{"id":675,"date":"2025-10-09T15:29:43","date_gmt":"2025-10-09T15:29:43","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=675"},"modified":"2026-05-26T14:11:42","modified_gmt":"2026-05-26T14:11:42","slug":"double-integration-and-applications-background-youll-need-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/double-integration-and-applications-background-youll-need-3\/","title":{"raw":"Double Integration and Applications: Background You'll Need 3","rendered":"Double Integration and Applications: Background You&#8217;ll Need 3"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Recognize indeterminate forms and apply L\u2019H\u00f4pital\u2019s rule for quotients, products, subtractions, and powers<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>L\u2019H\u00f4pital\u2019s Rule<\/h2>\r\nL'H\u00f4pital's Rule uses derivatives to evaluate limits that produce indeterminate forms [latex]\\frac{0}{0}[\/latex] or [latex]\\frac{\\infty}{\\infty}[\/latex].\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>L\u2019H\u00f4pital\u2019s rule<\/h3>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable near [latex]a[\/latex]. If [latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}[\/latex] produces [latex]\\frac{0}{0}[\/latex] or [latex]\\frac{\\infty}{\\infty}[\/latex], then:<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f'(x)}{g'(x)}[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">provided the limit on the right exists or equals [latex]\\pm\\infty[\/latex].<\/p>\r\n&nbsp;\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">This also applies to one-sided limits and when [latex]a = \\infty[\/latex] or [latex]-\\infty[\/latex].<\/p>\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Critical Points:<\/strong>\r\n<ul>\r\n \t<li>Always verify the limit produces [latex]\\frac{0}{0}[\/latex] or [latex]\\frac{\\infty}{\\infty}[\/latex] before applying L'H\u00f4pital's Rule<\/li>\r\n \t<li>\u00a0Differentiate numerator and denominator **separately** (this is not the quotient rule)<\/li>\r\n \t<li>\u00a0You may need to apply L'H\u00f4pital's Rule multiple times<\/li>\r\n \t<li>\u00a0If the limit doesn't produce an indeterminate form, do not use L'H\u00f4pital's Rule<\/li>\r\n<\/ul>\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><section class=\"textbox example\" aria-label=\"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 tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]325819[\/ohm_question]<\/section>\r\n<h3 class=\"font-claude-response-heading text-text-100 mt-1 -mb-0.5\">Indeterminate Form [latex]0 \\cdot \\infty[\/latex]<\/h3>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">When [latex]\\underset{x\\to a}{\\lim}f(x) = 0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x) = \\infty[\/latex], the product [latex]f(x)g(x)[\/latex] produces the indeterminate form [latex]0 \\cdot \\infty[\/latex]. To apply L'H\u00f4pital's Rule, rewrite the product as a quotient:<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]f(x) \\cdot g(x) = \\dfrac{f(x)}{\\frac{1}{g(x)}}[\/latex] or [latex]f(x) \\cdot g(x) = \\dfrac{g(x)}{\\frac{1}{f(x)}}[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Choose whichever form gives [latex]\\frac{0}{0}[\/latex] or [latex]\\frac{\\infty}{\\infty}[\/latex].<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate [latex]\\underset{x\\to 0^+}{\\lim}x \\ln x[\/latex].[reveal-answer q=\"805062\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"805062\"]\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][\/hidden-answer]<\/section>\r\n<h3>Indeterminate Form [latex]\\infty - \\infty[\/latex]<\/h3>\r\nWhen both [latex]f(x) \\to \\infty[\/latex] and [latex]g(x) \\to \\infty[\/latex] as [latex]x \\to a[\/latex], the difference [latex]f(x) - g(x)[\/latex] produces [latex]\\infty - \\infty[\/latex]. To use L'H\u00f4pital's Rule, combine the terms into a single quotient, typically by finding a common denominator.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\left(\\dfrac{1}{x^2}-\\dfrac{1}{\\tan x}\\right)[\/latex].\r\n[reveal-answer q=\"894347\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"894347\"]\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<p id=\"fs-id1165043281296\">[\/hidden-answer]<\/p>\r\n\r\n<\/section>\r\n<h3 class=\"font-claude-response-heading text-text-100 mt-1 -mb-0.5\">Indeterminate Forms Involving Exponents<\/h3>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The forms [latex]0^0[\/latex], [latex]\\infty^0[\/latex], and [latex]1^\\infty[\/latex] are indeterminate when they arise in limits. To handle limits of the form [latex]\\underset{x\\to a}{\\lim}f(x)^{g(x)}[\/latex], use logarithms.<\/p>\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>Strategy for Exponential Indeterminate Forms<\/strong>\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Let [latex]y = f(x)^{g(x)}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Take the natural logarithm: [latex]\\ln y = g(x) \\ln(f(x))[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Evaluate [latex]L = \\underset{x\\to a}{\\lim}g(x)\\ln(f(x))[\/latex] using techniques for [latex]0 \\cdot \\infty[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Then [latex]\\underset{x\\to a}{\\lim}f(x)^{g(x)} = e^L[\/latex]<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Evaluate [latex]\\underset{x\\to \\infty }{\\lim} x^{\\frac{1}{x}}[\/latex][reveal-answer q=\"fs-id1165043390866\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043390866\"]\r\n<p id=\"fs-id1165043390866\">Let [latex]y=x^{1\/x}[\/latex]. Then,<\/p>\r\n\r\n<div id=\"fs-id1165043281565\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln (x^{1\/x})=\\frac{1}{x} \\ln x=\\frac{\\ln x}{x}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043281615\">We need to evaluate [latex]\\underset{x\\to \\infty }{\\lim}\\frac{\\ln x}{x}[\/latex]. Applying L\u2019H\u00f4pital\u2019s rule, we obtain<\/p>\r\n\r\n<div id=\"fs-id1165043281645\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim} \\ln y=\\underset{x\\to \\infty}{\\lim}\\frac{\\ln x}{x}=\\underset{x\\to \\infty}{\\lim}\\frac{1\/x}{1}=0[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043173746\">Therefore, [latex]\\underset{x\\to \\infty }{\\lim}\\ln y=0[\/latex]. Since the natural logarithm function is continuous, we conclude that<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln (\\underset{x\\to \\infty}{\\lim} y)=0[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043427387\">which leads to<\/p>\r\n\r\n<div id=\"fs-id1165043427390\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim} y=\\underset{x\\to \\infty}{\\lim}\\frac{\\ln x}{x}=e^0=1[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042407320\">Hence,<\/p>\r\n\r\n<div id=\"fs-id1165042407323\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim} x^{1\/x}=1[\/latex]<\/div>\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=797&amp;end=919&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\/4.8LHopitalsRule797to919_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\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]288429[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Recognize indeterminate forms and apply L\u2019H\u00f4pital\u2019s rule for quotients, products, subtractions, and powers<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>L\u2019H\u00f4pital\u2019s Rule<\/h2>\n<p>L&#8217;H\u00f4pital&#8217;s Rule uses derivatives to evaluate limits that produce indeterminate forms [latex]\\frac{0}{0}[\/latex] or [latex]\\frac{\\infty}{\\infty}[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>L\u2019H\u00f4pital\u2019s rule<\/h3>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable near [latex]a[\/latex]. If [latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}[\/latex] produces [latex]\\frac{0}{0}[\/latex] or [latex]\\frac{\\infty}{\\infty}[\/latex], then:<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f'(x)}{g'(x)}[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">provided the limit on the right exists or equals [latex]\\pm\\infty[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">This also applies to one-sided limits and when [latex]a = \\infty[\/latex] or [latex]-\\infty[\/latex].<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Critical Points:<\/strong><\/p>\n<ul>\n<li>Always verify the limit produces [latex]\\frac{0}{0}[\/latex] or [latex]\\frac{\\infty}{\\infty}[\/latex] before applying L&#8217;H\u00f4pital&#8217;s Rule<\/li>\n<li>\u00a0Differentiate numerator and denominator **separately** (this is not the quotient rule)<\/li>\n<li>\u00a0You may need to apply L&#8217;H\u00f4pital&#8217;s Rule multiple times<\/li>\n<li>\u00a0If the limit doesn&#8217;t produce an indeterminate form, do not use L&#8217;H\u00f4pital&#8217;s Rule<\/li>\n<\/ul>\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<section class=\"textbox example\" aria-label=\"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 tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm325819\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=325819&theme=lumen&iframe_resize_id=ohm325819&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h3 class=\"font-claude-response-heading text-text-100 mt-1 -mb-0.5\">Indeterminate Form [latex]0 \\cdot \\infty[\/latex]<\/h3>\n<p class=\"font-claude-response-body whitespace-normal break-words\">When [latex]\\underset{x\\to a}{\\lim}f(x) = 0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x) = \\infty[\/latex], the product [latex]f(x)g(x)[\/latex] produces the indeterminate form [latex]0 \\cdot \\infty[\/latex]. To apply L&#8217;H\u00f4pital&#8217;s Rule, rewrite the product as a quotient:<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]f(x) \\cdot g(x) = \\dfrac{f(x)}{\\frac{1}{g(x)}}[\/latex] or [latex]f(x) \\cdot g(x) = \\dfrac{g(x)}{\\frac{1}{f(x)}}[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Choose whichever form gives [latex]\\frac{0}{0}[\/latex] or [latex]\\frac{\\infty}{\\infty}[\/latex].<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate [latex]\\underset{x\\to 0^+}{\\lim}x \\ln x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q805062\">Show Answer<\/button><\/p>\n<div id=\"q805062\" class=\"hidden-answer\" style=\"display: none\">\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<\/div>\n<\/div>\n<\/section>\n<h3>Indeterminate Form [latex]\\infty - \\infty[\/latex]<\/h3>\n<p>When both [latex]f(x) \\to \\infty[\/latex] and [latex]g(x) \\to \\infty[\/latex] as [latex]x \\to a[\/latex], the difference [latex]f(x) - g(x)[\/latex] produces [latex]\\infty - \\infty[\/latex]. To use L&#8217;H\u00f4pital&#8217;s Rule, combine the terms into a single quotient, typically by finding a common denominator.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\left(\\dfrac{1}{x^2}-\\dfrac{1}{\\tan x}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q894347\">Show Answer<\/button><\/p>\n<div id=\"q894347\" class=\"hidden-answer\" style=\"display: none\">\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<\/div>\n<\/div>\n<\/section>\n<h3 class=\"font-claude-response-heading text-text-100 mt-1 -mb-0.5\">Indeterminate Forms Involving Exponents<\/h3>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The forms [latex]0^0[\/latex], [latex]\\infty^0[\/latex], and [latex]1^\\infty[\/latex] are indeterminate when they arise in limits. To handle limits of the form [latex]\\underset{x\\to a}{\\lim}f(x)^{g(x)}[\/latex], use logarithms.<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>Strategy for Exponential Indeterminate Forms<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Let [latex]y = f(x)^{g(x)}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Take the natural logarithm: [latex]\\ln y = g(x) \\ln(f(x))[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Evaluate [latex]L = \\underset{x\\to a}{\\lim}g(x)\\ln(f(x))[\/latex] using techniques for [latex]0 \\cdot \\infty[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Then [latex]\\underset{x\\to a}{\\lim}f(x)^{g(x)} = e^L[\/latex]<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate [latex]\\underset{x\\to \\infty }{\\lim} x^{\\frac{1}{x}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043390866\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165043390866\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043390866\">Let [latex]y=x^{1\/x}[\/latex]. Then,<\/p>\n<div id=\"fs-id1165043281565\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln (x^{1\/x})=\\frac{1}{x} \\ln x=\\frac{\\ln x}{x}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043281615\">We need to evaluate [latex]\\underset{x\\to \\infty }{\\lim}\\frac{\\ln x}{x}[\/latex]. Applying L\u2019H\u00f4pital\u2019s rule, we obtain<\/p>\n<div id=\"fs-id1165043281645\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim} \\ln y=\\underset{x\\to \\infty}{\\lim}\\frac{\\ln x}{x}=\\underset{x\\to \\infty}{\\lim}\\frac{1\/x}{1}=0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043173746\">Therefore, [latex]\\underset{x\\to \\infty }{\\lim}\\ln y=0[\/latex]. Since the natural logarithm function is continuous, we conclude that<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln (\\underset{x\\to \\infty}{\\lim} y)=0[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043427387\">which leads to<\/p>\n<div id=\"fs-id1165043427390\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim} y=\\underset{x\\to \\infty}{\\lim}\\frac{\\ln x}{x}=e^0=1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042407320\">Hence,<\/p>\n<div id=\"fs-id1165042407323\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim} x^{1\/x}=1[\/latex]<\/div>\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=797&amp;end=919&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. 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