{"id":320,"date":"2023-09-20T22:49:04","date_gmt":"2023-09-20T22:49:04","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/applying-lhopitals-rule\/"},"modified":"2024-09-11T05:39:12","modified_gmt":"2024-09-11T05:39:12","slug":"lhopitals-rule-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/lhopitals-rule-learn-it-1\/","title":{"raw":"L\u2019H\u00f4pital\u2019s Rule: Learn It 1","rendered":"L\u2019H\u00f4pital\u2019s Rule: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Spot indeterminate forms like in calculations, and use L\u2019H\u00f4pital\u2019s rule to find precise values<\/li>\r\n\t<li>Explain how quickly different functions increase or decrease compared to each other<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>L\u2019H\u00f4pital\u2019s Rule<\/h2>\r\n<p><strong>L'H\u00f4pital's Rule<\/strong> is a powerful technique for evaluating limits of functions. It leverages derivatives to determine limits that are otherwise challenging to compute directly, providing a clear way to resolve indeterminate forms.<\/p>\r\n<h3>Applying L\u2019H\u00f4pital\u2019s Rule<\/h3>\r\n<p id=\"fs-id1165042941863\">L'H\u00f4pital's Rule comes into play when you're dealing with limits that result in indeterminate forms.<\/p>\r\n<p>Consider the quotient of two functions, represented as:<\/p>\r\n<div id=\"fs-id1165042458008\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}[\/latex].<\/div>\r\n<p id=\"fs-id1165042613148\">If [latex]\\underset{x\\to a}{\\lim}f(x)=L_1[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=L_2 \\ne 0[\/latex], then<\/p>\r\n<div id=\"fs-id1165043088102\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\dfrac{L_1}{L_2}[\/latex]<\/div>\r\n<p id=\"fs-id1165042330308\">However, what happens if [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex]?<\/p>\r\n<p>We call this one of the <strong>indeterminate forms<\/strong>, of type [latex]\\frac{0}{0}[\/latex]. This is considered an indeterminate form because we cannot determine the exact behavior of [latex]\\frac{f(x)}{g(x)}[\/latex] as [latex]x\\to a[\/latex] without further analysis. We have seen examples of this earlier in the text.<\/p>\r\n<section class=\"textbox example\">\r\n<ul>\r\n\t<li>[latex]\\underset{x\\to 2}{\\lim}\\dfrac{x^2-4}{x-2}[\/latex] can be solved by factoring the numerator and simplifying.<center>[latex]\\underset{x\\to 2}{\\lim}\\dfrac{x^2-4}{x-2}=\\underset{x\\to 2}{\\lim}\\dfrac{(x+2)(x-2)}{x-2}=\\underset{x\\to 2}{\\lim}(x+2)=2+2=4[\/latex]<\/center><\/li>\r\n\t<li>[latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x}{x}[\/latex] has been proven geometrically to equal [latex]1[\/latex]. Using L'H\u00f4pital's Rule, differentiating both numerator and denominator yields:<center>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{\\sin x}{x}=1[\/latex]<\/center><\/li>\r\n<\/ul>\r\n<\/section>\r\n<p id=\"fs-id1165043001194\">L'H\u00f4pital's Rule not only simplifies the calculation of certain limits but also provides insights into evaluating complex limits that would be difficult to handle by other methods. This rule is particularly useful in analyzing the behavior of functions as they approach critical points.<\/p>\r\n<h4 style=\"text-align: left;\">L\u2019H\u00f4pital\u2019s Rule (0\/0 Case)<\/h4>\r\n<p id=\"fs-id1165043062373\">The idea behind L\u2019H\u00f4pital\u2019s rule can be explained using local linear approximations.<\/p>\r\n<section class=\"textbox example\">\r\n<p>Consider two differentiable functions [latex]f[\/latex] and [latex]g[\/latex] such that [latex]\\underset{x\\to a}{\\lim}f(x)=0=\\underset{x\\to a}{\\lim}g(x)[\/latex] and such that [latex]g^{\\prime}(a)\\ne 0[\/latex] For [latex]x[\/latex] near [latex]a[\/latex], we can write<\/p>\r\n<div id=\"fs-id1165043199940\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)\\approx f(a)+f^{\\prime}(a)(x-a)[\/latex]<\/div>\r\n<p id=\"fs-id1165042355300\">and<\/p>\r\n<div id=\"fs-id1165042320393\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g(x)\\approx g(a)+g^{\\prime}(a)(x-a)[\/latex].<\/div>\r\n<p id=\"fs-id1165043178271\">Therefore,<\/p>\r\n<div id=\"fs-id1165042331440\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(x)}{g(x)}\\approx \\dfrac{f(a)+f^{\\prime}(a)(x-a)}{g(a)+g^{\\prime}(a)(x-a)}[\/latex]<\/div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"618\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211304\/CNX_Calc_Figure_04_08_003.jpg\" alt=\"Two functions y = f(x) and y = g(x) are drawn such that they cross at a point above x = a. The linear approximations of these two functions y = f(a) + f\u2019(a)(x \u2013 a) and y = g(a) + g\u2019(a)(x \u2013 a) are also drawn.\" width=\"618\" height=\"390\" \/> Figure 1. If [latex]\\underset{x\\to a}{\\lim}f(x)=\\underset{x\\to a}{\\lim}g(x)[\/latex], then the ratio [latex]f(x)\/g(x)[\/latex] is approximately equal to the ratio of their linear approximations near [latex]a[\/latex].[\/caption]\r\n\r\n<p>Since [latex]f[\/latex] is differentiable at [latex]a[\/latex], then [latex]f[\/latex] is continuous at [latex]a[\/latex], and therefore [latex]f(a)=\\underset{x\\to a}{\\lim}f(x)=0[\/latex]. Similarly, [latex]g(a)=\\underset{x\\to a}{\\lim}g(x)=0[\/latex].\u00a0 If we also assume that [latex]f^{\\prime}[\/latex] and [latex]g^{\\prime}[\/latex] are continuous at [latex]x=a[\/latex], then [latex]f^{\\prime}(a)=\\underset{x\\to a}{\\lim}f^{\\prime}(x)[\/latex] and [latex]g^{\\prime}(a)=\\underset{x\\to a}{\\lim}g^{\\prime}(x)[\/latex].<\/p>\r\n<p>Using these ideas, we conclude that:<\/p>\r\n<div id=\"fs-id1165042373953\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)(x-a)}{g^{\\prime}(x)(x-a)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex]<span style=\"background-color: #f4f3ef;\">\u00a0<\/span><\/div>\r\n\r\nNote that the assumption that [latex]f^{\\prime}[\/latex] and [latex]g^{\\prime}[\/latex] are continuous at [latex]a[\/latex] and [latex]g^{\\prime}(a)\\ne 0[\/latex] can be loosened.<\/section>\r\n<section class=\"textbox proTip\">\r\n<p>The notation [latex]\\frac{0}{0}[\/latex] does not mean we are actually dividing zero by zero. Rather, we are using the notation [latex]\\frac{0}{0}[\/latex] to represent a quotient of limits, each of which is zero.<\/p>\r\n<\/section>\r\n<p>We state L\u2019H\u00f4pital\u2019s rule formally for the indeterminate form [latex]\\frac{0}{0}[\/latex].\u00a0<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">L\u2019H\u00f4pital\u2019s rule (0\/0 case)<\/h3>\r\n<p id=\"fs-id1165043276140\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval containing [latex]a[\/latex], except possibly at [latex]a[\/latex]. If [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex], then<\/p>\r\n<div id=\"fs-id1165043020148\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex],<\/div>\r\n<p id=\"fs-id1165043035673\">assuming the limit on the right exists or is [latex]\\infty [\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if we are considering one-sided limits, or if [latex]a=\\infty[\/latex] or [latex]-\\infty[\/latex].<\/p>\r\n<\/section>\r\n<section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n<hr \/>\r\n<p id=\"fs-id1165043429940\">We provide a proof of this theorem in the special case when [latex]f, \\, g, \\, f^{\\prime}[\/latex], and [latex]g^{\\prime}[\/latex] are all continuous over an open interval containing [latex]a[\/latex]. In that case, since [latex]\\underset{x\\to a}{\\lim}f(x)=0=\\underset{x\\to a}{\\lim}g(x)[\/latex] and [latex]f[\/latex] and [latex]g[\/latex] are continuous at [latex]a[\/latex], it follows that [latex]f(a)=0=g(a)[\/latex]. Therefore,<\/p>\r\n<div id=\"fs-id1165043353664\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} \\underset{x\\to a}{\\lim}\\frac{f(x)}{g(x)} &amp; =\\underset{x\\to a}{\\lim}\\frac{f(x)-f(a)}{g(x)-g(a)} &amp; &amp; &amp; \\text{since} \\, f(a)=0=g(a) \\\\ &amp; =\\underset{x\\to a}{\\lim}\\frac{\\frac{f(x)-f(a)}{x-a}}{\\frac{g(x)-g(a)}{x-a}} &amp; &amp; &amp; \\text{multiplying numerator and denominator by} \\, \\frac{1}{x-a} \\\\ &amp; =\\frac{\\underset{x\\to a}{\\lim}\\frac{f(x)-f(a)}{x-a}}{\\underset{x\\to a}{\\lim}\\frac{g(x)-g(a)}{x-a}} &amp; &amp; &amp; \\text{limit of a quotient} \\\\ &amp; =\\frac{f^{\\prime}(a)}{g^{\\prime}(a)} &amp; &amp; &amp; \\text{definition of the derivative} \\\\ &amp; =\\frac{\\underset{x\\to a}{\\lim}f^{\\prime}(x)}{\\underset{x\\to a}{\\lim}g^{\\prime}(x)} &amp; &amp; &amp; \\text{continuity of} \\, f^{\\prime} \\, \\text{and} \\, g^{\\prime} \\\\ &amp; =\\underset{x\\to a}{\\lim}\\frac{f^{\\prime}(x)}{g^{\\prime}(x)} &amp; &amp; &amp; \\text{limit of a quotient} \\end{array}[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1165043092431\">Note that L\u2019H\u00f4pital\u2019s rule states we can calculate the limit of a quotient [latex]\\frac{f}{g}[\/latex] by considering the limit of the quotient of the derivatives [latex]\\frac{f^{\\prime}}{g^{\\prime}}[\/latex]. It is important to realize that we are not calculating the derivative of the quotient [latex]\\frac{f}{g}[\/latex].<\/p>\r\n<p id=\"fs-id1165042988256\">[latex]_\\blacksquare[\/latex]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1165042456913\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n\t<li>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{1- \\cos x}{x}[\/latex]<\/li>\r\n\t<li>[latex]\\underset{x\\to 1}{\\lim}\\dfrac{\\sin (\\pi x)}{\\ln x}[\/latex]<\/li>\r\n\t<li>[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{e^{\\frac{1}{x}}-1}{\\frac{1}{x}}[\/latex]<\/li>\r\n\t<li>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{\\sin x-x}{x^2}[\/latex]<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"fs-id1165042535045\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165042535045\"]<\/p>\r\n<ol id=\"fs-id1165042535045\" style=\"list-style-type: lower-alpha;\">\r\n\t<li>Since the numerator [latex]1- \\cos x\\to 0[\/latex] and the denominator [latex]x\\to 0[\/latex], we can apply L\u2019H\u00f4pital\u2019s rule to evaluate this limit. We have<br \/>\r\n<div id=\"fs-id1165042328696\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to 0}{\\lim}\\frac{1- \\cos x}{x} &amp; =\\underset{x\\to 0}{\\lim}\\frac{\\frac{d}{dx}(1- \\cos x)}{\\frac{d}{dx}(x)} \\\\ &amp; =\\underset{x\\to 0}{\\lim}\\frac{\\sin x}{1} \\\\ &amp; =\\frac{\\underset{x\\to 0}{\\lim}(\\sin x)}{\\underset{x\\to 0}{\\lim}(1)} \\\\ &amp; =\\frac{0}{1}=0 \\end{array}[\/latex]<\/div>\r\n<\/li>\r\n\t<li>As [latex]x\\to 1[\/latex], the numerator [latex]\\sin (\\pi x)\\to 0[\/latex] and the denominator [latex]\\ln x \\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain<br \/>\r\n<div id=\"fs-id1165043116372\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to 1}{\\lim}\\frac{\\sin (\\pi x)}{\\ln x} &amp; =\\underset{x\\to 1}{\\lim}\\frac{\\pi \\cos (\\pi x)}{1\/x} \\\\ &amp; =\\underset{x\\to 1}{\\lim}(\\pi x) \\cos (\\pi x) \\\\ &amp; =(\\pi \\cdot 1)(-1)=\u2212\\pi \\end{array}[\/latex]<\/div>\r\n<\/li>\r\n\t<li>As [latex]x\\to \\infty[\/latex], the numerator [latex]e^{1\/x}-1\\to 0[\/latex] and the denominator [latex](\\frac{1}{x})\\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain<br \/>\r\n<div id=\"fs-id1165042327460\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\frac{e^{1\/x}-1}{\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\frac{e^{1\/x}(\\frac{-1}{x^2})}{(\\frac{-1}{x^2})}=\\underset{x\\to \\infty}{\\lim} e^{1\/x}=e^0=1[\/latex]<\/div>\r\n<\/li>\r\n\t<li>As [latex]x\\to 0[\/latex], both the numerator and denominator approach zero. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain<br \/>\r\n<div id=\"fs-id1165042562974\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x-x}{x^2}=\\underset{x\\to 0}{\\lim}\\frac{\\cos x-1}{2x}[\/latex].<\/div>\r\n<p>Since the numerator and denominator of this new quotient both approach zero as [latex]x\\to 0[\/latex], we apply L\u2019H\u00f4pital\u2019s rule again. In doing so, we see that<\/p>\r\n<div id=\"fs-id1165043281524\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\cos x-1}{2x}=\\underset{x\\to 0}{\\lim}\\frac{\u2212\\sin x}{2}=0[\/latex].<\/div>\r\n<p>Therefore, we conclude that<\/p>\r\n<div id=\"fs-id1165043284142\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x-x}{x^2}=0[\/latex].<\/div>\r\n<\/li>\r\n<\/ol>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/e58sGIZe1wU?controls=0&amp;start=535&amp;end=587&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\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\/4.8LHopitalsRule49to346_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.8 L'Hopital's Rule\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Spot indeterminate forms like in calculations, and use L\u2019H\u00f4pital\u2019s rule to find precise values<\/li>\n<li>Explain how quickly different functions increase or decrease compared to each other<\/li>\n<\/ul>\n<\/section>\n<h2>L\u2019H\u00f4pital\u2019s Rule<\/h2>\n<p><strong>L&#8217;H\u00f4pital&#8217;s Rule<\/strong> is a powerful technique for evaluating limits of functions. It leverages derivatives to determine limits that are otherwise challenging to compute directly, providing a clear way to resolve indeterminate forms.<\/p>\n<h3>Applying L\u2019H\u00f4pital\u2019s Rule<\/h3>\n<p id=\"fs-id1165042941863\">L&#8217;H\u00f4pital&#8217;s Rule comes into play when you&#8217;re dealing with limits that result in indeterminate forms.<\/p>\n<p>Consider the quotient of two functions, represented as:<\/p>\n<div id=\"fs-id1165042458008\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}[\/latex].<\/div>\n<p id=\"fs-id1165042613148\">If [latex]\\underset{x\\to a}{\\lim}f(x)=L_1[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=L_2 \\ne 0[\/latex], then<\/p>\n<div id=\"fs-id1165043088102\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\dfrac{L_1}{L_2}[\/latex]<\/div>\n<p id=\"fs-id1165042330308\">However, what happens if [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex]?<\/p>\n<p>We call this one of the <strong>indeterminate forms<\/strong>, of type [latex]\\frac{0}{0}[\/latex]. This is considered an indeterminate form because we cannot determine the exact behavior of [latex]\\frac{f(x)}{g(x)}[\/latex] as [latex]x\\to a[\/latex] without further analysis. We have seen examples of this earlier in the text.<\/p>\n<section class=\"textbox example\">\n<ul>\n<li>[latex]\\underset{x\\to 2}{\\lim}\\dfrac{x^2-4}{x-2}[\/latex] can be solved by factoring the numerator and simplifying.\n<div style=\"text-align: center;\">[latex]\\underset{x\\to 2}{\\lim}\\dfrac{x^2-4}{x-2}=\\underset{x\\to 2}{\\lim}\\dfrac{(x+2)(x-2)}{x-2}=\\underset{x\\to 2}{\\lim}(x+2)=2+2=4[\/latex]<\/div>\n<\/li>\n<li>[latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x}{x}[\/latex] has been proven geometrically to equal [latex]1[\/latex]. Using L&#8217;H\u00f4pital&#8217;s Rule, differentiating both numerator and denominator yields:\n<div style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\dfrac{\\sin x}{x}=1[\/latex]<\/div>\n<\/li>\n<\/ul>\n<\/section>\n<p id=\"fs-id1165043001194\">L&#8217;H\u00f4pital&#8217;s Rule not only simplifies the calculation of certain limits but also provides insights into evaluating complex limits that would be difficult to handle by other methods. This rule is particularly useful in analyzing the behavior of functions as they approach critical points.<\/p>\n<h4 style=\"text-align: left;\">L\u2019H\u00f4pital\u2019s Rule (0\/0 Case)<\/h4>\n<p id=\"fs-id1165043062373\">The idea behind L\u2019H\u00f4pital\u2019s rule can be explained using local linear approximations.<\/p>\n<section class=\"textbox example\">\n<p>Consider two differentiable functions [latex]f[\/latex] and [latex]g[\/latex] such that [latex]\\underset{x\\to a}{\\lim}f(x)=0=\\underset{x\\to a}{\\lim}g(x)[\/latex] and such that [latex]g^{\\prime}(a)\\ne 0[\/latex] For [latex]x[\/latex] near [latex]a[\/latex], we can write<\/p>\n<div id=\"fs-id1165043199940\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)\\approx f(a)+f^{\\prime}(a)(x-a)[\/latex]<\/div>\n<p id=\"fs-id1165042355300\">and<\/p>\n<div id=\"fs-id1165042320393\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g(x)\\approx g(a)+g^{\\prime}(a)(x-a)[\/latex].<\/div>\n<p id=\"fs-id1165043178271\">Therefore,<\/p>\n<div id=\"fs-id1165042331440\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(x)}{g(x)}\\approx \\dfrac{f(a)+f^{\\prime}(a)(x-a)}{g(a)+g^{\\prime}(a)(x-a)}[\/latex]<\/div>\n<figure style=\"width: 618px\" 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\/11211304\/CNX_Calc_Figure_04_08_003.jpg\" alt=\"Two functions y = f(x) and y = g(x) are drawn such that they cross at a point above x = a. The linear approximations of these two functions y = f(a) + f\u2019(a)(x \u2013 a) and y = g(a) + g\u2019(a)(x \u2013 a) are also drawn.\" width=\"618\" height=\"390\" \/><figcaption class=\"wp-caption-text\">Figure 1. If [latex]\\underset{x\\to a}{\\lim}f(x)=\\underset{x\\to a}{\\lim}g(x)[\/latex], then the ratio [latex]f(x)\/g(x)[\/latex] is approximately equal to the ratio of their linear approximations near [latex]a[\/latex].<\/figcaption><\/figure>\n<p>Since [latex]f[\/latex] is differentiable at [latex]a[\/latex], then [latex]f[\/latex] is continuous at [latex]a[\/latex], and therefore [latex]f(a)=\\underset{x\\to a}{\\lim}f(x)=0[\/latex]. Similarly, [latex]g(a)=\\underset{x\\to a}{\\lim}g(x)=0[\/latex].\u00a0 If we also assume that [latex]f^{\\prime}[\/latex] and [latex]g^{\\prime}[\/latex] are continuous at [latex]x=a[\/latex], then [latex]f^{\\prime}(a)=\\underset{x\\to a}{\\lim}f^{\\prime}(x)[\/latex] and [latex]g^{\\prime}(a)=\\underset{x\\to a}{\\lim}g^{\\prime}(x)[\/latex].<\/p>\n<p>Using these ideas, we conclude that:<\/p>\n<div id=\"fs-id1165042373953\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)(x-a)}{g^{\\prime}(x)(x-a)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex]<span style=\"background-color: #f4f3ef;\">\u00a0<\/span><\/div>\n<p>Note that the assumption that [latex]f^{\\prime}[\/latex] and [latex]g^{\\prime}[\/latex] are continuous at [latex]a[\/latex] and [latex]g^{\\prime}(a)\\ne 0[\/latex] can be loosened.<\/section>\n<section class=\"textbox proTip\">\n<p>The notation [latex]\\frac{0}{0}[\/latex] does not mean we are actually dividing zero by zero. Rather, we are using the notation [latex]\\frac{0}{0}[\/latex] to represent a quotient of limits, each of which is zero.<\/p>\n<\/section>\n<p>We state L\u2019H\u00f4pital\u2019s rule formally for the indeterminate form [latex]\\frac{0}{0}[\/latex].\u00a0<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">L\u2019H\u00f4pital\u2019s rule (0\/0 case)<\/h3>\n<p id=\"fs-id1165043276140\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval containing [latex]a[\/latex], except possibly at [latex]a[\/latex]. If [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex], then<\/p>\n<div id=\"fs-id1165043020148\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex],<\/div>\n<p id=\"fs-id1165043035673\">assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if we are considering one-sided limits, or if [latex]a=\\infty[\/latex] or [latex]-\\infty[\/latex].<\/p>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<p id=\"fs-id1165043429940\">We provide a proof of this theorem in the special case when [latex]f, \\, g, \\, f^{\\prime}[\/latex], and [latex]g^{\\prime}[\/latex] are all continuous over an open interval containing [latex]a[\/latex]. In that case, since [latex]\\underset{x\\to a}{\\lim}f(x)=0=\\underset{x\\to a}{\\lim}g(x)[\/latex] and [latex]f[\/latex] and [latex]g[\/latex] are continuous at [latex]a[\/latex], it follows that [latex]f(a)=0=g(a)[\/latex]. Therefore,<\/p>\n<div id=\"fs-id1165043353664\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} \\underset{x\\to a}{\\lim}\\frac{f(x)}{g(x)} & =\\underset{x\\to a}{\\lim}\\frac{f(x)-f(a)}{g(x)-g(a)} & & & \\text{since} \\, f(a)=0=g(a) \\\\ & =\\underset{x\\to a}{\\lim}\\frac{\\frac{f(x)-f(a)}{x-a}}{\\frac{g(x)-g(a)}{x-a}} & & & \\text{multiplying numerator and denominator by} \\, \\frac{1}{x-a} \\\\ & =\\frac{\\underset{x\\to a}{\\lim}\\frac{f(x)-f(a)}{x-a}}{\\underset{x\\to a}{\\lim}\\frac{g(x)-g(a)}{x-a}} & & & \\text{limit of a quotient} \\\\ & =\\frac{f^{\\prime}(a)}{g^{\\prime}(a)} & & & \\text{definition of the derivative} \\\\ & =\\frac{\\underset{x\\to a}{\\lim}f^{\\prime}(x)}{\\underset{x\\to a}{\\lim}g^{\\prime}(x)} & & & \\text{continuity of} \\, f^{\\prime} \\, \\text{and} \\, g^{\\prime} \\\\ & =\\underset{x\\to a}{\\lim}\\frac{f^{\\prime}(x)}{g^{\\prime}(x)} & & & \\text{limit of a quotient} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043092431\">Note that L\u2019H\u00f4pital\u2019s rule states we can calculate the limit of a quotient [latex]\\frac{f}{g}[\/latex] by considering the limit of the quotient of the derivatives [latex]\\frac{f^{\\prime}}{g^{\\prime}}[\/latex]. It is important to realize that we are not calculating the derivative of the quotient [latex]\\frac{f}{g}[\/latex].<\/p>\n<p id=\"fs-id1165042988256\">[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1165042456913\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{1- \\cos x}{x}[\/latex]<\/li>\n<li>[latex]\\underset{x\\to 1}{\\lim}\\dfrac{\\sin (\\pi x)}{\\ln x}[\/latex]<\/li>\n<li>[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{e^{\\frac{1}{x}}-1}{\\frac{1}{x}}[\/latex]<\/li>\n<li>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{\\sin x-x}{x^2}[\/latex]<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165042535045\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165042535045\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165042535045\" style=\"list-style-type: lower-alpha;\">\n<li>Since the numerator [latex]1- \\cos x\\to 0[\/latex] and the denominator [latex]x\\to 0[\/latex], we can apply L\u2019H\u00f4pital\u2019s rule to evaluate this limit. We have\n<div id=\"fs-id1165042328696\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to 0}{\\lim}\\frac{1- \\cos x}{x} & =\\underset{x\\to 0}{\\lim}\\frac{\\frac{d}{dx}(1- \\cos x)}{\\frac{d}{dx}(x)} \\\\ & =\\underset{x\\to 0}{\\lim}\\frac{\\sin x}{1} \\\\ & =\\frac{\\underset{x\\to 0}{\\lim}(\\sin x)}{\\underset{x\\to 0}{\\lim}(1)} \\\\ & =\\frac{0}{1}=0 \\end{array}[\/latex]<\/div>\n<\/li>\n<li>As [latex]x\\to 1[\/latex], the numerator [latex]\\sin (\\pi x)\\to 0[\/latex] and the denominator [latex]\\ln x \\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\n<div id=\"fs-id1165043116372\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to 1}{\\lim}\\frac{\\sin (\\pi x)}{\\ln x} & =\\underset{x\\to 1}{\\lim}\\frac{\\pi \\cos (\\pi x)}{1\/x} \\\\ & =\\underset{x\\to 1}{\\lim}(\\pi x) \\cos (\\pi x) \\\\ & =(\\pi \\cdot 1)(-1)=\u2212\\pi \\end{array}[\/latex]<\/div>\n<\/li>\n<li>As [latex]x\\to \\infty[\/latex], the numerator [latex]e^{1\/x}-1\\to 0[\/latex] and the denominator [latex](\\frac{1}{x})\\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\n<div id=\"fs-id1165042327460\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\frac{e^{1\/x}-1}{\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\frac{e^{1\/x}(\\frac{-1}{x^2})}{(\\frac{-1}{x^2})}=\\underset{x\\to \\infty}{\\lim} e^{1\/x}=e^0=1[\/latex]<\/div>\n<\/li>\n<li>As [latex]x\\to 0[\/latex], both the numerator and denominator approach zero. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\n<div id=\"fs-id1165042562974\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x-x}{x^2}=\\underset{x\\to 0}{\\lim}\\frac{\\cos x-1}{2x}[\/latex].<\/div>\n<p>Since the numerator and denominator of this new quotient both approach zero as [latex]x\\to 0[\/latex], we apply L\u2019H\u00f4pital\u2019s rule again. In doing so, we see that<\/p>\n<div id=\"fs-id1165043281524\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\cos x-1}{2x}=\\underset{x\\to 0}{\\lim}\\frac{\u2212\\sin x}{2}=0[\/latex].<\/div>\n<p>Therefore, we conclude that<\/p>\n<div id=\"fs-id1165043284142\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x-x}{x^2}=0[\/latex].<\/div>\n<\/li>\n<\/ol>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/e58sGIZe1wU?controls=0&amp;start=535&amp;end=587&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\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\/4.8LHopitalsRule49to346_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.8 L&#8217;Hopital&#8217;s Rule&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":17,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.8 L\\'Hopital\\'s Rule\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","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":null,"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\/320"}],"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\/6"}],"version-history":[{"count":9,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/320\/revisions"}],"predecessor-version":[{"id":4856,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/320\/revisions\/4856"}],"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\/320\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=320"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=320"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=320"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=320"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}