title":{"raw":"Properties of Limits: Apply It","rendered":"Properties of Limits: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find the limit of a sum, a difference, and a product.<\/li>\r\n \t<li>Find the limit of a polynomial.<\/li>\r\n \t<li>Find the limit of a power or a root.<\/li>\r\n \t<li>Find the limit of a quotient.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<div data-test-render-count=\"1\">\r\n<div class=\"group relative pb-3\" data-is-streaming=\"false\">\r\n<div class=\"font-claude-response relative leading-[1.65rem] [&amp;_pre&gt;div]:bg-bg-000\/50 [&amp;_pre&gt;div]:border-0.5 [&amp;_pre&gt;div]:border-border-400 [&amp;_.ignore-pre-bg&gt;div]:bg-transparent [&amp;_.standard-markdown_:is(p,blockquote,h1,h2,h3,h4,h5,h6)]:pl-2 [&amp;_.standard-markdown_:is(p,blockquote,ul,ol,h1,h2,h3,h4,h5,h6)]:pr-8 [&amp;_.progressive-markdown_:is(p,blockquote,h1,h2,h3,h4,h5,h6)]:pl-2 [&amp;_.progressive-markdown_:is(p,blockquote,ul,ol,h1,h2,h3,h4,h5,h6)]:pr-8\">\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0 standard-markdown\">\r\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Analyzing Business Functions<\/h2>\r\n<p class=\"whitespace-normal break-words\">Business models often use polynomial and rational functions to predict profit, cost, and revenue. Understanding limits helps us analyze what happens as production levels change, even at points where functions have discontinuities.<\/p>\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<p class=\"whitespace-normal break-words\"><strong>Question Help: Evaluating Limits Using Properties<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">For polynomial functions, use direct substitution: [latex]\\lim_{x \\to a} p(x) = p(a)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For quotients that give [latex]\\frac{0}{0}[\/latex], try factoring and simplifying first<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For roots in quotients, consider multiplying by a conjugate<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Always try direct substitution first to see if it works<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\">A company's monthly profit in thousands of dollars is modeled by [latex]P(x) = 2x^3 - 3x + 1[\/latex], where [latex]x[\/latex] represents the number of units produced (in hundreds). As production approaches 500 units ([latex]x[\/latex] approaches 5), what profit does the model predict?<\/p>\r\n<p class=\"whitespace-normal break-words\">Find [latex]\\lim_{x \\to 5} (2x^3 - 3x + 1)[\/latex].<\/p>\r\n[reveal-answer q=\"689696\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"689696\"]Since this is a polynomial function, we can use direct substitution: [latex]\\begin{aligned} \\lim_{x \\to 5} (2x^3 - 3x + 1) &amp;= 2(5)^3 - 3(5) + 1 \\ &amp;= 2(125) - 15 + 1 \\ &amp;= 250 - 15 + 1 \\ &amp;= 236 \\end{aligned}[\/latex] As production approaches 500 units, the model predicts the profit will approach $236,000.[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">Polynomial functions are continuous everywhere, meaning the limit as [latex]x[\/latex] approaches any value equals the function value at that point. This means we can always use direct substitution to evaluate their limits !<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<p class=\"whitespace-normal break-words\">A company's quarterly revenue in thousands of dollars is modeled by [latex]R(x) = x^4 - 4x^3 + 5[\/latex], where [latex]x[\/latex] represents thousands of units sold. Find [latex]\\lim_{x \\to -1} R(x)[\/latex] to determine what the model predicts as sales approach -1 thousand units (though negative sales don't make physical sense, the mathematical limit exists).<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]\\lim_{x \\to -1} (x^4 - 4x^3 + 5) =[\/latex] [response area]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Correct answer:<\/strong> 10<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for correct answer:<\/strong> Excellent! You used direct substitution: [latex]-1^4 - 4(-1)^3 + 5 = 1 - 4(-1) + 5 = 1 + 4 + 5 = 10[\/latex] thousand dollars. Since this is a polynomial, the limit equals the function value at [latex]x = -1[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for incorrect answer:<\/strong> Since this is a polynomial function, you can find the limit by substituting [latex]x = -1[\/latex] directly into the expression. Calculate [latex]-1^4 - 4(-1)^3 + 5[\/latex], being careful with signs and exponents.<\/p>\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<p class=\"whitespace-normal break-words\">A company's average cost function (in dollars per unit) is [latex]C(x) = \\frac{x^2 - 6x + 8}{x - 2}[\/latex] for [latex]x[\/latex] hundred units produced. The function is undefined at [latex]x = 2[\/latex] due to a discontinuity in the cost data. Find [latex]\\lim_{x \\to 2} C(x)[\/latex] to determine what the average cost approaches as production nears 200 units.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Solution:<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">First, try direct substitution:<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]\\frac{(2)^2 - 6(2) + 8}{2 - 2} = \\frac{4 - 12 + 8}{0} = \\frac{0}{0}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This is indeterminate, so we need to factor and simplify:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">[latex]\\begin{aligned} \\lim_{x \\to 2} \\frac{x^2 - 6x + 8}{x - 2} &amp;= \\lim_{x \\to 2} \\frac{(x - 2)(x - 4)}{x - 2} &amp;&amp; \\text{factor the numerator} \\ &amp;= \\lim_{x \\to 2} \\frac{\\cancel{(x - 2)}(x - 4)}{\\cancel{x - 2}} &amp;&amp; \\text{cancel common factors} \\ &amp;= \\lim_{x \\to 2} (x - 4) &amp;&amp; \\text{simplify} \\ &amp;= 2 - 4 \\ &amp;= -2 \\end{aligned}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">As production approaches 200 units, the average cost approaches -$2 per unit. The negative value suggests the model breaks down near this production level, but mathematically the limit exists.<\/p>\r\n\r\n<\/section><section class=\"textbox recall\" aria-label=\"Recall\">When direct substitution gives [latex]\\frac{0}{0}[\/latex], this is called an indeterminate form. It doesn't mean the limit doesn't exist\u2014it means we need to simplify the expression algebraically first. The function [latex]\\frac{x^2 - 6x + 8}{x - 2}[\/latex] is equivalent to [latex]x - 4[\/latex] everywhere except at [latex]x = 2[\/latex] where it's undefined.<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<p class=\"whitespace-normal break-words\">A manufacturer's efficiency ratio is given by [latex]E(x) = \\frac{x^2 - 11x + 28}{7 - x}[\/latex] when producing [latex]x[\/latex] hundred units. Find [latex]\\lim_{x \\to 7} E(x)[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]\\lim_{x \\to 7} \\frac{x^2 - 11x + 28}{7 - x} =[\/latex] [response area]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Correct answer:<\/strong> -3<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for correct answer:<\/strong> Great work! You factored the numerator as [latex](x - 4)(x - 7)[\/latex], then noticed that [latex]7 - x = -(x - 7)[\/latex]. After factoring out -1 from the denominator, you could cancel [latex](x - 7)[\/latex] to get [latex]\\lim_{x \\to 7} -(x - 4) = -(7 - 4) = -3[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for incorrect answer:<\/strong> Direct substitution gives [latex]\\frac{0}{0}[\/latex], so factor first. The numerator factors as [latex](x - 4)(x - 7)[\/latex]. Notice that [latex]7 - x = -(x - 7)[\/latex]. Factor out the -1, then cancel common factors before evaluating.<\/p>\r\n\r\n<\/section><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find the limit of a sum, a difference, and a product.<\/li>\n<li>Find the limit of a polynomial.<\/li>\n<li>Find the limit of a power or a root.<\/li>\n<li>Find the limit of a quotient.<\/li>\n<\/ul>\n<\/section>\n<div data-test-render-count=\"1\">\n<div class=\"group relative pb-3\" data-is-streaming=\"false\">\n<div class=\"font-claude-response relative leading-[1.65rem] [&amp;_pre&gt;div]:bg-bg-000\/50 [&amp;_pre&gt;div]:border-0.5 [&amp;_pre&gt;div]:border-border-400 [&amp;_.ignore-pre-bg&gt;div]:bg-transparent [&amp;_.standard-markdown_:is(p,blockquote,h1,h2,h3,h4,h5,h6)]:pl-2 [&amp;_.standard-markdown_:is(p,blockquote,ul,ol,h1,h2,h3,h4,h5,h6)]:pr-8 [&amp;_.progressive-markdown_:is(p,blockquote,h1,h2,h3,h4,h5,h6)]:pl-2 [&amp;_.progressive-markdown_:is(p,blockquote,ul,ol,h1,h2,h3,h4,h5,h6)]:pr-8\">\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0 standard-markdown\">\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Analyzing Business Functions<\/h2>\n<p class=\"whitespace-normal break-words\">Business models often use polynomial and rational functions to predict profit, cost, and revenue. Understanding limits helps us analyze what happens as production levels change, even at points where functions have discontinuities.<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<p class=\"whitespace-normal break-words\"><strong>Question Help: Evaluating Limits Using Properties<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">For polynomial functions, use direct substitution: [latex]\\lim_{x \\to a} p(x) = p(a)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">For quotients that give [latex]\\frac{0}{0}[\/latex], try factoring and simplifying first<\/li>\n<li class=\"whitespace-normal break-words\">For roots in quotients, consider multiplying by a conjugate<\/li>\n<li class=\"whitespace-normal break-words\">Always try direct substitution first to see if it works<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\">A company&#8217;s monthly profit in thousands of dollars is modeled by [latex]P(x) = 2x^3 - 3x + 1[\/latex], where [latex]x[\/latex] represents the number of units produced (in hundreds). As production approaches 500 units ([latex]x[\/latex] approaches 5), what profit does the model predict?<\/p>\n<p class=\"whitespace-normal break-words\">Find [latex]\\lim_{x \\to 5} (2x^3 - 3x + 1)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q689696\">Show Solution<\/button><\/p>\n<div id=\"q689696\" class=\"hidden-answer\" style=\"display: none\">Since this is a polynomial function, we can use direct substitution: [latex]\\begin{aligned} \\lim_{x \\to 5} (2x^3 - 3x + 1) &= 2(5)^3 - 3(5) + 1 \\ &= 2(125) - 15 + 1 \\ &= 250 - 15 + 1 \\ &= 236 \\end{aligned}[\/latex] As production approaches 500 units, the model predicts the profit will approach $236,000.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Polynomial functions are continuous everywhere, meaning the limit as [latex]x[\/latex] approaches any value equals the function value at that point. This means we can always use direct substitution to evaluate their limits !<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<p class=\"whitespace-normal break-words\">A company&#8217;s quarterly revenue in thousands of dollars is modeled by [latex]R(x) = x^4 - 4x^3 + 5[\/latex], where [latex]x[\/latex] represents thousands of units sold. Find [latex]\\lim_{x \\to -1} R(x)[\/latex] to determine what the model predicts as sales approach -1 thousand units (though negative sales don&#8217;t make physical sense, the mathematical limit exists).<\/p>\n<p class=\"whitespace-normal break-words\">[latex]\\lim_{x \\to -1} (x^4 - 4x^3 + 5) =[\/latex] [response area]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Correct answer:<\/strong> 10<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for correct answer:<\/strong> Excellent! You used direct substitution: [latex]-1^4 - 4(-1)^3 + 5 = 1 - 4(-1) + 5 = 1 + 4 + 5 = 10[\/latex] thousand dollars. Since this is a polynomial, the limit equals the function value at [latex]x = -1[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for incorrect answer:<\/strong> Since this is a polynomial function, you can find the limit by substituting [latex]x = -1[\/latex] directly into the expression. Calculate [latex]-1^4 - 4(-1)^3 + 5[\/latex], being careful with signs and exponents.<\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<p class=\"whitespace-normal break-words\">A company&#8217;s average cost function (in dollars per unit) is [latex]C(x) = \\frac{x^2 - 6x + 8}{x - 2}[\/latex] for [latex]x[\/latex] hundred units produced. The function is undefined at [latex]x = 2[\/latex] due to a discontinuity in the cost data. Find [latex]\\lim_{x \\to 2} C(x)[\/latex] to determine what the average cost approaches as production nears 200 units.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Solution:<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">First, try direct substitution:<\/p>\n<p class=\"whitespace-normal break-words\">[latex]\\frac{(2)^2 - 6(2) + 8}{2 - 2} = \\frac{4 - 12 + 8}{0} = \\frac{0}{0}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This is indeterminate, so we need to factor and simplify:<\/p>\n<p class=\"whitespace-pre-wrap break-words\">[latex]\\begin{aligned} \\lim_{x \\to 2} \\frac{x^2 - 6x + 8}{x - 2} &= \\lim_{x \\to 2} \\frac{(x - 2)(x - 4)}{x - 2} && \\text{factor the numerator} \\ &= \\lim_{x \\to 2} \\frac{\\cancel{(x - 2)}(x - 4)}{\\cancel{x - 2}} && \\text{cancel common factors} \\ &= \\lim_{x \\to 2} (x - 4) && \\text{simplify} \\ &= 2 - 4 \\ &= -2 \\end{aligned}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">As production approaches 200 units, the average cost approaches -$2 per unit. The negative value suggests the model breaks down near this production level, but mathematically the limit exists.<\/p>\n<\/section>\n<section class=\"textbox recall\" aria-label=\"Recall\">When direct substitution gives [latex]\\frac{0}{0}[\/latex], this is called an indeterminate form. It doesn&#8217;t mean the limit doesn&#8217;t exist\u2014it means we need to simplify the expression algebraically first. The function [latex]\\frac{x^2 - 6x + 8}{x - 2}[\/latex] is equivalent to [latex]x - 4[\/latex] everywhere except at [latex]x = 2[\/latex] where it&#8217;s undefined.<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<p class=\"whitespace-normal break-words\">A manufacturer&#8217;s efficiency ratio is given by [latex]E(x) = \\frac{x^2 - 11x + 28}{7 - x}[\/latex] when producing [latex]x[\/latex] hundred units. Find [latex]\\lim_{x \\to 7} E(x)[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">[latex]\\lim_{x \\to 7} \\frac{x^2 - 11x + 28}{7 - x} =[\/latex] [response area]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Correct answer:<\/strong> -3<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for correct answer:<\/strong> Great work! You factored the numerator as [latex](x - 4)(x - 7)[\/latex], then noticed that [latex]7 - x = -(x - 7)[\/latex]. After factoring out -1 from the denominator, you could cancel [latex](x - 7)[\/latex] to get [latex]\\lim_{x \\to 7} -(x - 4) = -(7 - 4) = -3[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for incorrect answer:<\/strong> Direct substitution gives [latex]\\frac{0}{0}[\/latex], so factor first. The numerator factors as [latex](x - 4)(x - 7)[\/latex]. Notice that [latex]7 - x = -(x - 7)[\/latex]. Factor out the -1, then cancel common factors before evaluating.<\/p>\n<\/section>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":67,"menu_order":16,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":263,"module-header":"apply_it","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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3308"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/67"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3308\/revisions"}],"predecessor-version":[{"id":6013,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3308\/revisions\/6013"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/263"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3308\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=3308"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=3308"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=3308"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=3308"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}