{"id":3084,"date":"2024-06-12T16:47:14","date_gmt":"2024-06-12T16:47:14","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=3084"},"modified":"2024-08-05T02:24:57","modified_gmt":"2024-08-05T02:24:57","slug":"limits-at-infinity-and-asymptotes-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/limits-at-infinity-and-asymptotes-learn-it-4\/","title":{"raw":"Limits at Infinity and Asymptotes: Learn It 4","rendered":"Limits at Infinity and Asymptotes: Learn It 4"},"content":{"raw":"<h2>End Behavior Cont.<\/h2>\r\n<h3>End Behavior for Algebraic Functions<\/h3>\r\n<p id=\"fs-id1165042638493\">The end behavior for rational functions and functions involving radicals is a little more complicated than for polynomials.<\/p>\r\n<section class=\"textbox recall\">\r\n<p>Note that this is not your first encounter with horizontal asymptotes. It may be helpful to recall what you already know about them.<\/p>\r\n<p>The <strong>horizontal asymptote<\/strong> of a rational function can be determined by looking at the degrees of the numerator and denominator.<\/p>\r\n<ul id=\"fs-id1165137722720\">\r\n\t<li><strong>Case 1:<\/strong> Degree of numerator <em>is less than<\/em> degree of denominator: horizontal asymptote at\u00a0[latex]y=0[\/latex]<\/li>\r\n\t<li><strong>Case 2<\/strong>: Degree of numerator <em>is greater than degree of denominator by one<\/em>: no horizontal asymptote; slant asymptote.\r\n\r\n<ul>\r\n\t<li>If the degree of the numerator is greater than the degree of the denominator by\u00a0<em>more than one<\/em>, the end behavior of the function's graph will mimic that of the graph of the reduced ratio of leading terms.<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li><strong>Case 3<\/strong>: Degree of numerator <em>is equal to<\/em> degree of denominator: horizontal asymptote at ratio of leading coefficients.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<p>In the example below, we show that the limits at infinity of a rational function [latex]f(x)=\\frac{p(x)}{q(x)}[\/latex] depend on the relationship between the degree of the numerator and the degree of the denominator.<\/p>\r\n<p>To evaluate the limits at infinity for a rational function, we divide the numerator and denominator by the highest power of [latex]x[\/latex] appearing in the denominator. This determines which term in the overall expression dominates the behavior of the function at large values of [latex]x[\/latex].<\/p>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1165042638562\">For each of the following functions, determine the limits as [latex]x\\to \\infty [\/latex] and [latex]x\\to \u2212\\infty[\/latex]. Then, use this information to describe the end behavior of the function.<\/p>\r\n<ol id=\"fs-id1165043390828\" style=\"list-style-type: lower-alpha;\">\r\n\t<li>[latex]f(x)=\\frac{3x-1}{2x+5}[\/latex] (<em>Note:<\/em> The degree of the numerator and the denominator are the same.)<\/li>\r\n\t<li>[latex]f(x)=\\frac{3x^2+2x}{4x^3-5x+7}[\/latex] (<em>Note:<\/em> The degree of numerator is less than the degree of the denominator.)<\/li>\r\n\t<li>[latex]f(x)=\\frac{3x^2+4x}{x+2}[\/latex] (<em>Note:<\/em> The degree of numerator is greater than the degree of the denominator.)<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"fs-id1165042708379\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165042708379\"]<\/p>\r\n<ol id=\"fs-id1165042708379\" style=\"list-style-type: lower-alpha;\">\r\n\t<li>The highest power of [latex]x[\/latex] in the denominator is [latex]x[\/latex]. Therefore, dividing the numerator and denominator by [latex]x[\/latex] and applying the algebraic limit laws, we see that,<br \/>\r\n<div id=\"fs-id1165043281584\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to \\pm \\infty }{\\lim}\\frac{3x-1}{2x+5} &amp; =\\underset{x\\to \\pm \\infty }{\\lim}\\frac{3-1\/x}{2+5\/x} \\\\ &amp; =\\frac{\\underset{x\\to \\pm \\infty }{\\lim}(3-1\/x)}{\\underset{x\\to \\pm \\infty }{\\lim}(2+5\/x)} \\\\ &amp; =\\frac{\\underset{x\\to \\pm \\infty }{\\lim}3-\\underset{x\\to \\pm \\infty }{\\lim}1\/x}{\\underset{x\\to \\pm \\infty }{\\lim}2+\\underset{x\\to \\pm \\infty }{\\lim}5\/x} \\\\ &amp; =\\frac{3-0}{2+0}=\\frac{3}{2}. \\end{array}[\/latex]<\/div>\r\n<p>Since [latex]\\underset{x\\to \\pm \\infty }{\\lim}f(x)=\\frac{3}{2}[\/latex], we know that [latex]y=\\frac{3}{2}[\/latex] is a horizontal asymptote for this function as shown in the following graph.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"342\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211104\/CNX_Calc_Figure_04_06_007.jpg\" alt=\"The function f(x) = (3x + 1)\/(2x + 5) is plotted as is its horizontal asymptote at y = 3\/2.\" width=\"342\" height=\"347\" \/> Figure 14. The graph of this rational function approaches a horizontal asymptote as [latex]x\\to \\pm \\infty[\/latex].[\/caption]\r\n<\/li>\r\n\t<li>Since the largest power of [latex]x[\/latex] appearing in the denominator is [latex]x^3[\/latex], divide the numerator and denominator by [latex]x^3[\/latex]. After doing so and applying algebraic limit laws, we obtain,<br \/>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\pm \\infty }{\\lim}\\frac{3x^2+2x}{4x^3-5x+7}=\\underset{x\\to \\pm \\infty }{\\lim}\\frac{3\/x+2\/x^2}{4-5\/x^2+7\/x^3}=\\frac{3(0)+2(0)}{4-5(0)+7(0)}=0[\/latex]<\/div>\r\n<p>Therefore [latex]f[\/latex] has a horizontal asymptote of [latex]y=0[\/latex] as shown in the following graph.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"417\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211107\/CNX_Calc_Figure_04_06_008.jpg\" alt=\"The function f(x) = (3x2 + 2x)\/(4x2 \u2013 5x + 7) is plotted as is its horizontal asymptote at y = 0.\" width=\"417\" height=\"422\" \/> Figure 15. The graph of this rational function approaches the horizontal asymptote [latex]y=0[\/latex] as [latex]x\\to \\pm \\infty[\/latex].[\/caption]\r\n<\/li>\r\n\t<li>Dividing the numerator and denominator by [latex]x[\/latex], we have,<br \/>\r\n<div id=\"fs-id1165042333346\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\pm \\infty }{\\lim}\\frac{3x^2+4x}{x+2}=\\underset{x\\to \\pm \\infty }{\\lim}\\frac{3x+4}{1+2\/x}[\/latex].<\/div>\r\n<p>As [latex]x\\to \\pm \\infty[\/latex], the denominator approaches 1. As [latex]x\\to \\infty[\/latex], the numerator approaches [latex]+\\infty[\/latex]. As [latex]x\\to \u2212\\infty[\/latex], the numerator approaches [latex]\u2212\\infty[\/latex]. Therefore [latex]\\underset{x\\to \\infty }{\\lim}f(x)=\\infty[\/latex], whereas [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=\u2212\\infty [\/latex] as shown in the following figure.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"569\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211110\/CNX_Calc_Figure_04_06_027.jpg\" alt=\"The function f(x) = (3x2 + 4x)\/(x + 2) is plotted. It appears to have a diagonal asymptote as well as a vertical asymptote at x = \u22122.\" width=\"569\" height=\"497\" \/> Figure 16. As [latex]x\\to \\infty[\/latex], the values [latex]f(x)\\to \\infty[\/latex]. As [latex]x\\to \u2212\\infty[\/latex], the values [latex]f(x)\\to \u2212\\infty[\/latex].[\/caption]\r\n<\/li>\r\n<\/ol>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<p><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/0OVSQCWCzqc?controls=0&amp;start=984&amp;end=1278&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.6LimitsAtInfinityAndAsymptotes984to1278_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.6 Limits at Infinity and Asymptotes\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p id=\"fs-id1165042374905\">Before proceeding, consider the graph of [latex]f(x)=\\frac{(3x^2+4x)}{(x+2)}[\/latex] shown below.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"267\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211113\/CNX_Calc_Figure_04_06_009.jpg\" alt=\"The function f(x) = (3x2 + 4x)\/(x + 2) is plotted as is its diagonal asymptote y = 3x \u2013 2.\" width=\"267\" height=\"347\" \/> Figure 17. The graph of the rational function [latex]f(x)=(3x^2+4x)\/(x+2)[\/latex] approaches the oblique asymptote [latex]y=3x-2[\/latex] as [latex]x\\to \\pm \\infty[\/latex].[\/caption]\r\n\r\n<p>As [latex]x\\to \\infty [\/latex] and [latex]x\\to \u2212\\infty[\/latex], the graph of [latex]f[\/latex] appears almost linear. Although [latex]f[\/latex] is certainly not a linear function, we now investigate why the graph of [latex]f[\/latex] seems to be approaching a linear function.<\/p>\r\n<p>First, using long division of polynomials, we can write,<\/p>\r\n<div id=\"fs-id1165043219202\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\frac{3x^2+4x}{x+2}=3x-2+\\frac{4}{x+2}[\/latex]<\/div>\r\n<p id=\"fs-id1165043219268\">Since [latex]\\frac{4}{(x+2)}\\to 0[\/latex] as [latex]x\\to \\pm \\infty[\/latex], we conclude that,<\/p>\r\n<div id=\"fs-id1165042465555\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\pm \\infty }{\\lim}(f(x)-(3x-2))=\\underset{x\\to \\pm \\infty }{\\lim}\\frac{4}{x+2}=0[\/latex]<\/div>\r\n<p id=\"fs-id1165042465646\">Therefore, the graph of [latex]f[\/latex] approaches the line [latex]y=3x-2[\/latex] as [latex]x\\to \\pm \\infty[\/latex]. This line is known as an <strong>oblique asymptote<\/strong> for [latex]f[\/latex].<\/p>\r\n<p id=\"fs-id1165042461217\">We can summarize the results of the example above\u00a0to make the following conclusion regarding end behavior for rational functions.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>end behavior for rational functions<\/h3>\r\n<p>Consider a rational function<\/p>\r\n<div id=\"fs-id1165042461226\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\frac{p(x)}{q(x)}=\\frac{a_n x^n + a_{n-1} x^{n-1} + \\cdots + a_1 x + a_0}{b_m x^m + b_{m-1} x^{m-1} + \\cdots + b_1 x + b_0}[\/latex],<\/div>\r\n<p id=\"fs-id1165042315666\">where [latex]a_n\\ne 0[\/latex] and [latex]b_m \\ne 0[\/latex].<\/p>\r\n<ol id=\"fs-id1165043422338\">\r\n\t<li>If the degree of the numerator is the same as the degree of the denominator [latex](n=m)[\/latex], then [latex]f[\/latex] has a horizontal asymptote of [latex]y=a_n\/b_m[\/latex] as [latex]x\\to \\pm \\infty[\/latex].<\/li>\r\n\t<li>If the degree of the numerator is less than the degree of the denominator [latex](n &lt; m)[\/latex], then [latex]f[\/latex] has a horizontal asymptote of [latex]y=0[\/latex] as [latex]x\\to \\pm \\infty[\/latex].<\/li>\r\n\t<li>If the degree of the numerator is greater than the degree of the denominator [latex](n&gt;m)[\/latex], then [latex]f[\/latex] does not have a horizontal asymptote. The limits at infinity are either positive or negative infinity, depending on the signs of the leading terms.<\/li>\r\n<\/ol>\r\n<\/section>\r\n<ol id=\"fs-id1165043422338\"><\/ol>\r\n<ol id=\"fs-id1165043422338\"><\/ol>\r\n<p>In addition, using long division, the function can be rewritten as<\/p>\r\n<div id=\"fs-id1165043422484\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\frac{p(x)}{q(x)}=g(x)+\\frac{r(x)}{q(x)}[\/latex],<\/div>\r\n<p>where the degree of [latex]r(x)[\/latex] is less than the degree of [latex]q(x)[\/latex]. As a result, [latex]\\underset{x\\to \\pm \\infty }{\\lim}r(x)\/q(x)=0[\/latex].<\/p>\r\n<p>Therefore, the values of [latex][f(x)-g(x)][\/latex] approach zero as [latex]x\\to \\pm \\infty[\/latex].<\/p>\r\n<p>If the degree of [latex]p(x)[\/latex] is exactly one more than the degree of [latex]q(x)[\/latex] [latex](n=m+1)[\/latex], the function [latex]g(x)[\/latex] is a linear function. In this case, we call [latex]g(x)[\/latex] an oblique asymptote.<\/p>\r\n<section class=\"textbox example\">\r\n<p>Find the limits as [latex]x\\to \\infty [\/latex] and [latex]x\\to \u2212\\infty [\/latex] for [latex]f(x)=\\frac{3x-2}{\\sqrt{4x^2+5}}[\/latex] and describe the end behavior of [latex]f[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1165042631905\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165042631905\"]<\/p>\r\n<p id=\"fs-id1165042631905\">Let\u2019s use the same strategy as we did for rational functions: divide the numerator and denominator by a power of [latex]x[\/latex]. To determine the appropriate power of [latex]x[\/latex], consider the expression [latex]\\sqrt{4x^2+5}[\/latex] in the denominator. Since,<\/p>\r\n<div id=\"fs-id1165042418061\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sqrt{4x^2+5}\\approx \\sqrt{4x^2}=2|x|[\/latex]<\/div>\r\n<p id=\"fs-id1165042418106\">for large values of [latex]x[\/latex] in effect [latex]x[\/latex] appears just to the first power in the denominator.<\/p>\r\n<p>Therefore, we divide the numerator and denominator by [latex]|x|[\/latex].<\/p>\r\n<p>Then, using the fact that [latex]|x|=x[\/latex] for [latex]x&gt;0[\/latex], [latex]|x|=\u2212x[\/latex] for [latex]x&lt;0[\/latex], and [latex]|x|=\\sqrt{x^2}[\/latex] for all [latex]x[\/latex], we calculate the limits as follows:<\/p>\r\n<div id=\"fs-id1165042418216\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lll} \\underset{x\\to \\infty }{\\lim}\\frac{3x-2}{\\sqrt{4x^2+5}} &amp; = &amp; \\underset{x\\to \\infty }{\\lim}\\frac{(1\/|x|)(3x-2)}{(1\/|x|)\\sqrt{4x^2+5}} \\\\ &amp; = &amp; \\underset{x\\to \\infty }{\\lim}\\frac{(1\/x)(3x-2)}{\\sqrt{(1\/x^2)(4x^2+5)}} \\\\ &amp; = &amp; \\underset{x\\to \\infty }{\\lim}\\frac{3-2\/x}{\\sqrt{4+5\/x^2}}=\\frac{3}{\\sqrt{4}}=\\frac{3}{2} \\\\ \\underset{x\\to \u2212\\infty }{\\lim}\\frac{3x-2}{\\sqrt{4x^2+5}} &amp; = &amp; \\underset{x\\to \u2212\\infty }{\\lim}\\frac{(1\/|x|)(3x-2)}{(1\/|x|)\\sqrt{4x^2+5}} \\\\ &amp; = &amp; \\underset{x\\to \u2212\\infty }{\\lim}\\frac{(-1\/x)(3x-2)}{\\sqrt{(1\/x^2)(4x^2+5)}} \\\\ &amp; = &amp; \\underset{x\\to \u2212\\infty }{\\lim}\\frac{-3+2\/x}{\\sqrt{4+5\/x^2}}=\\frac{-3}{\\sqrt{4}}=\\frac{-3}{2}. \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165042463819\">Therefore, [latex]f(x)[\/latex] approaches the horizontal asymptote [latex]y=\\frac{3}{2}[\/latex] as [latex]x\\to \\infty [\/latex] and the horizontal asymptote [latex]y=-\\frac{3}{2}[\/latex] as [latex]x\\to \u2212\\infty [\/latex] as shown in the following graph.<\/p>\r\n<p>&nbsp;<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"592\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211116\/CNX_Calc_Figure_04_06_010.jpg\" alt=\"The function f(x) = (3x \u2212 2)\/(the square root of the quantity (4x2 + 5)) is plotted. It has two horizontal asymptotes at y = \u00b13\/2, and it crosses y = \u22123\/2 before converging toward it from below.\" width=\"592\" height=\"197\" \/> Figure 18. This function has two horizontal asymptotes and it crosses one of the asymptotes.[\/caption]\r\n\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<h2>End Behavior Cont.<\/h2>\n<h3>End Behavior for Algebraic Functions<\/h3>\n<p id=\"fs-id1165042638493\">The end behavior for rational functions and functions involving radicals is a little more complicated than for polynomials.<\/p>\n<section class=\"textbox recall\">\n<p>Note that this is not your first encounter with horizontal asymptotes. It may be helpful to recall what you already know about them.<\/p>\n<p>The <strong>horizontal asymptote<\/strong> of a rational function can be determined by looking at the degrees of the numerator and denominator.<\/p>\n<ul id=\"fs-id1165137722720\">\n<li><strong>Case 1:<\/strong> Degree of numerator <em>is less than<\/em> degree of denominator: horizontal asymptote at\u00a0[latex]y=0[\/latex]<\/li>\n<li><strong>Case 2<\/strong>: Degree of numerator <em>is greater than degree of denominator by one<\/em>: no horizontal asymptote; slant asymptote.\n<ul>\n<li>If the degree of the numerator is greater than the degree of the denominator by\u00a0<em>more than one<\/em>, the end behavior of the function&#8217;s graph will mimic that of the graph of the reduced ratio of leading terms.<\/li>\n<\/ul>\n<\/li>\n<li><strong>Case 3<\/strong>: Degree of numerator <em>is equal to<\/em> degree of denominator: horizontal asymptote at ratio of leading coefficients.<\/li>\n<\/ul>\n<\/section>\n<p>In the example below, we show that the limits at infinity of a rational function [latex]f(x)=\\frac{p(x)}{q(x)}[\/latex] depend on the relationship between the degree of the numerator and the degree of the denominator.<\/p>\n<p>To evaluate the limits at infinity for a rational function, we divide the numerator and denominator by the highest power of [latex]x[\/latex] appearing in the denominator. This determines which term in the overall expression dominates the behavior of the function at large values of [latex]x[\/latex].<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1165042638562\">For each of the following functions, determine the limits as [latex]x\\to \\infty[\/latex] and [latex]x\\to \u2212\\infty[\/latex]. Then, use this information to describe the end behavior of the function.<\/p>\n<ol id=\"fs-id1165043390828\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]f(x)=\\frac{3x-1}{2x+5}[\/latex] (<em>Note:<\/em> The degree of the numerator and the denominator are the same.)<\/li>\n<li>[latex]f(x)=\\frac{3x^2+2x}{4x^3-5x+7}[\/latex] (<em>Note:<\/em> The degree of numerator is less than the degree of the denominator.)<\/li>\n<li>[latex]f(x)=\\frac{3x^2+4x}{x+2}[\/latex] (<em>Note:<\/em> The degree of numerator is greater than the degree of the denominator.)<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165042708379\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165042708379\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165042708379\" style=\"list-style-type: lower-alpha;\">\n<li>The highest power of [latex]x[\/latex] in the denominator is [latex]x[\/latex]. Therefore, dividing the numerator and denominator by [latex]x[\/latex] and applying the algebraic limit laws, we see that,\n<div id=\"fs-id1165043281584\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to \\pm \\infty }{\\lim}\\frac{3x-1}{2x+5} & =\\underset{x\\to \\pm \\infty }{\\lim}\\frac{3-1\/x}{2+5\/x} \\\\ & =\\frac{\\underset{x\\to \\pm \\infty }{\\lim}(3-1\/x)}{\\underset{x\\to \\pm \\infty }{\\lim}(2+5\/x)} \\\\ & =\\frac{\\underset{x\\to \\pm \\infty }{\\lim}3-\\underset{x\\to \\pm \\infty }{\\lim}1\/x}{\\underset{x\\to \\pm \\infty }{\\lim}2+\\underset{x\\to \\pm \\infty }{\\lim}5\/x} \\\\ & =\\frac{3-0}{2+0}=\\frac{3}{2}. \\end{array}[\/latex]<\/div>\n<p>Since [latex]\\underset{x\\to \\pm \\infty }{\\lim}f(x)=\\frac{3}{2}[\/latex], we know that [latex]y=\\frac{3}{2}[\/latex] is a horizontal asymptote for this function as shown in the following graph.<\/p>\n<figure style=\"width: 342px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211104\/CNX_Calc_Figure_04_06_007.jpg\" alt=\"The function f(x) = (3x + 1)\/(2x + 5) is plotted as is its horizontal asymptote at y = 3\/2.\" width=\"342\" height=\"347\" \/><figcaption class=\"wp-caption-text\">Figure 14. The graph of this rational function approaches a horizontal asymptote as [latex]x\\to \\pm \\infty[\/latex].<\/figcaption><\/figure>\n<\/li>\n<li>Since the largest power of [latex]x[\/latex] appearing in the denominator is [latex]x^3[\/latex], divide the numerator and denominator by [latex]x^3[\/latex]. After doing so and applying algebraic limit laws, we obtain,\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\pm \\infty }{\\lim}\\frac{3x^2+2x}{4x^3-5x+7}=\\underset{x\\to \\pm \\infty }{\\lim}\\frac{3\/x+2\/x^2}{4-5\/x^2+7\/x^3}=\\frac{3(0)+2(0)}{4-5(0)+7(0)}=0[\/latex]<\/div>\n<p>Therefore [latex]f[\/latex] has a horizontal asymptote of [latex]y=0[\/latex] as shown in the following graph.<\/p>\n<figure style=\"width: 417px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211107\/CNX_Calc_Figure_04_06_008.jpg\" alt=\"The function f(x) = (3x2 + 2x)\/(4x2 \u2013 5x + 7) is plotted as is its horizontal asymptote at y = 0.\" width=\"417\" height=\"422\" \/><figcaption class=\"wp-caption-text\">Figure 15. The graph of this rational function approaches the horizontal asymptote [latex]y=0[\/latex] as [latex]x\\to \\pm \\infty[\/latex].<\/figcaption><\/figure>\n<\/li>\n<li>Dividing the numerator and denominator by [latex]x[\/latex], we have,\n<div id=\"fs-id1165042333346\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\pm \\infty }{\\lim}\\frac{3x^2+4x}{x+2}=\\underset{x\\to \\pm \\infty }{\\lim}\\frac{3x+4}{1+2\/x}[\/latex].<\/div>\n<p>As [latex]x\\to \\pm \\infty[\/latex], the denominator approaches 1. As [latex]x\\to \\infty[\/latex], the numerator approaches [latex]+\\infty[\/latex]. As [latex]x\\to \u2212\\infty[\/latex], the numerator approaches [latex]\u2212\\infty[\/latex]. Therefore [latex]\\underset{x\\to \\infty }{\\lim}f(x)=\\infty[\/latex], whereas [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=\u2212\\infty[\/latex] as shown in the following figure.<\/p>\n<figure style=\"width: 569px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211110\/CNX_Calc_Figure_04_06_027.jpg\" alt=\"The function f(x) = (3x2 + 4x)\/(x + 2) is plotted. It appears to have a diagonal asymptote as well as a vertical asymptote at x = \u22122.\" width=\"569\" height=\"497\" \/><figcaption class=\"wp-caption-text\">Figure 16. As [latex]x\\to \\infty[\/latex], the values [latex]f(x)\\to \\infty[\/latex]. As [latex]x\\to \u2212\\infty[\/latex], the values [latex]f(x)\\to \u2212\\infty[\/latex].<\/figcaption><\/figure>\n<\/li>\n<\/ol>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/0OVSQCWCzqc?controls=0&amp;start=984&amp;end=1278&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.6LimitsAtInfinityAndAsymptotes984to1278_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.6 Limits at Infinity and Asymptotes&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1165042374905\">Before proceeding, consider the graph of [latex]f(x)=\\frac{(3x^2+4x)}{(x+2)}[\/latex] shown below.<\/p>\n<figure style=\"width: 267px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211113\/CNX_Calc_Figure_04_06_009.jpg\" alt=\"The function f(x) = (3x2 + 4x)\/(x + 2) is plotted as is its diagonal asymptote y = 3x \u2013 2.\" width=\"267\" height=\"347\" \/><figcaption class=\"wp-caption-text\">Figure 17. The graph of the rational function [latex]f(x)=(3x^2+4x)\/(x+2)[\/latex] approaches the oblique asymptote [latex]y=3x-2[\/latex] as [latex]x\\to \\pm \\infty[\/latex].<\/figcaption><\/figure>\n<p>As [latex]x\\to \\infty[\/latex] and [latex]x\\to \u2212\\infty[\/latex], the graph of [latex]f[\/latex] appears almost linear. Although [latex]f[\/latex] is certainly not a linear function, we now investigate why the graph of [latex]f[\/latex] seems to be approaching a linear function.<\/p>\n<p>First, using long division of polynomials, we can write,<\/p>\n<div id=\"fs-id1165043219202\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\frac{3x^2+4x}{x+2}=3x-2+\\frac{4}{x+2}[\/latex]<\/div>\n<p id=\"fs-id1165043219268\">Since [latex]\\frac{4}{(x+2)}\\to 0[\/latex] as [latex]x\\to \\pm \\infty[\/latex], we conclude that,<\/p>\n<div id=\"fs-id1165042465555\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\pm \\infty }{\\lim}(f(x)-(3x-2))=\\underset{x\\to \\pm \\infty }{\\lim}\\frac{4}{x+2}=0[\/latex]<\/div>\n<p id=\"fs-id1165042465646\">Therefore, the graph of [latex]f[\/latex] approaches the line [latex]y=3x-2[\/latex] as [latex]x\\to \\pm \\infty[\/latex]. This line is known as an <strong>oblique asymptote<\/strong> for [latex]f[\/latex].<\/p>\n<p id=\"fs-id1165042461217\">We can summarize the results of the example above\u00a0to make the following conclusion regarding end behavior for rational functions.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>end behavior for rational functions<\/h3>\n<p>Consider a rational function<\/p>\n<div id=\"fs-id1165042461226\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\frac{p(x)}{q(x)}=\\frac{a_n x^n + a_{n-1} x^{n-1} + \\cdots + a_1 x + a_0}{b_m x^m + b_{m-1} x^{m-1} + \\cdots + b_1 x + b_0}[\/latex],<\/div>\n<p id=\"fs-id1165042315666\">where [latex]a_n\\ne 0[\/latex] and [latex]b_m \\ne 0[\/latex].<\/p>\n<ol id=\"fs-id1165043422338\">\n<li>If the degree of the numerator is the same as the degree of the denominator [latex](n=m)[\/latex], then [latex]f[\/latex] has a horizontal asymptote of [latex]y=a_n\/b_m[\/latex] as [latex]x\\to \\pm \\infty[\/latex].<\/li>\n<li>If the degree of the numerator is less than the degree of the denominator [latex](n < m)[\/latex], then [latex]f[\/latex] has a horizontal asymptote of [latex]y=0[\/latex] as [latex]x\\to \\pm \\infty[\/latex].<\/li>\n<li>If the degree of the numerator is greater than the degree of the denominator [latex](n>m)[\/latex], then [latex]f[\/latex] does not have a horizontal asymptote. The limits at infinity are either positive or negative infinity, depending on the signs of the leading terms.<\/li>\n<\/ol>\n<\/section>\n<ol><\/ol>\n<ol><\/ol>\n<p>In addition, using long division, the function can be rewritten as<\/p>\n<div id=\"fs-id1165043422484\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\frac{p(x)}{q(x)}=g(x)+\\frac{r(x)}{q(x)}[\/latex],<\/div>\n<p>where the degree of [latex]r(x)[\/latex] is less than the degree of [latex]q(x)[\/latex]. As a result, [latex]\\underset{x\\to \\pm \\infty }{\\lim}r(x)\/q(x)=0[\/latex].<\/p>\n<p>Therefore, the values of [latex][f(x)-g(x)][\/latex] approach zero as [latex]x\\to \\pm \\infty[\/latex].<\/p>\n<p>If the degree of [latex]p(x)[\/latex] is exactly one more than the degree of [latex]q(x)[\/latex] [latex](n=m+1)[\/latex], the function [latex]g(x)[\/latex] is a linear function. In this case, we call [latex]g(x)[\/latex] an oblique asymptote.<\/p>\n<section class=\"textbox example\">\n<p>Find the limits as [latex]x\\to \\infty[\/latex] and [latex]x\\to \u2212\\infty[\/latex] for [latex]f(x)=\\frac{3x-2}{\\sqrt{4x^2+5}}[\/latex] and describe the end behavior of [latex]f[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165042631905\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165042631905\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042631905\">Let\u2019s use the same strategy as we did for rational functions: divide the numerator and denominator by a power of [latex]x[\/latex]. To determine the appropriate power of [latex]x[\/latex], consider the expression [latex]\\sqrt{4x^2+5}[\/latex] in the denominator. Since,<\/p>\n<div id=\"fs-id1165042418061\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sqrt{4x^2+5}\\approx \\sqrt{4x^2}=2|x|[\/latex]<\/div>\n<p id=\"fs-id1165042418106\">for large values of [latex]x[\/latex] in effect [latex]x[\/latex] appears just to the first power in the denominator.<\/p>\n<p>Therefore, we divide the numerator and denominator by [latex]|x|[\/latex].<\/p>\n<p>Then, using the fact that [latex]|x|=x[\/latex] for [latex]x>0[\/latex], [latex]|x|=\u2212x[\/latex] for [latex]x<0[\/latex], and [latex]|x|=\\sqrt{x^2}[\/latex] for all [latex]x[\/latex], we calculate the limits as follows:<\/p>\n<div id=\"fs-id1165042418216\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lll} \\underset{x\\to \\infty }{\\lim}\\frac{3x-2}{\\sqrt{4x^2+5}} & = & \\underset{x\\to \\infty }{\\lim}\\frac{(1\/|x|)(3x-2)}{(1\/|x|)\\sqrt{4x^2+5}} \\\\ & = & \\underset{x\\to \\infty }{\\lim}\\frac{(1\/x)(3x-2)}{\\sqrt{(1\/x^2)(4x^2+5)}} \\\\ & = & \\underset{x\\to \\infty }{\\lim}\\frac{3-2\/x}{\\sqrt{4+5\/x^2}}=\\frac{3}{\\sqrt{4}}=\\frac{3}{2} \\\\ \\underset{x\\to \u2212\\infty }{\\lim}\\frac{3x-2}{\\sqrt{4x^2+5}} & = & \\underset{x\\to \u2212\\infty }{\\lim}\\frac{(1\/|x|)(3x-2)}{(1\/|x|)\\sqrt{4x^2+5}} \\\\ & = & \\underset{x\\to \u2212\\infty }{\\lim}\\frac{(-1\/x)(3x-2)}{\\sqrt{(1\/x^2)(4x^2+5)}} \\\\ & = & \\underset{x\\to \u2212\\infty }{\\lim}\\frac{-3+2\/x}{\\sqrt{4+5\/x^2}}=\\frac{-3}{\\sqrt{4}}=\\frac{-3}{2}. \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165042463819\">Therefore, [latex]f(x)[\/latex] approaches the horizontal asymptote [latex]y=\\frac{3}{2}[\/latex] as [latex]x\\to \\infty[\/latex] and the horizontal asymptote [latex]y=-\\frac{3}{2}[\/latex] as [latex]x\\to \u2212\\infty[\/latex] as shown in the following graph.<\/p>\n<p>&nbsp;<\/p>\n<figure style=\"width: 592px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211116\/CNX_Calc_Figure_04_06_010.jpg\" alt=\"The function f(x) = (3x \u2212 2)\/(the square root of the quantity (4x2 + 5)) is plotted. It has two horizontal asymptotes at y = \u00b13\/2, and it crosses y = \u22123\/2 before converging toward it from below.\" width=\"592\" height=\"197\" \/><figcaption class=\"wp-caption-text\">Figure 18. This function has two horizontal asymptotes and it crosses one of the asymptotes.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":8,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":292,"module-header":"learn_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\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3084"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3084\/revisions"}],"predecessor-version":[{"id":4143,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3084\/revisions\/4143"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/292"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3084\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=3084"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=3084"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=3084"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=3084"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}