{"id":850,"date":"2025-06-20T17:17:40","date_gmt":"2025-06-20T17:17:40","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=850"},"modified":"2025-08-28T13:13:13","modified_gmt":"2025-08-28T13:13:13","slug":"sequences-and-series-foundations-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/sequences-and-series-foundations-background-youll-need-1\/","title":{"raw":"Sequences and Series Foundations: Background You'll Need 1","rendered":"Sequences and Series Foundations: Background You&#8217;ll Need 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Spot indeterminate forms like 0\/0 in calculations, and use L\u2019H\u00f4pital\u2019s rule to find precise values<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>L\u2019H\u00f4pital\u2019s Rule<\/h2>\r\n<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.\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. Consider the quotient of two functions, represented as:<\/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<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<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\nWe 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.\r\n\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\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\r\n<section class=\"textbox example\">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\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\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\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[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]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].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]<span style=\"background-color: #f4f3ef;\">\u00a0<\/span><\/div>\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.\r\n\r\n<\/section><section class=\"textbox proTip\">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.<\/section>We state L\u2019H\u00f4pital\u2019s rule formally for the indeterminate form [latex]\\frac{0}{0}[\/latex].\r\n\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\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\r\n<\/section><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\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[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\nWatch the following video to see the worked solution to this example.\r\n\r\nFor 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\r\n<\/section>\r\n<h4 style=\"text-align: left;\">L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/ \\infty[\/latex] Case)<\/h4>\r\nL'H\u00f4pital's Rule provides a method to resolve indeterminate forms in calculus, specifically those that result in [latex]\\infty \/ \\infty[\/latex] when calculating limits.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">L\u2019H\u00f4pital\u2019s rule ([latex]\\infty \/ \\infty[\/latex] case)<\/h3>\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<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<\/section><section class=\"textbox example\">\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[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<\/section><section class=\"textbox proTip\">To correctly apply L'H\u00f4pital's Rule to a quotient [latex]\\frac{f(x)}{g(x)}[\/latex], it is essential that the original limit of the quotient is an indeterminate form, either [latex]0\/0[\/latex] or [latex]\\infty \/ \\infty[\/latex]. This is crucial because applying the rule outside these conditions does not yield valid results.<\/section>While L'H\u00f4pital's Rule is an invaluable tool for calculus, its application should be carefully considered. Not all limits of the form [latex]\\infty \/ \\infty[\/latex] are suitable for L'H\u00f4pital's Rule without additional analysis or transformation of the function.\r\n\r\nConsider the following non-applicable example to better understand the limitations:\r\n\r\n<section class=\"textbox example\">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.\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<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<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<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<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<\/section><section class=\"textbox example\">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.[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\nWatch the following video to see the worked solution to this example.\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\" data-mce-fragment=\"1\"><\/iframe><\/center>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\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]209328[\/ohm_question]<\/section><section>\r\n<h3>Indeterminate Form of Type [latex]0 \\cdot \\infty[\/latex]<\/h3>\r\n<p id=\"fs-id1165042320301\">Suppose we want to evaluate [latex]\\underset{x\\to a}{\\lim}(f(x) \\cdot g(x))[\/latex], where [latex]f(x)\\to 0[\/latex] and [latex]g(x)\\to \\infty[\/latex] (or [latex]\u2212\\infty[\/latex]) as [latex]x\\to a[\/latex].<\/p>\r\nSince one term in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen to the product. We use the notation [latex]0 \\cdot \\infty[\/latex] to denote the form that arises in this situation.\r\n\r\nThe expression [latex]0 \\cdot \\infty[\/latex] is considered indeterminate because we cannot determine without further analysis the exact behavior of the product [latex]f(x)g(x)[\/latex] as [latex]x\\to {a}[\/latex]. For example, let [latex]n[\/latex] be a positive integer and consider\r\n<div id=\"fs-id1165043259749\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\dfrac{1}{(x^n+1)}[\/latex] and [latex]g(x)=3x^2[\/latex].<\/div>\r\n<p id=\"fs-id1165042323522\">As [latex]x\\to \\infty[\/latex], [latex]f(x)\\to 0[\/latex] and [latex]g(x)\\to \\infty[\/latex].<\/p>\r\nHowever, the limit as [latex]x\\to \\infty [\/latex] of [latex]f(x)g(x)=\\frac{3x^2}{(x^n+1)}[\/latex] varies, depending on [latex]n[\/latex]. If [latex]n=2[\/latex], then [latex]\\underset{x\\to \\infty }{\\lim}f(x)g(x)=3[\/latex]. If [latex]n=1[\/latex], then [latex]\\underset{x\\to \\infty }{\\lim}f(x)g(x)=\\infty[\/latex]. If [latex]n=3[\/latex], then [latex]\\underset{x\\to \\infty }{\\lim}f(x)g(x)=0[\/latex].\r\n\r\nHere we consider another limit involving the indeterminate form [latex]0 \\cdot \\infty[\/latex] and show how to rewrite the function as a quotient to use L\u2019H\u00f4pital\u2019s rule.\r\n\r\n<section class=\"textbox example\">Evaluate [latex]\\underset{x\\to 0^+}{\\lim}x \\ln x[\/latex].\r\n<p id=\"fs-id1165042545829\">First, rewrite the function [latex]x \\ln x[\/latex] as a quotient to apply L\u2019H\u00f4pital\u2019s rule. If we write<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x \\ln x=\\frac{\\ln x}{1\/x}[\/latex],<\/div>\r\n<p id=\"fs-id1165042383922\">we see that [latex]\\ln x\\to \u2212\\infty [\/latex] as [latex]x\\to 0^+[\/latex] and [latex]\\frac{1}{x}\\to \\infty [\/latex] as [latex]x\\to 0^+[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule and obtain<\/p>\r\n\r\n<div id=\"fs-id1165042705715\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{\\ln x}{1\/x}=\\underset{x\\to 0^+}{\\lim}\\frac{\\frac{d}{dx}(\\ln x)}{\\frac{d}{dx}(1\/x)}=\\underset{x\\to 0^+}{\\lim}\\frac{1\/x}{-1\/x^2}=\\underset{x\\to 0^+}{\\lim}(\u2212x)=0[\/latex].<\/div>\r\n<p id=\"fs-id1165042318647\">We conclude that<\/p>\r\n\r\n<div id=\"fs-id1165042318650\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}x \\ln x=0[\/latex].<\/div>\r\n[caption id=\"\" align=\"aligncenter\" width=\"358\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211307\/CNX_Calc_Figure_04_08_004.jpg\" alt=\"The function y = x ln(x) is graphed for values x \u2265 0. At x = 0, the value of the function is 0.\" width=\"358\" height=\"347\" \/> Figure 2. Finding the limit at [latex]x=0[\/latex] of the function [latex]f(x)=x \\ln x[\/latex].[\/caption]<\/section><section class=\"textbox example\">Evaluate [latex]\\underset{x\\to 0}{\\lim}x \\cot x[\/latex].\r\n\r\n[reveal-answer q=\"404616\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"404616\"]\r\n\r\nWrite [latex]x \\cot x=\\frac{x \\cos x}{\\sin x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043286669\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043286669\"]\r\n\r\n[latex]1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3>Indeterminate Form of Type [latex]\\infty -\\infty[\/latex]<\/h3>\r\nAnother type of indeterminate form is [latex]\\infty -\\infty[\/latex]. Consider the following example:\r\n\r\n<section class=\"textbox example\">Let [latex]n[\/latex] be a positive integer and let [latex]f(x)=3x^n[\/latex] and [latex]g(x)=3x^2+5[\/latex].\r\n\r\nAs [latex]x\\to \\infty[\/latex], [latex]f(x)\\to \\infty [\/latex] and [latex]g(x)\\to \\infty [\/latex]. We are interested in [latex]\\underset{x\\to \\infty}{\\lim}(f(x)-g(x))[\/latex].\r\n\r\nDepending on whether [latex]f(x)[\/latex] grows faster, [latex]g(x)[\/latex] grows faster, or they grow at the same rate, as we see next, anything can happen in this limit. Since [latex]f(x)\\to \\infty [\/latex] and [latex]g(x)\\to \\infty[\/latex], we write [latex]\\infty -\\infty [\/latex] to denote the form of this limit.\r\n\r\n<\/section>As with our other indeterminate forms, [latex]\\infty -\\infty [\/latex] has no meaning on its own and we must do more analysis to determine the value of the limit.\r\n\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1165042318816\">Suppose the exponent [latex]n[\/latex] in the function [latex]f(x)=3x^n[\/latex] is [latex]n=3[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1165043430807\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x^3-3x^2-5)=\\infty[\/latex].<\/div>\r\n<p id=\"fs-id1165042333220\">On the other hand, if [latex]n=2[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1165042333235\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x^2-3x^2-5)=-5[\/latex].<\/div>\r\n<p id=\"fs-id1165043395183\">However, if [latex]n=1[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1165043254251\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x-3x^2-5)=\u2212\\infty[\/latex].<\/div>\r\n<p id=\"fs-id1165043323851\">Therefore, the limit cannot be determined by considering only [latex]\\infty -\\infty[\/latex].<\/p>\r\n\r\n<\/section>\r\n<div>Next we see how to rewrite an expression involving the indeterminate form [latex]\\infty -\\infty [\/latex] as a fraction to apply L\u2019H\u00f4pital\u2019s rule.<\/div>\r\n<section class=\"textbox example\">Evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\left(\\dfrac{1}{x^2}-\\dfrac{1}{\\tan x}\\right)[\/latex].\r\n<p id=\"fs-id1165043281296\">By combining the fractions, we can write the function as a quotient. Since the least common denominator is [latex]x^2 \\tan x[\/latex], we have<\/p>\r\n\r\n<div id=\"fs-id1165043281316\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{x^2}-\\frac{1}{\\tan x}=\\frac{(\\tan x)-x^2}{x^2 \\tan x}[\/latex]<\/div>\r\n<p id=\"fs-id1165043259808\">As [latex]x\\to 0^+[\/latex], the numerator [latex]\\tan x-x^2 \\to 0[\/latex] and the denominator [latex]x^2 \\tan x \\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. Taking the derivatives of the numerator and the denominator, we have<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{(\\tan x)-x^2}{x^2 \\tan x}=\\underset{x\\to 0^+}{\\lim}\\frac{(\\sec^2 x)-2x}{x^2 \\sec^2 x+2x \\tan x}[\/latex]<\/div>\r\n<p id=\"fs-id1165043327626\">As [latex]x\\to 0^+[\/latex], [latex](\\sec^2 x)-2x \\to 1[\/latex] and [latex]x^2 \\sec^2 x+2x \\tan x \\to 0[\/latex]. Since the denominator is positive as [latex]x[\/latex] approaches zero from the right, we conclude that<\/p>\r\n\r\n<div id=\"fs-id1165042710940\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{(\\sec^2 x)-2x}{x^2 \\sec^2 x+2x \\tan x}=\\infty[\/latex]<\/div>\r\n<p id=\"fs-id1165043396304\">Therefore,<\/p>\r\n\r\n<div id=\"fs-id1165043396307\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}(\\frac{1}{x^2}-\\frac{1}{ tan x})=\\infty[\/latex]<span style=\"background-color: #f4f3ef;\">\u00a0<\/span><\/div>\r\n<\/section><section class=\"textbox watchIt\">Watch the following video to see the worked solution to this example.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/e58sGIZe1wU?controls=0&amp;start=671&amp;end=790&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/4.8LHopitalsRule671to790_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\r\n<\/section><section class=\"textbox example\">Evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\left(\\dfrac{1}{x}-\\dfrac{1}{\\sin x}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"1238807\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"1238807\"]\r\n\r\nRewrite the difference of fractions as a single fraction.\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043317356\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043317356\"]\r\n\r\n[latex]0[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Spot indeterminate forms like 0\/0 in calculations, and use L\u2019H\u00f4pital\u2019s rule to find precise values<\/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<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. 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\">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].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.<\/p>\n<\/section>\n<section class=\"textbox proTip\">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.<\/section>\n<p>We state L\u2019H\u00f4pital\u2019s rule formally for the indeterminate form [latex]\\frac{0}{0}[\/latex].<\/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 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<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<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<h4 style=\"text-align: left;\">L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/ \\infty[\/latex] Case)<\/h4>\n<p>L&#8217;H\u00f4pital&#8217;s Rule provides a method to resolve indeterminate forms in calculus, specifically those that result in [latex]\\infty \/ \\infty[\/latex] when calculating limits.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">L\u2019H\u00f4pital\u2019s rule ([latex]\\infty \/ \\infty[\/latex] case)<\/h3>\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 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<\/section>\n<section class=\"textbox example\">\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 class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165042376758\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox proTip\">To correctly apply L&#8217;H\u00f4pital&#8217;s Rule to a quotient [latex]\\frac{f(x)}{g(x)}[\/latex], it is essential that the original limit of the quotient is an indeterminate form, either [latex]0\/0[\/latex] or [latex]\\infty \/ \\infty[\/latex]. This is crucial because applying the rule outside these conditions does not yield valid results.<\/section>\n<p>While L&#8217;H\u00f4pital&#8217;s Rule is an invaluable tool for calculus, its application should be carefully considered. Not all limits of the form [latex]\\infty \/ \\infty[\/latex] are suitable for L&#8217;H\u00f4pital&#8217;s Rule without additional analysis or transformation of the function.<\/p>\n<p>Consider the following non-applicable example to better understand the limitations:<\/p>\n<section class=\"textbox example\">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<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 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 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 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 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].[\/hidden-answer]<\/div>\n<\/section>\n<section class=\"textbox example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q6688909\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165042632518\">Show Solution<\/button><\/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<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\/4.8LHopitalsRule535to587_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<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm209328\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=209328&theme=lumen&iframe_resize_id=ohm209328&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section>\n<h3>Indeterminate Form of Type [latex]0 \\cdot \\infty[\/latex]<\/h3>\n<p id=\"fs-id1165042320301\">Suppose we want to evaluate [latex]\\underset{x\\to a}{\\lim}(f(x) \\cdot g(x))[\/latex], where [latex]f(x)\\to 0[\/latex] and [latex]g(x)\\to \\infty[\/latex] (or [latex]\u2212\\infty[\/latex]) as [latex]x\\to a[\/latex].<\/p>\n<p>Since one term in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen to the product. We use the notation [latex]0 \\cdot \\infty[\/latex] to denote the form that arises in this situation.<\/p>\n<p>The expression [latex]0 \\cdot \\infty[\/latex] is considered indeterminate because we cannot determine without further analysis the exact behavior of the product [latex]f(x)g(x)[\/latex] as [latex]x\\to {a}[\/latex]. For example, let [latex]n[\/latex] be a positive integer and consider<\/p>\n<div id=\"fs-id1165043259749\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\dfrac{1}{(x^n+1)}[\/latex] and [latex]g(x)=3x^2[\/latex].<\/div>\n<p id=\"fs-id1165042323522\">As [latex]x\\to \\infty[\/latex], [latex]f(x)\\to 0[\/latex] and [latex]g(x)\\to \\infty[\/latex].<\/p>\n<p>However, the limit as [latex]x\\to \\infty[\/latex] of [latex]f(x)g(x)=\\frac{3x^2}{(x^n+1)}[\/latex] varies, depending on [latex]n[\/latex]. If [latex]n=2[\/latex], then [latex]\\underset{x\\to \\infty }{\\lim}f(x)g(x)=3[\/latex]. If [latex]n=1[\/latex], then [latex]\\underset{x\\to \\infty }{\\lim}f(x)g(x)=\\infty[\/latex]. If [latex]n=3[\/latex], then [latex]\\underset{x\\to \\infty }{\\lim}f(x)g(x)=0[\/latex].<\/p>\n<p>Here we consider another limit involving the indeterminate form [latex]0 \\cdot \\infty[\/latex] and show how to rewrite the function as a quotient to use L\u2019H\u00f4pital\u2019s rule.<\/p>\n<section class=\"textbox example\">Evaluate [latex]\\underset{x\\to 0^+}{\\lim}x \\ln x[\/latex].<\/p>\n<p id=\"fs-id1165042545829\">First, rewrite the function [latex]x \\ln x[\/latex] as a quotient to apply L\u2019H\u00f4pital\u2019s rule. If we write<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x \\ln x=\\frac{\\ln x}{1\/x}[\/latex],<\/div>\n<p id=\"fs-id1165042383922\">we see that [latex]\\ln x\\to \u2212\\infty[\/latex] as [latex]x\\to 0^+[\/latex] and [latex]\\frac{1}{x}\\to \\infty[\/latex] as [latex]x\\to 0^+[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule and obtain<\/p>\n<div id=\"fs-id1165042705715\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{\\ln x}{1\/x}=\\underset{x\\to 0^+}{\\lim}\\frac{\\frac{d}{dx}(\\ln x)}{\\frac{d}{dx}(1\/x)}=\\underset{x\\to 0^+}{\\lim}\\frac{1\/x}{-1\/x^2}=\\underset{x\\to 0^+}{\\lim}(\u2212x)=0[\/latex].<\/div>\n<p id=\"fs-id1165042318647\">We conclude that<\/p>\n<div id=\"fs-id1165042318650\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}x \\ln x=0[\/latex].<\/div>\n<figure style=\"width: 358px\" 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\/11211307\/CNX_Calc_Figure_04_08_004.jpg\" alt=\"The function y = x ln(x) is graphed for values x \u2265 0. At x = 0, the value of the function is 0.\" width=\"358\" height=\"347\" \/><figcaption class=\"wp-caption-text\">Figure 2. Finding the limit at [latex]x=0[\/latex] of the function [latex]f(x)=x \\ln x[\/latex].<\/figcaption><\/figure>\n<\/section>\n<section class=\"textbox example\">Evaluate [latex]\\underset{x\\to 0}{\\lim}x \\cot x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q404616\">Hint<\/button><\/p>\n<div id=\"q404616\" class=\"hidden-answer\" style=\"display: none\">\n<p>Write [latex]x \\cot x=\\frac{x \\cos x}{\\sin x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043286669\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165043286669\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h3>Indeterminate Form of Type [latex]\\infty -\\infty[\/latex]<\/h3>\n<p>Another type of indeterminate form is [latex]\\infty -\\infty[\/latex]. Consider the following example:<\/p>\n<section class=\"textbox example\">Let [latex]n[\/latex] be a positive integer and let [latex]f(x)=3x^n[\/latex] and [latex]g(x)=3x^2+5[\/latex].<\/p>\n<p>As [latex]x\\to \\infty[\/latex], [latex]f(x)\\to \\infty[\/latex] and [latex]g(x)\\to \\infty[\/latex]. We are interested in [latex]\\underset{x\\to \\infty}{\\lim}(f(x)-g(x))[\/latex].<\/p>\n<p>Depending on whether [latex]f(x)[\/latex] grows faster, [latex]g(x)[\/latex] grows faster, or they grow at the same rate, as we see next, anything can happen in this limit. Since [latex]f(x)\\to \\infty[\/latex] and [latex]g(x)\\to \\infty[\/latex], we write [latex]\\infty -\\infty[\/latex] to denote the form of this limit.<\/p>\n<\/section>\n<p>As with our other indeterminate forms, [latex]\\infty -\\infty[\/latex] has no meaning on its own and we must do more analysis to determine the value of the limit.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1165042318816\">Suppose the exponent [latex]n[\/latex] in the function [latex]f(x)=3x^n[\/latex] is [latex]n=3[\/latex], then<\/p>\n<div id=\"fs-id1165043430807\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x^3-3x^2-5)=\\infty[\/latex].<\/div>\n<p id=\"fs-id1165042333220\">On the other hand, if [latex]n=2[\/latex], then<\/p>\n<div id=\"fs-id1165042333235\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x^2-3x^2-5)=-5[\/latex].<\/div>\n<p id=\"fs-id1165043395183\">However, if [latex]n=1[\/latex], then<\/p>\n<div id=\"fs-id1165043254251\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x-3x^2-5)=\u2212\\infty[\/latex].<\/div>\n<p id=\"fs-id1165043323851\">Therefore, the limit cannot be determined by considering only [latex]\\infty -\\infty[\/latex].<\/p>\n<\/section>\n<div>Next we see how to rewrite an expression involving the indeterminate form [latex]\\infty -\\infty[\/latex] as a fraction to apply L\u2019H\u00f4pital\u2019s rule.<\/div>\n<section class=\"textbox example\">Evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\left(\\dfrac{1}{x^2}-\\dfrac{1}{\\tan x}\\right)[\/latex].<\/p>\n<p id=\"fs-id1165043281296\">By combining the fractions, we can write the function as a quotient. Since the least common denominator is [latex]x^2 \\tan x[\/latex], we have<\/p>\n<div id=\"fs-id1165043281316\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{x^2}-\\frac{1}{\\tan x}=\\frac{(\\tan x)-x^2}{x^2 \\tan x}[\/latex]<\/div>\n<p id=\"fs-id1165043259808\">As [latex]x\\to 0^+[\/latex], the numerator [latex]\\tan x-x^2 \\to 0[\/latex] and the denominator [latex]x^2 \\tan x \\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. Taking the derivatives of the numerator and the denominator, we have<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{(\\tan x)-x^2}{x^2 \\tan x}=\\underset{x\\to 0^+}{\\lim}\\frac{(\\sec^2 x)-2x}{x^2 \\sec^2 x+2x \\tan x}[\/latex]<\/div>\n<p id=\"fs-id1165043327626\">As [latex]x\\to 0^+[\/latex], [latex](\\sec^2 x)-2x \\to 1[\/latex] and [latex]x^2 \\sec^2 x+2x \\tan x \\to 0[\/latex]. Since the denominator is positive as [latex]x[\/latex] approaches zero from the right, we conclude that<\/p>\n<div id=\"fs-id1165042710940\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{(\\sec^2 x)-2x}{x^2 \\sec^2 x+2x \\tan x}=\\infty[\/latex]<\/div>\n<p id=\"fs-id1165043396304\">Therefore,<\/p>\n<div id=\"fs-id1165043396307\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}(\\frac{1}{x^2}-\\frac{1}{ tan x})=\\infty[\/latex]<span style=\"background-color: #f4f3ef;\">\u00a0<\/span><\/div>\n<\/section>\n<section class=\"textbox watchIt\">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=671&amp;end=790&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/button><\/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\/4.8LHopitalsRule671to790_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<section class=\"textbox example\">Evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\left(\\dfrac{1}{x}-\\dfrac{1}{\\sin x}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q1238807\">Hint<\/button><\/p>\n<div id=\"q1238807\" class=\"hidden-answer\" style=\"display: none\">\n<p>Rewrite the difference of fractions as a single fraction.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043317356\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165043317356\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/section>\n","protected":false},"author":15,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":671,"module-header":"- Select Header -","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\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/850"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":4,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/850\/revisions"}],"predecessor-version":[{"id":2045,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/850\/revisions\/2045"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/671"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/850\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=850"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=850"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=850"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=850"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}