{"id":413,"date":"2021-02-04T02:04:30","date_gmt":"2021-02-04T02:04:30","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=413"},"modified":"2022-03-16T05:49:54","modified_gmt":"2022-03-16T05:49:54","slug":"applying-lhopitals-rule","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/applying-lhopitals-rule\/","title":{"raw":"Applying L\u2019H\u00f4pital\u2019s Rule","rendered":"Applying L\u2019H\u00f4pital\u2019s Rule"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Recognize when to apply L\u2019H\u00f4pital\u2019s rule<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165043085155\" class=\"bc-section section\">\r\n<h2>Applying L\u2019H\u00f4pital\u2019s Rule<\/h2>\r\n<p id=\"fs-id1165042941863\">L\u2019H\u00f4pital\u2019s rule can be used to evaluate limits involving the quotient of two functions. Consider<\/p>\r\n\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&nbsp;\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\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&nbsp;\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]? 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. For example, consider<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 2}{\\lim}\\dfrac{x^2-4}{x-2}[\/latex] and [latex]\\underset{x\\to 0}{\\lim}\\dfrac{ \\sin x}{x}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043036412\">For the first of these examples, we can evaluate the limit by factoring the numerator and writing<\/p>\r\n\r\n<div id=\"fs-id1165043067705\" class=\"equation unnumbered\" 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>\r\nFor [latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x}{x}[\/latex] we were able to show, using a geometric argument, that\r\n<div id=\"fs-id1165042954700\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\dfrac{\\sin x}{x}=1[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043001194\">Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could not calculate previously.<\/p>\r\n<p id=\"fs-id1165043062373\">The idea behind L\u2019H\u00f4pital\u2019s rule can be explained using local linear approximations. 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\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&nbsp;\r\n<p id=\"fs-id1165042355300\">and<\/p>\r\n\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&nbsp;\r\n<p id=\"fs-id1165043178271\">Therefore,<\/p>\r\n\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&nbsp;\r\n<div><\/div>\r\n<div>[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]<\/div>\r\nSince [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]. 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]. Using these ideas, we conclude that\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]<\/div>\r\n<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1165043085155\" class=\"bc-section section\">\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. We state L\u2019H\u00f4pital\u2019s rule formally for the indeterminate form [latex]\\frac{0}{0}[\/latex]. Also note that 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.\r\n<div id=\"fs-id1165043352593\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">L\u2019H\u00f4pital\u2019s Rule (0\/0 Case)<\/h3>\r\n\r\n<hr \/>\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\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&nbsp;\r\n<div><\/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\r\n<\/div>\r\n<div id=\"fs-id1165042712928\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\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\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&nbsp;\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\r\n<div id=\"fs-id1165043397578\" class=\"textbook exercises\">\r\n<h3>Example: Applying L\u2019H\u00f4pital\u2019s Rule (0\/0 Case)<\/h3>\r\n<p id=\"fs-id1165042456913\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\r\n\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<div id=\"fs-id1165043104016\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"fs-id1165042535045\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042535045\"]\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\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><\/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\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><\/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\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><\/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\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\nSince 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\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\nTherefore, we conclude that\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><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Applying L\u2019H\u00f4pital\u2019s Rule (0\/0 Case).\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/e58sGIZe1wU?controls=0&amp;start=49&amp;end=346&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]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.\r\n\r\nYou 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]\r\n<div class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\underset{x\\to 0}{\\lim}\\dfrac{x}{\\tan x}[\/latex].\r\n\r\n[reveal-answer q=\"37002811\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"37002811\"]\r\n\r\n[latex]\\frac{d}{dx} \\tan x= \\sec ^2 x[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042377480\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042377480\"]\r\n\r\n[latex]1[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165042318672\">We can also use L\u2019H\u00f4pital\u2019s rule to evaluate limits of quotients [latex]\\frac{f(x)}{g(x)}[\/latex] in which [latex]f(x)\\to \\pm \\infty [\/latex] and [latex]g(x)\\to \\pm \\infty[\/latex]. Limits of this form are classified as <em>indeterminate forms of type<\/em> [latex]\\infty \/ \\infty[\/latex]. Again, note that we are not actually dividing [latex]\\infty[\/latex] by [latex]\\infty[\/latex]. Since [latex]\\infty[\/latex] is not a real number, that is impossible; rather, [latex]\\infty \/ \\infty[\/latex] is used to represent a quotient of limits, each of which is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165042330826\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/ \\infty[\/latex] Case)<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165043426174\">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]. Suppose [latex]\\underset{x\\to a}{\\lim}f(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]) and [latex]\\underset{x\\to a}{\\lim}g(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]). Then,<\/p>\r\n\r\n<div id=\"fs-id1165042373703\" 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&nbsp;\r\n<div><\/div>\r\n<p id=\"fs-id1165042707280\">assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if the limit is infinite, if [latex]a=\\infty[\/latex] or [latex]\u2212\\infty[\/latex], or the limit is one-sided.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043427623\" class=\"textbook exercises\">\r\n<h3>Example: Applying L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/\\infty[\/latex] Case)<\/h3>\r\n<p id=\"fs-id1165042709794\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\r\n\r\n<ol id=\"fs-id1165042709798\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1165043427625\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"fs-id1165042376758\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042376758\"]\r\n<ol id=\"fs-id1165042376758\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Since [latex]3x+5[\/latex] and [latex]2x+1[\/latex] are first-degree polynomials with positive leading coefficients, [latex]\\underset{x\\to \\infty }{\\lim}(3x+5)=\\infty [\/latex] and [latex]\\underset{x\\to \\infty }{\\lim}(2x+1)=\\infty[\/latex]. Therefore, we apply L\u2019H\u00f4pital\u2019s rule and obtain\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim}\\dfrac{3x+5}{2x+\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3}{2}=\\dfrac{3}{2}[\/latex].<\/div>\r\nNote that this limit can also be calculated without invoking L\u2019H\u00f4pital\u2019s rule. Earlier in the module we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of [latex]x[\/latex] in the denominator. In doing so, we saw that\r\n<div id=\"fs-id1165042319986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3+\\frac{5}{x}}{2+\\frac{1}{x}}=\\dfrac{3}{2}[\/latex].<\/div>\r\nL\u2019H\u00f4pital\u2019s rule provides us with an alternative means of evaluating this type of limit.<\/li>\r\n \t<li>Here, [latex]\\underset{x\\to 0^+}{\\lim} \\ln x=\u2212\\infty [\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} \\cot x=\\infty[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule and obtain\r\n<div id=\"fs-id1165042374803\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\frac{1}{x}}{\u2212\\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}[\/latex].<\/div>\r\nNow as [latex]x\\to 0^+[\/latex], [latex]\\csc^2 x\\to \\infty[\/latex]. Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of [latex]\\csc x[\/latex] to write\r\n<div id=\"fs-id1165042376466\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}[\/latex].<\/div>\r\nNow [latex]\\underset{x\\to 0^+}{\\lim} \\sin^2 x=0[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} x=0[\/latex], so we apply L\u2019H\u00f4pital\u2019s rule again. We find\r\n<div id=\"fs-id1165043249678\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{2 \\sin x \\cos x}{-1}=\\dfrac{0}{-1}=0[\/latex].<\/div>\r\nWe conclude that\r\n<div id=\"fs-id1165042315721\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=0[\/latex].<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042333392\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{\\ln x}{5x}[\/latex]\r\n\r\n[reveal-answer q=\"30011179\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"30011179\"]\r\n\r\n[latex]\\frac{d}{dx}\\ln x=\\frac{1}{x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042367881\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042367881\"]\r\n\r\n0\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165042320327\">As mentioned, L\u2019H\u00f4pital\u2019s rule is an extremely useful tool for evaluating limits. It is important to remember, however, that to apply L\u2019H\u00f4pital\u2019s rule to a quotient [latex]\\frac{f(x)}{g(x)}[\/latex], it is essential that the limit of [latex]\\frac{f(x)}{g(x)}[\/latex] be of the form [latex]0\/0[\/latex] or [latex]\\infty \/ \\infty[\/latex] Consider the following example.<\/p>\r\n\r\n<div id=\"fs-id1165042350228\" class=\"textbook exercises\">\r\n<h3>Example: When L\u2019H\u00f4pital\u2019s Rule Does Not Apply<\/h3>\r\nConsider [latex]\\underset{x\\to 1}{\\lim}\\dfrac{x^2+5}{3x+4}[\/latex]. Show that the limit cannot be evaluated by applying L\u2019H\u00f4pital\u2019s rule.\r\n<div id=\"fs-id1165043219076\" class=\"exercise\">[reveal-answer q=\"fs-id1165042327372\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042327372\"]\r\n<p id=\"fs-id1165042327372\">Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply L\u2019H\u00f4pital\u2019s rule. If we try to do so, we get<\/p>\r\n\r\n<div id=\"fs-id1165042398961\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^2+5)=2x[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043395079\">and<\/p>\r\n\r\n<div id=\"fs-id1165042318693\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(3x+4)=3[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042364614\">At which point we would conclude erroneously that<\/p>\r\n\r\n<div id=\"fs-id1165042364618\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1}{\\lim}\\frac{x^2+5}{3x+4}=\\underset{x\\to 1}{\\lim}\\frac{2x}{3}=\\frac{2}{3}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043222028\">However, since [latex]\\underset{x\\to 1}{\\lim}(x^2+5)=6[\/latex] and [latex]\\underset{x\\to 1}{\\lim}(3x+4)=7[\/latex], we actually have<\/p>\r\n\r\n<div id=\"fs-id1165042970462\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1}{\\lim}\\frac{x^2+5}{3x+4}=\\frac{6}{7}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042383150\">We can conclude that<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1}{\\lim}\\frac{x^2+5}{3x+4}\\ne \\underset{x\\to 1}{\\lim}\\frac{\\frac{d}{dx}(x^2+5)}{\\frac{d}{dx}(3x+4)}[\/latex].[\/hidden-answer]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165043174646\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nExplain why we cannot apply L\u2019H\u00f4pital\u2019s rule to evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\cos x}{x}[\/latex]. Evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\cos x}{x}[\/latex] by other means.\r\n\r\n[reveal-answer q=\"6688909\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"6688909\"]\r\n\r\nDetermine the limits of the numerator and denominator separately.\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042632518\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042632518\"]\r\n\r\n[latex]\\underset{x\\to 0^+}{\\lim} \\cos x=1[\/latex]. Therefore, we cannot apply L\u2019H\u00f4pital\u2019s rule. The limit of the quotient is [latex]\\infty [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\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\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]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.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.8LHopitalsRule535to587_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]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]209328[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Recognize when to apply L\u2019H\u00f4pital\u2019s rule<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165043085155\" class=\"bc-section section\">\n<h2>Applying L\u2019H\u00f4pital\u2019s Rule<\/h2>\n<p id=\"fs-id1165042941863\">L\u2019H\u00f4pital\u2019s rule can be used to evaluate limits involving the quotient of two functions. Consider<\/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>&nbsp;<\/p>\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>&nbsp;<\/p>\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]? 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. For example, consider<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 2}{\\lim}\\dfrac{x^2-4}{x-2}[\/latex] and [latex]\\underset{x\\to 0}{\\lim}\\dfrac{ \\sin x}{x}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043036412\">For the first of these examples, we can evaluate the limit by factoring the numerator and writing<\/p>\n<div id=\"fs-id1165043067705\" class=\"equation unnumbered\" 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<p>For [latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x}{x}[\/latex] we were able to show, using a geometric argument, that<\/p>\n<div id=\"fs-id1165042954700\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\dfrac{\\sin x}{x}=1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043001194\">Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could not calculate previously.<\/p>\n<p id=\"fs-id1165043062373\">The idea behind L\u2019H\u00f4pital\u2019s rule can be explained using local linear approximations. 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>&nbsp;<\/p>\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>&nbsp;<\/p>\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<p>&nbsp;<\/p>\n<div><\/div>\n<div>\n<div style=\"width: 628px\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\n<\/div>\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]. 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]. 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]<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1165043085155\" class=\"bc-section section\">\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. We state L\u2019H\u00f4pital\u2019s rule formally for the indeterminate form [latex]\\frac{0}{0}[\/latex]. Also note that 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<div id=\"fs-id1165043352593\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">L\u2019H\u00f4pital\u2019s Rule (0\/0 Case)<\/h3>\n<hr \/>\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>&nbsp;<\/p>\n<div><\/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<\/div>\n<div id=\"fs-id1165042712928\" class=\"bc-section section\">\n<h3>Proof<\/h3>\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<div id=\"fs-id1165043397578\" class=\"textbook exercises\">\n<h3>Example: Applying L\u2019H\u00f4pital\u2019s Rule (0\/0 Case)<\/h3>\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<div id=\"fs-id1165043104016\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042535045\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Applying L\u2019H\u00f4pital\u2019s Rule (0\/0 Case).<\/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=49&amp;end=346&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">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<div class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\underset{x\\to 0}{\\lim}\\dfrac{x}{\\tan x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q37002811\">Hint<\/span><\/p>\n<div id=\"q37002811\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{d}{dx} \\tan x= \\sec ^2 x[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042377480\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042377480\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]1[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165042318672\">We can also use L\u2019H\u00f4pital\u2019s rule to evaluate limits of quotients [latex]\\frac{f(x)}{g(x)}[\/latex] in which [latex]f(x)\\to \\pm \\infty[\/latex] and [latex]g(x)\\to \\pm \\infty[\/latex]. Limits of this form are classified as <em>indeterminate forms of type<\/em> [latex]\\infty \/ \\infty[\/latex]. Again, note that we are not actually dividing [latex]\\infty[\/latex] by [latex]\\infty[\/latex]. Since [latex]\\infty[\/latex] is not a real number, that is impossible; rather, [latex]\\infty \/ \\infty[\/latex] is used to represent a quotient of limits, each of which is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex].<\/p>\n<div id=\"fs-id1165042330826\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/ \\infty[\/latex] Case)<\/h3>\n<hr \/>\n<p id=\"fs-id1165043426174\">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]. Suppose [latex]\\underset{x\\to a}{\\lim}f(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]) and [latex]\\underset{x\\to a}{\\lim}g(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]). Then,<\/p>\n<div id=\"fs-id1165042373703\" 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>&nbsp;<\/p>\n<div><\/div>\n<p id=\"fs-id1165042707280\">assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if the limit is infinite, if [latex]a=\\infty[\/latex] or [latex]\u2212\\infty[\/latex], or the limit is one-sided.<\/p>\n<\/div>\n<div id=\"fs-id1165043427623\" class=\"textbook exercises\">\n<h3>Example: Applying L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/\\infty[\/latex] Case)<\/h3>\n<p id=\"fs-id1165042709794\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\n<ol id=\"fs-id1165042709798\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}[\/latex]<\/li>\n<li>[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1165043427625\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042376758\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042376758\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165042376758\" style=\"list-style-type: lower-alpha;\">\n<li>Since [latex]3x+5[\/latex] and [latex]2x+1[\/latex] are first-degree polynomials with positive leading coefficients, [latex]\\underset{x\\to \\infty }{\\lim}(3x+5)=\\infty[\/latex] and [latex]\\underset{x\\to \\infty }{\\lim}(2x+1)=\\infty[\/latex]. Therefore, we apply L\u2019H\u00f4pital\u2019s rule and obtain\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim}\\dfrac{3x+5}{2x+\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3}{2}=\\dfrac{3}{2}[\/latex].<\/div>\n<p>Note that this limit can also be calculated without invoking L\u2019H\u00f4pital\u2019s rule. Earlier in the module we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of [latex]x[\/latex] in the denominator. In doing so, we saw that<\/p>\n<div id=\"fs-id1165042319986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3+\\frac{5}{x}}{2+\\frac{1}{x}}=\\dfrac{3}{2}[\/latex].<\/div>\n<p>L\u2019H\u00f4pital\u2019s rule provides us with an alternative means of evaluating this type of limit.<\/li>\n<li>Here, [latex]\\underset{x\\to 0^+}{\\lim} \\ln x=\u2212\\infty[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} \\cot x=\\infty[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule and obtain\n<div id=\"fs-id1165042374803\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\frac{1}{x}}{\u2212\\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}[\/latex].<\/div>\n<p>Now as [latex]x\\to 0^+[\/latex], [latex]\\csc^2 x\\to \\infty[\/latex]. Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of [latex]\\csc x[\/latex] to write<\/p>\n<div id=\"fs-id1165042376466\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}[\/latex].<\/div>\n<p>Now [latex]\\underset{x\\to 0^+}{\\lim} \\sin^2 x=0[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} x=0[\/latex], so we apply L\u2019H\u00f4pital\u2019s rule again. We find<\/p>\n<div id=\"fs-id1165043249678\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{2 \\sin x \\cos x}{-1}=\\dfrac{0}{-1}=0[\/latex].<\/div>\n<p>We conclude that<\/p>\n<div id=\"fs-id1165042315721\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=0[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042333392\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{\\ln x}{5x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q30011179\">Hint<\/span><\/p>\n<div id=\"q30011179\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{d}{dx}\\ln x=\\frac{1}{x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042367881\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042367881\" class=\"hidden-answer\" style=\"display: none\">\n<p>0\n<\/p><\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165042320327\">As mentioned, L\u2019H\u00f4pital\u2019s rule is an extremely useful tool for evaluating limits. It is important to remember, however, that to apply L\u2019H\u00f4pital\u2019s rule to a quotient [latex]\\frac{f(x)}{g(x)}[\/latex], it is essential that the limit of [latex]\\frac{f(x)}{g(x)}[\/latex] be of the form [latex]0\/0[\/latex] or [latex]\\infty \/ \\infty[\/latex] Consider the following example.<\/p>\n<div id=\"fs-id1165042350228\" class=\"textbook exercises\">\n<h3>Example: When L\u2019H\u00f4pital\u2019s Rule Does Not Apply<\/h3>\n<p>Consider [latex]\\underset{x\\to 1}{\\lim}\\dfrac{x^2+5}{3x+4}[\/latex]. Show that the limit cannot be evaluated by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\n<div id=\"fs-id1165043219076\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042327372\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042327372\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042327372\">Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply L\u2019H\u00f4pital\u2019s rule. If we try to do so, we get<\/p>\n<div id=\"fs-id1165042398961\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^2+5)=2x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043395079\">and<\/p>\n<div id=\"fs-id1165042318693\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(3x+4)=3[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042364614\">At which point we would conclude erroneously that<\/p>\n<div id=\"fs-id1165042364618\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1}{\\lim}\\frac{x^2+5}{3x+4}=\\underset{x\\to 1}{\\lim}\\frac{2x}{3}=\\frac{2}{3}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043222028\">However, since [latex]\\underset{x\\to 1}{\\lim}(x^2+5)=6[\/latex] and [latex]\\underset{x\\to 1}{\\lim}(3x+4)=7[\/latex], we actually have<\/p>\n<div id=\"fs-id1165042970462\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1}{\\lim}\\frac{x^2+5}{3x+4}=\\frac{6}{7}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042383150\">We can conclude that<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1}{\\lim}\\frac{x^2+5}{3x+4}\\ne \\underset{x\\to 1}{\\lim}\\frac{\\frac{d}{dx}(x^2+5)}{\\frac{d}{dx}(3x+4)}[\/latex].<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043174646\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Explain why we cannot apply L\u2019H\u00f4pital\u2019s rule to evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\cos x}{x}[\/latex]. Evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\cos x}{x}[\/latex] by other means.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q6688909\">Hint<\/span><\/p>\n<div id=\"q6688909\" class=\"hidden-answer\" style=\"display: none\">\n<p>Determine the limits of the numerator and denominator separately.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042632518\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042632518\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\underset{x\\to 0^+}{\\lim} \\cos x=1[\/latex]. Therefore, we cannot apply L\u2019H\u00f4pital\u2019s rule. The limit of the quotient is [latex]\\infty[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/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\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">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\/4.8LHopitalsRule535to587_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<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm209328\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=209328&theme=oea&iframe_resize_id=ohm209328&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-413\">\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>4.8 L&#039;Hopital&#039;s Rule. <strong>Authored by<\/strong>: Ryan Melton. <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>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/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":28,"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-413","chapter","type-chapter","status-publish","hentry"],"part":48,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/413","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":29,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/413\/revisions"}],"predecessor-version":[{"id":4847,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/413\/revisions\/4847"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/48"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/413\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=413"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=413"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=413"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=413"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}