{"id":1989,"date":"2021-08-19T16:07:00","date_gmt":"2021-08-19T16:07:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-improper-integrals-2\/"},"modified":"2021-11-19T03:09:09","modified_gmt":"2021-11-19T03:09:09","slug":"skills-review-for-improper-integrals-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-improper-integrals-2\/","title":{"raw":"Skills Review for Improper Integrals","rendered":"Skills Review for Improper Integrals"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate the limit of a function as \ud835\udc65 increases or decreases without bound<\/li>\r\n \t<li>Recognize when to apply L\u2019H\u00f4pital\u2019s rule<\/li>\r\n \t<li>Identify indeterminate forms produced by quotients, products, subtractions, and powers, and apply L\u2019H\u00f4pital\u2019s rule in each case<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Improper Integrals section, we will be faced with evaluating integrals that contain bounds that are either infinite or where the function we are integrating is undefined. As a result, here we will review how to find limits of functions at infinity and apply L'Hopital's Rule.\r\n<h2>Limits at Infinity<\/h2>\r\n<div id=\"fs-id1165043145047\" class=\"bc-section section\">\r\n<p id=\"fs-id1165043107285\">Recall that [latex]\\underset{x \\to a}{\\lim}f(x)=L[\/latex] means [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] as long as [latex]x[\/latex] is sufficiently close to [latex]a[\/latex]. We can extend this idea to limits at infinity. For example, consider the function [latex]f(x)=2+\\frac{1}{x}[\/latex]. As can be seen graphically below, as the values of [latex]x[\/latex] get larger, the values of [latex]f(x)[\/latex] approach 2. We say the limit as [latex]x[\/latex] approaches [latex]\\infty [\/latex] of [latex]f(x)[\/latex] is 2 and write [latex]\\underset{x\\to \\infty }{\\lim}f(x)=2[\/latex]. Similarly, for [latex]x&lt;0[\/latex], as the values [latex]|x|[\/latex] get larger, the values of [latex]f(x)[\/latex] approaches 2. We say the limit as [latex]x[\/latex] approaches [latex]\u2212\\infty [\/latex] of [latex]f(x)[\/latex] is 2 and write [latex]\\underset{x\\to a}{\\lim}f(x)=2[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"717\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211025\/CNX_Calc_Figure_04_06_019.jpg\" alt=\"The function f(x) 2 + 1\/x is graphed. The function starts negative near y = 2 but then decreases to \u2212\u221e near x = 0. The function then decreases from \u221e near x = 0 and gets nearer to y = 2 as x increases. There is a horizontal line denoting the asymptote y = 2.\" width=\"717\" height=\"423\" \/> The function approaches the asymptote [latex]y=2[\/latex] as [latex]x[\/latex] approaches [latex]\\pm \\infty[\/latex].[\/caption]\r\n<p id=\"fs-id1165042936244\">More generally, for any function [latex]f[\/latex], we say the limit as [latex]x \\to \\infty [\/latex] of [latex]f(x)[\/latex] is [latex]L[\/latex] if [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] as long as [latex]x[\/latex] is sufficiently large. In that case, we write [latex]\\underset{x\\to \\infty}{\\lim}f(x)=L[\/latex]. Similarly, we say the limit as [latex]x\\to \u2212\\infty [\/latex] of [latex]f(x)[\/latex] is [latex]L[\/latex] if [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] as long as [latex]x&lt;0[\/latex] and [latex]|x|[\/latex] is sufficiently large. In that case, we write [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=L[\/latex]. We now look at the definition of a function having a limit at infinity.<\/p>\r\n\r\n<div id=\"fs-id1165042331960\" class=\"textbox shaded\">\r\n<div class=\"title\">\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1.2em; font-weight: 600; text-align: center; text-transform: uppercase; background-color: #eeeeee;\">Definition<\/span><\/p>\r\n\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\n<p id=\"fs-id1165042970725\">(Informal) If the values of [latex]f(x)[\/latex] become arbitrarily close to [latex]L[\/latex] as [latex]x[\/latex] becomes sufficiently large, we say the function [latex]f[\/latex] has a limit at infinity and write<\/p>\r\n\r\n<div id=\"fs-id1165042986551\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}f(x)=L[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042374662\">If the values of [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] for [latex]x&lt;0[\/latex] as [latex]|x|[\/latex] becomes sufficiently large, we say that the function [latex]f[\/latex] has a limit at negative infinity and write<\/p>\r\n\r\n<div id=\"fs-id1165043105208\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to -\\infty }{\\lim}f(x)=L[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbook exercises\">\r\n<h3>Example: Computing Limits at Infinity<\/h3>\r\n<p id=\"fs-id1165043262623\">For each of the following functions [latex]f[\/latex], evaluate [latex]\\underset{x\\to \\infty }{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1165042356111\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]f(x)=5-\\frac{2}{x^2}[\/latex]<\/li>\r\n \t<li>[latex]f(x)=\\dfrac{\\sin x}{x}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1165043183885\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043183885\"]\r\n<ol id=\"fs-id1165043183885\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Using the algebraic limit laws, we have [latex]\\underset{x\\to \\infty }{\\lim}(5-\\frac{2}{x^2})=\\underset{x\\to \\infty }{\\lim}5-2(\\underset{x\\to \\infty }{\\lim}\\frac{1}{x})(\\underset{x\\to \\infty }{\\lim}\\frac{1}{x})=5-2 \\cdot 0=5[\/latex]. Similarly, [latex]\\underset{x\\to -\\infty }{\\lim}f(x)=5[\/latex].\r\n<div class=\"mceTemp\"><\/div><\/li>\r\n \t<li>nce [latex]-1\\le \\sin x\\le 1[\/latex] for all [latex]x[\/latex], we have\r\n<div id=\"fs-id1165043093355\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{-1}{x}\\le \\frac{\\sin x}{x}\\le \\frac{1}{x}[\/latex]<\/div>\r\nfor all [latex]x \\ne 0[\/latex]. Also, since\r\n<div id=\"fs-id1165043197153\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\frac{-1}{x}=0=\\underset{x\\to \\infty }{\\lim}\\frac{1}{x}[\/latex],<\/div>\r\nwe can apply the squeeze theorem to conclude that\r\n<div id=\"fs-id1165043036581\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\frac{\\sin x}{x}=0[\/latex]<\/div>\r\nSimilarly,\r\n<div id=\"fs-id1165043122536\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \u2212\\infty}{\\lim}\\frac{\\sin x}{x}=0[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042320881\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165043315935\">Evaluate [latex]\\underset{x\\to \u2212\\infty}{\\lim}\\left(3+\\frac{4}{x}\\right)[\/latex] and [latex]\\underset{x\\to \\infty }{\\lim}\\left(3+\\frac{4}{x}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"2473508\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"2473508\"]\r\n<p id=\"fs-id1165042318511\">[latex]\\underset{x\\to \\pm \\infty }{\\lim}1\/x=0[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043390798\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043390798\"]\r\n<p id=\"fs-id1165043390798\">Both limits are 3.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042333169\" class=\"bc-section section\">\r\n<p id=\"fs-id1165042333174\">Sometimes the values of a function [latex]f[\/latex] become arbitrarily large as [latex]x\\to \\infty [\/latex] (or as [latex]x\\to \u2212\\infty )[\/latex]. In this case, we write [latex]\\underset{x\\to \\infty }{\\lim}f(x)=\\infty [\/latex] (or [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=\\infty )[\/latex]. On the other hand, if the values of [latex]f[\/latex] are negative but become arbitrarily large in magnitude as [latex]x\\to \\infty [\/latex] (or as [latex]x\\to \u2212\\infty )[\/latex], we write [latex]\\underset{x\\to \\infty }{\\lim}f(x)=\u2212\\infty [\/latex] (or [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=\u2212\\infty )[\/latex].<\/p>\r\n<p id=\"fs-id1165042606820\">For example, consider the function [latex]f(x)=x^3[\/latex]. The [latex]\\underset{x\\to \\infty }{\\lim}x^3=\\infty[\/latex]. On the other hand, as [latex]x\\to \u2212\\infty[\/latex], the values of [latex]f(x)=x^3[\/latex] are negative but become arbitrarily large in magnitude. Consequently, [latex]\\underset{x\\to \u2212\\infty }{\\lim}x^3=\u2212\\infty[\/latex].<\/p>\r\n\r\n<table id=\"fs-id1165042406634\" class=\"column-header\" summary=\"The table has four rows and six columns. The first column is a header column and it reads x, x3, x, and x3. After the header, the first row reads 10, 20, 50, 100, and 1000. The second row reads 1000, 8000, 125000, 1,000,000, and 1,000,000,000. The third row reads \u221210, \u221220, \u221250, \u2212100, and \u22121000. The forth row reads \u22121000, \u22128000, \u2212125,000, \u22121,000,000, and \u22121,000,000,000.\"><caption>Values of a power function as [latex]x\\to \\pm \\infty [\/latex]<\/caption>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td>10<\/td>\r\n<td>20<\/td>\r\n<td>50<\/td>\r\n<td>100<\/td>\r\n<td>1000<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td><strong>[latex]x^3[\/latex]<\/strong><\/td>\r\n<td>1000<\/td>\r\n<td>8000<\/td>\r\n<td>125,000<\/td>\r\n<td>1,000,000<\/td>\r\n<td>1,000,000,000<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td>-10<\/td>\r\n<td>-20<\/td>\r\n<td>-50<\/td>\r\n<td>-100<\/td>\r\n<td>-1000<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td><strong>[latex]x^3[\/latex]<\/strong><\/td>\r\n<td>-1000<\/td>\r\n<td>-8000<\/td>\r\n<td>-125,000<\/td>\r\n<td>-1,000,000<\/td>\r\n<td>-1,000,000,000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"642\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211045\/CNX_Calc_Figure_04_06_022.jpg\" alt=\"The function f(x) = x3 is graphed. It is apparent that this function rapidly approaches infinity as x approaches infinity.\" width=\"642\" height=\"272\" \/> For this function, the functional values approach infinity as [latex]x\\to \\pm \\infty[\/latex].[\/caption]<\/div>\r\n<div id=\"fs-id1165043276353\" class=\"textbox shaded\">\r\n<div class=\"title\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\n<p id=\"fs-id1165043276356\">(Informal) We say a function [latex]f[\/latex] has an infinite limit at infinity and write<\/p>\r\n\r\n<div id=\"fs-id1165043276364\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}f(x)=\\infty [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042709557\">if [latex]f(x)[\/latex] becomes arbitrarily large for [latex]x[\/latex] sufficiently large. We say a function has a negative infinite limit at infinity and write<\/p>\r\n\r\n<div id=\"fs-id1165042647077\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}f(x)=\u2212\\infty[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042327355\">if [latex]f(x)&lt;0[\/latex] and [latex]|f(x)|[\/latex] becomes arbitrarily large for [latex]x[\/latex] sufficiently large. Similarly, we can define infinite limits as [latex]x\\to \u2212\\infty[\/latex].<\/p>\r\n\r\n<\/div>\r\n<h2>Apply L'Hopital's Rule<\/h2>\r\n<h3>Indeterminate Form of Type [latex]\\frac{0}{0}[\/latex]<\/h3>\r\n<div id=\"fs-id1165043085155\" class=\"bc-section section\">\r\n<p id=\"fs-id1165042941863\">L\u2019H\u00f4pital\u2019s rule can be used to evaluate limits involving the quotient of two functions. Consider<\/p>\r\n\r\n<div id=\"fs-id1165042458008\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042613148\">If [latex]\\underset{x\\to a}{\\lim}f(x)=L_1[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=L_2 \\ne 0[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1165043088102\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\dfrac{L_1}{L_2}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042330308\">However, what happens if [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex]? We call this one of the <strong>indeterminate forms<\/strong>, of type [latex]\\frac{0}{0}[\/latex]. This is considered an indeterminate form because we cannot determine the exact behavior of [latex]\\frac{f(x)}{g(x)}[\/latex] as [latex]x\\to a[\/latex] without further analysis.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043085155\" class=\"bc-section section\">\r\n<div id=\"fs-id1165043352593\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">L\u2019H\u00f4pital\u2019s Rule (0\/0 Case)<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165043276140\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval containing [latex]a[\/latex], except possibly at [latex]a[\/latex]. If [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1165043020148\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex],<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<p id=\"fs-id1165043035673\">assuming the limit on the right exists or is [latex]\\infty [\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if we are considering one-sided limits, or if [latex]a=\\infty[\/latex] or [latex]-\\infty[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042712928\" class=\"bc-section section\">\r\n<div id=\"fs-id1165043397578\" class=\"textbook exercises\">\r\n<h3>Example: Applying L\u2019H\u00f4pital\u2019s Rule (0\/0 Case)<\/h3>\r\n<p id=\"fs-id1165042456913\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\r\n\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{1- \\cos x}{x}[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to 1}{\\lim}\\dfrac{\\sin (\\pi x)}{\\ln x}[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{e^{\\frac{1}{x}}-1}{\\frac{1}{x}}[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{\\sin x-x}{x^2}[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1165043104016\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"fs-id1165042535045\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042535045\"]\r\n<ol id=\"fs-id1165042535045\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Since the numerator [latex]1- \\cos x\\to 0[\/latex] and the denominator [latex]x\\to 0[\/latex], we can apply L\u2019H\u00f4pital\u2019s rule to evaluate this limit. We have\r\n<div id=\"fs-id1165042328696\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to 0}{\\lim}\\frac{1- \\cos x}{x} &amp; =\\underset{x\\to 0}{\\lim}\\frac{\\frac{d}{dx}(1- \\cos x)}{\\frac{d}{dx}(x)} \\\\ &amp; =\\underset{x\\to 0}{\\lim}\\frac{\\sin x}{1} \\\\ &amp; =\\frac{\\underset{x\\to 0}{\\lim}(\\sin x)}{\\underset{x\\to 0}{\\lim}(1)} \\\\ &amp; =\\frac{0}{1}=0 \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>As [latex]x\\to 1[\/latex], the numerator [latex]\\sin (\\pi x)\\to 0[\/latex] and the denominator [latex]\\ln x \\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\r\n<div id=\"fs-id1165043116372\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to 1}{\\lim}\\frac{\\sin (\\pi x)}{\\ln x} &amp; =\\underset{x\\to 1}{\\lim}\\frac{\\pi \\cos (\\pi x)}{1\/x} \\\\ &amp; =\\underset{x\\to 1}{\\lim}(\\pi x) \\cos (\\pi x) \\\\ &amp; =(\\pi \\cdot 1)(-1)=\u2212\\pi \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>As [latex]x\\to \\infty[\/latex], the numerator [latex]e^{1\/x}-1\\to 0[\/latex] and the denominator [latex](\\frac{1}{x})\\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\r\n<div id=\"fs-id1165042327460\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\frac{e^{1\/x}-1}{\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\frac{e^{1\/x}(\\frac{-1}{x^2})}{(\\frac{-1}{x^2})}=\\underset{x\\to \\infty}{\\lim} e^{1\/x}=e^0=1[\/latex]<\/div><\/li>\r\n \t<li>As [latex]x\\to 0[\/latex], both the numerator and denominator approach zero. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\r\n<div id=\"fs-id1165042562974\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x-x}{x^2}=\\underset{x\\to 0}{\\lim}\\frac{\\cos x-1}{2x}[\/latex].<\/div>\r\nSince the numerator and denominator of this new quotient both approach zero as [latex]x\\to 0[\/latex], we apply L\u2019H\u00f4pital\u2019s rule again. In doing so, we see that\r\n<div id=\"fs-id1165043281524\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\cos x-1}{2x}=\\underset{x\\to 0}{\\lim}\\frac{\u2212\\sin x}{2}=0[\/latex].<\/div>\r\nTherefore, we conclude that\r\n<div id=\"fs-id1165043284142\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x-x}{x^2}=0[\/latex].<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\underset{x\\to 0}{\\lim}\\dfrac{x}{\\tan x}[\/latex].\r\n\r\n[reveal-answer q=\"37002811\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"37002811\"]\r\n\r\n[latex]\\frac{d}{dx} \\tan x= \\sec ^2 x[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042377480\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042377480\"]\r\n\r\n[latex]1[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<h3>Indeterminate Form of Type [latex]\\frac{\\infty}{\\infty}[\/latex]<\/h3>\r\n<p id=\"fs-id1165042318672\">We can also use L\u2019H\u00f4pital\u2019s rule to evaluate limits of quotients [latex]\\frac{f(x)}{g(x)}[\/latex] in which [latex]f(x)\\to \\pm \\infty [\/latex] and [latex]g(x)\\to \\pm \\infty[\/latex]. Limits of this form are classified as <em>indeterminate forms of type<\/em> [latex]\\infty \/ \\infty[\/latex]. Again, note that we are not actually dividing [latex]\\infty[\/latex] by [latex]\\infty[\/latex]. Since [latex]\\infty[\/latex] is not a real number, that is impossible; rather, [latex]\\infty \/ \\infty[\/latex] is used to represent a quotient of limits, each of which is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165042330826\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/ \\infty[\/latex] Case)<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165043426174\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval containing [latex]a[\/latex], except possibly at [latex]a[\/latex]. Suppose [latex]\\underset{x\\to a}{\\lim}f(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]) and [latex]\\underset{x\\to a}{\\lim}g(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]). Then,<\/p>\r\n\r\n<div id=\"fs-id1165042373703\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex],<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<p id=\"fs-id1165042707280\">assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if the limit is infinite, if [latex]a=\\infty[\/latex] or [latex]\u2212\\infty[\/latex], or the limit is one-sided.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043427623\" class=\"textbook exercises\">\r\n<h3>Example: Applying L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/\\infty[\/latex] Case)<\/h3>\r\n<p id=\"fs-id1165042709794\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\r\n\r\n<ol id=\"fs-id1165042709798\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1165043427625\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"fs-id1165042376758\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042376758\"]\r\n<ol id=\"fs-id1165042376758\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Since [latex]3x+5[\/latex] and [latex]2x+1[\/latex] are first-degree polynomials with positive leading coefficients, [latex]\\underset{x\\to \\infty }{\\lim}(3x+5)=\\infty [\/latex] and [latex]\\underset{x\\to \\infty }{\\lim}(2x+1)=\\infty[\/latex]. Therefore, we apply L\u2019H\u00f4pital\u2019s rule and obtain\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim}\\dfrac{3x+5}{2x+\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3}{2}=\\dfrac{3}{2}[\/latex].<\/div>\r\nNote that this limit can also be calculated without invoking L\u2019H\u00f4pital\u2019s rule. Earlier in the module we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of [latex]x[\/latex] in the denominator. In doing so, we saw that\r\n<div id=\"fs-id1165042319986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3+\\frac{5}{x}}{2+\\frac{1}{x}}=\\dfrac{3}{2}[\/latex].<\/div>\r\nL\u2019H\u00f4pital\u2019s rule provides us with an alternative means of evaluating this type of limit.<\/li>\r\n \t<li>Here, [latex]\\underset{x\\to 0^+}{\\lim} \\ln x=\u2212\\infty [\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} \\cot x=\\infty[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule and obtain\r\n<div id=\"fs-id1165042374803\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\frac{1}{x}}{\u2212\\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}[\/latex].<\/div>\r\nNow as [latex]x\\to 0^+[\/latex], [latex]\\csc^2 x\\to \\infty[\/latex]. Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of [latex]\\csc x[\/latex] to write\r\n<div id=\"fs-id1165042376466\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}[\/latex].<\/div>\r\nNow [latex]\\underset{x\\to 0^+}{\\lim} \\sin^2 x=0[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} x=0[\/latex], so we apply L\u2019H\u00f4pital\u2019s rule again. We find\r\n<div id=\"fs-id1165043249678\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{2 \\sin x \\cos x}{-1}=\\dfrac{0}{-1}=0[\/latex].<\/div>\r\nWe conclude that\r\n<div id=\"fs-id1165042315721\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=0[\/latex].<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042333392\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{\\ln x}{5x}[\/latex]\r\n\r\n[reveal-answer q=\"30011179\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"30011179\"]\r\n\r\n[latex]\\frac{d}{dx}\\ln x=\\frac{1}{x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042367881\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042367881\"]\r\n\r\n0\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165042331798\">L\u2019H\u00f4pital\u2019s rule is very useful for evaluating limits involving the indeterminate forms [latex]\\frac{0}{0}[\/latex] and [latex]\\frac{\\infty}{\\infty}[\/latex]. However, we can also use L\u2019H\u00f4pital\u2019s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits. The expressions [latex]0 \\cdot \\infty[\/latex], [latex]\\infty - \\infty[\/latex], [latex]1^{\\infty}[\/latex], [latex]\\infty^0[\/latex], and [latex]0^0[\/latex] are all considered indeterminate forms. These expressions are not real numbers. Rather, they represent forms that arise when trying to evaluate certain limits. Next we realize why these are indeterminate forms and then understand how to use L\u2019H\u00f4pital\u2019s rule in these cases. The key idea is that we must rewrite the indeterminate forms in such a way that we arrive at the indeterminate form [latex]\\frac{0}{0}[\/latex] or [latex]\\frac{\\infty}{\\infty}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165042320288\" class=\"bc-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]. 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. 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>\r\n\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&nbsp;\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]. 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]. 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>\r\n\r\n<div id=\"fs-id1165042368495\" class=\"textbook exercises\">\r\n<h3>Example: Indeterminate Form of Type [latex]0\u00b7\\infty [\/latex]<\/h3>\r\nEvaluate [latex]\\underset{x\\to 0^+}{\\lim}x \\ln x[\/latex]\r\n<div id=\"fs-id1165042368497\" class=\"exercise\">[reveal-answer q=\"fs-id1165042545829\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042545829\"]\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<div><\/div>\r\n<div>[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\" \/> Finding the limit at [latex]x=0[\/latex] of the function [latex]f(x)=x \\ln x[\/latex].[\/caption]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165043430905\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [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\n1\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042318802\" class=\"bc-section section\">\r\n<h3>Indeterminate Form of Type [latex]\\infty -\\infty[\/latex]<\/h3>\r\n<p id=\"fs-id1165042318816\">Another type of indeterminate form is [latex]\\infty -\\infty[\/latex]. Consider the following 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]. 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]. 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. 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. For example, 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&nbsp;\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&nbsp;\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&nbsp;\r\n<p id=\"fs-id1165043323851\">Therefore, the limit cannot be determined by considering only [latex]\\infty -\\infty[\/latex]. 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.<\/p>\r\n\r\n<div id=\"fs-id1165043323875\" class=\"textbook exercises\">\r\n<h3>Example: Indeterminate Form of Type [latex]\\infty -\\infty[\/latex]<\/h3>\r\nEvaluate [latex]\\underset{x\\to 0^+}{\\lim}\\left(\\dfrac{1}{x^2}-\\dfrac{1}{\\tan x}\\right)[\/latex].\r\n<div id=\"fs-id1165043323877\" class=\"exercise\">[reveal-answer q=\"fs-id1165043281296\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043281296\"]\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&nbsp;\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&nbsp;\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&nbsp;\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]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165043348549\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [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\n0\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Other Types of Indeterminate Form<\/h3>\r\n<p id=\"fs-id1165042364139\">Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions [latex]0^0[\/latex], [latex]\\infty^0[\/latex], and [latex]1^{\\infty}[\/latex] are all indeterminate forms. On their own, these expressions are meaningless because we cannot actually evaluate these expressions as we would evaluate an expression involving real numbers. Rather, these expressions represent forms that arise when finding limits. Now we examine how L\u2019H\u00f4pital\u2019s rule can be used to evaluate limits involving these indeterminate forms.<\/p>\r\n<p id=\"fs-id1165042364178\">Since L\u2019H\u00f4pital\u2019s rule applies to quotients, we use the natural logarithm function and its properties to reduce a problem evaluating a limit involving exponents to a related problem involving a limit of a quotient. For example, suppose we want to evaluate [latex]\\underset{x\\to a}{\\lim}f(x)^{g(x)}[\/latex] and we arrive at the indeterminate form [latex]\\infty^0[\/latex]. (The indeterminate forms [latex]0^0[\/latex] and [latex]1^{\\infty}[\/latex] can be handled similarly.)<\/p>\r\n\r\n<div id=\"fs-id1165043390815\" class=\"textbook exercises\">\r\n<h3>Example: Indeterminate Form of Type [latex]\\infty^0[\/latex]<\/h3>\r\nEvaluate [latex]\\underset{x\\to \\infty }{\\lim} x^{\\frac{1}{x}}[\/latex]\r\n<div id=\"fs-id1165043390817\" class=\"exercise\">[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\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042407362\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\underset{x\\to \\infty}{\\lim} x^{\\frac{1}{\\ln x}}[\/latex]\r\n\r\n[reveal-answer q=\"3762844\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3762844\"]\r\n\r\nLet [latex]y=x^{1\/ \\ln x}[\/latex] and apply the natural logarithm to both sides of the equation.\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043108248\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043108248\"]\r\n\r\n[latex]e[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043108292\" class=\"textbook exercises\">\r\n<h3>Example: Indeterminate Form of Type [latex]0^0[\/latex]<\/h3>\r\nEvaluate [latex]\\underset{x\\to 0^+}{\\lim} x^{\\sin x}[\/latex]\r\n<div id=\"fs-id1165043108295\" class=\"exercise\">[reveal-answer q=\"fs-id1165042657755\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042657755\"]\r\n<p id=\"fs-id1165042657755\">Let<\/p>\r\n\r\n<div id=\"fs-id1165042657759\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=x^{\\sin x}[\/latex]<\/div>\r\n<p id=\"fs-id1165042657780\">Therefore,<\/p>\r\n\r\n<div id=\"fs-id1165042657783\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln y=\\ln (x^{\\sin x})= \\sin x \\ln x[\/latex]<\/div>\r\n<p id=\"fs-id1165042707171\">We now evaluate [latex]\\underset{x\\to 0^+}{\\lim} \\sin x \\ln x[\/latex]. Since [latex]\\underset{x\\to 0^+}{\\lim} \\sin x=0[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} \\ln x=\u2212\\infty[\/latex], we have the indeterminate form [latex]0 \\cdot \\infty[\/latex]. To apply L\u2019H\u00f4pital\u2019s rule, we need to rewrite [latex] \\sin x \\ln x[\/latex] as a fraction. We could write<\/p>\r\n\r\n<div id=\"fs-id1165043173832\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex] \\sin x \\ln x=\\frac{\\sin x}{1\/ \\ln x}[\/latex]<\/div>\r\n<p id=\"fs-id1165043173865\">or<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex] \\sin x \\ln x=\\frac{\\ln x}{1\/ \\sin x}=\\frac{\\ln x}{\\csc x}[\/latex]<\/div>\r\n<p id=\"fs-id1165042364489\">Let\u2019s consider the first option. In this case, applying L\u2019H\u00f4pital\u2019s rule, we would obtain<\/p>\r\n\r\n<div id=\"fs-id1165042364494\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} \\sin x \\ln x=\\underset{x\\to 0^+}{\\lim}\\frac{\\sin x}{1\/ \\ln x}=\\underset{x\\to 0^+}{\\lim}\\frac{\\cos x}{-1\/(x(\\ln x)^2)}=\\underset{x\\to 0^+}{\\lim}(\u2212x(\\ln x)^2 \\cos x)[\/latex]<\/div>\r\n<p id=\"fs-id1165043317274\">Unfortunately, we not only have another expression involving the indeterminate form [latex]0 \\cdot \\infty[\/latex], but the new limit is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By writing<\/p>\r\n\r\n<div id=\"fs-id1165043131555\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex] \\sin x \\ln x=\\frac{\\ln x}{1\/ \\sin x}=\\frac{\\ln x}{\\csc x}[\/latex]<\/div>\r\n<p id=\"fs-id1165043131604\">and applying L\u2019H\u00f4pital\u2019s rule, we obtain<\/p>\r\n\r\n<div id=\"fs-id1165043131609\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} \\sin x \\ln x=\\underset{x\\to 0^+}{\\lim}\\frac{\\ln x}{\\csc x}=\\underset{x\\to 0^+}{\\lim}\\frac{1\/x}{\u2212 \\csc x \\cot x}=\\underset{x\\to 0^+}{\\lim}\\frac{-1}{x \\csc x \\cot x}[\/latex]<\/div>\r\n<p id=\"fs-id1165042651533\">Using the fact that [latex]\\csc x=\\frac{1}{\\sin x}[\/latex] and [latex]\\cot x=\\frac{\\cos x}{\\sin x}[\/latex], we can rewrite the expression on the right-hand side as<\/p>\r\n\r\n<div id=\"fs-id1165043251999\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{\u2212\\sin^2 x}{x \\cos x}=\\underset{x\\to 0^+}{\\lim}[\\frac{\\sin x}{x} \\cdot (\u2212\\tan x)]=(\\underset{x\\to 0^+}{\\lim}\\frac{\\sin x}{x}) \\cdot (\\underset{x\\to 0^+}{\\lim}(\u2212\\tan x))=1 \\cdot 0=0[\/latex]<\/div>\r\n<p id=\"fs-id1165042676314\">We conclude that [latex]\\underset{x\\to 0^+}{\\lim} \\ln y=0[\/latex]. Therefore, [latex]\\ln (\\underset{x\\to 0^+}{\\lim} y)=0[\/latex] and we have<\/p>\r\n\r\n<div id=\"fs-id1165042327527\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} y=\\underset{x\\to 0^+}{\\lim} x^{\\sin x}=e^0=1[\/latex]<\/div>\r\n<p id=\"fs-id1165042327592\">Hence,<\/p>\r\n\r\n<div id=\"fs-id1165042327595\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} x^{\\sin x}=1[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042660254\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\underset{x\\to 0^+}{\\lim} x^x[\/latex]\r\n\r\n[reveal-answer q=\"929037\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"929037\"]\r\n\r\nLet [latex]y=x^x[\/latex] and take the natural logarithm of both sides of the equation.\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042660293\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042660293\"]\r\n\r\n1\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate the limit of a function as \ud835\udc65 increases or decreases without bound<\/li>\n<li>Recognize when to apply L\u2019H\u00f4pital\u2019s rule<\/li>\n<li>Identify indeterminate forms produced by quotients, products, subtractions, and powers, and apply L\u2019H\u00f4pital\u2019s rule in each case<\/li>\n<\/ul>\n<\/div>\n<p>In the Improper Integrals section, we will be faced with evaluating integrals that contain bounds that are either infinite or where the function we are integrating is undefined. As a result, here we will review how to find limits of functions at infinity and apply L&#8217;Hopital&#8217;s Rule.<\/p>\n<h2>Limits at Infinity<\/h2>\n<div id=\"fs-id1165043145047\" class=\"bc-section section\">\n<p id=\"fs-id1165043107285\">Recall that [latex]\\underset{x \\to a}{\\lim}f(x)=L[\/latex] means [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] as long as [latex]x[\/latex] is sufficiently close to [latex]a[\/latex]. We can extend this idea to limits at infinity. For example, consider the function [latex]f(x)=2+\\frac{1}{x}[\/latex]. As can be seen graphically below, as the values of [latex]x[\/latex] get larger, the values of [latex]f(x)[\/latex] approach 2. We say the limit as [latex]x[\/latex] approaches [latex]\\infty[\/latex] of [latex]f(x)[\/latex] is 2 and write [latex]\\underset{x\\to \\infty }{\\lim}f(x)=2[\/latex]. Similarly, for [latex]x<0[\/latex], as the values [latex]|x|[\/latex] get larger, the values of [latex]f(x)[\/latex] approaches 2. We say the limit as [latex]x[\/latex] approaches [latex]\u2212\\infty[\/latex] of [latex]f(x)[\/latex] is 2 and write [latex]\\underset{x\\to a}{\\lim}f(x)=2[\/latex].<\/p>\n<div style=\"width: 727px\" 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\/11211025\/CNX_Calc_Figure_04_06_019.jpg\" alt=\"The function f(x) 2 + 1\/x is graphed. The function starts negative near y = 2 but then decreases to \u2212\u221e near x = 0. The function then decreases from \u221e near x = 0 and gets nearer to y = 2 as x increases. There is a horizontal line denoting the asymptote y = 2.\" width=\"717\" height=\"423\" \/><\/p>\n<p class=\"wp-caption-text\">The function approaches the asymptote [latex]y=2[\/latex] as [latex]x[\/latex] approaches [latex]\\pm \\infty[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1165042936244\">More generally, for any function [latex]f[\/latex], we say the limit as [latex]x \\to \\infty[\/latex] of [latex]f(x)[\/latex] is [latex]L[\/latex] if [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] as long as [latex]x[\/latex] is sufficiently large. In that case, we write [latex]\\underset{x\\to \\infty}{\\lim}f(x)=L[\/latex]. Similarly, we say the limit as [latex]x\\to \u2212\\infty[\/latex] of [latex]f(x)[\/latex] is [latex]L[\/latex] if [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] as long as [latex]x<0[\/latex] and [latex]|x|[\/latex] is sufficiently large. In that case, we write [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=L[\/latex]. We now look at the definition of a function having a limit at infinity.<\/p>\n<div id=\"fs-id1165042331960\" class=\"textbox shaded\">\n<div class=\"title\">\n<p style=\"text-align: center;\"><span style=\"font-size: 1.2em; font-weight: 600; text-align: center; text-transform: uppercase; background-color: #eeeeee;\">Definition<\/span><\/p>\n<hr \/>\n<\/div>\n<p id=\"fs-id1165042970725\">(Informal) If the values of [latex]f(x)[\/latex] become arbitrarily close to [latex]L[\/latex] as [latex]x[\/latex] becomes sufficiently large, we say the function [latex]f[\/latex] has a limit at infinity and write<\/p>\n<div id=\"fs-id1165042986551\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}f(x)=L[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042374662\">If the values of [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] for [latex]x<0[\/latex] as [latex]|x|[\/latex] becomes sufficiently large, we say that the function [latex]f[\/latex] has a limit at negative infinity and write<\/p>\n<div id=\"fs-id1165043105208\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to -\\infty }{\\lim}f(x)=L[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbook exercises\">\n<h3>Example: Computing Limits at Infinity<\/h3>\n<p id=\"fs-id1165043262623\">For each of the following functions [latex]f[\/latex], evaluate [latex]\\underset{x\\to \\infty }{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)[\/latex].<\/p>\n<ol id=\"fs-id1165042356111\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]f(x)=5-\\frac{2}{x^2}[\/latex]<\/li>\n<li>[latex]f(x)=\\dfrac{\\sin x}{x}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043183885\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043183885\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165043183885\" style=\"list-style-type: lower-alpha;\">\n<li>Using the algebraic limit laws, we have [latex]\\underset{x\\to \\infty }{\\lim}(5-\\frac{2}{x^2})=\\underset{x\\to \\infty }{\\lim}5-2(\\underset{x\\to \\infty }{\\lim}\\frac{1}{x})(\\underset{x\\to \\infty }{\\lim}\\frac{1}{x})=5-2 \\cdot 0=5[\/latex]. Similarly, [latex]\\underset{x\\to -\\infty }{\\lim}f(x)=5[\/latex].\n<div class=\"mceTemp\"><\/div>\n<\/li>\n<li>nce [latex]-1\\le \\sin x\\le 1[\/latex] for all [latex]x[\/latex], we have\n<div id=\"fs-id1165043093355\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{-1}{x}\\le \\frac{\\sin x}{x}\\le \\frac{1}{x}[\/latex]<\/div>\n<p>for all [latex]x \\ne 0[\/latex]. Also, since<\/p>\n<div id=\"fs-id1165043197153\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\frac{-1}{x}=0=\\underset{x\\to \\infty }{\\lim}\\frac{1}{x}[\/latex],<\/div>\n<p>we can apply the squeeze theorem to conclude that<\/p>\n<div id=\"fs-id1165043036581\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\frac{\\sin x}{x}=0[\/latex]<\/div>\n<p>Similarly,<\/p>\n<div id=\"fs-id1165043122536\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \u2212\\infty}{\\lim}\\frac{\\sin x}{x}=0[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042320881\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165043315935\">Evaluate [latex]\\underset{x\\to \u2212\\infty}{\\lim}\\left(3+\\frac{4}{x}\\right)[\/latex] and [latex]\\underset{x\\to \\infty }{\\lim}\\left(3+\\frac{4}{x}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q2473508\">Hint<\/span><\/p>\n<div id=\"q2473508\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042318511\">[latex]\\underset{x\\to \\pm \\infty }{\\lim}1\/x=0[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043390798\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043390798\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043390798\">Both limits are 3.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042333169\" class=\"bc-section section\">\n<p id=\"fs-id1165042333174\">Sometimes the values of a function [latex]f[\/latex] become arbitrarily large as [latex]x\\to \\infty[\/latex] (or as [latex]x\\to \u2212\\infty )[\/latex]. In this case, we write [latex]\\underset{x\\to \\infty }{\\lim}f(x)=\\infty[\/latex] (or [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=\\infty )[\/latex]. On the other hand, if the values of [latex]f[\/latex] are negative but become arbitrarily large in magnitude as [latex]x\\to \\infty[\/latex] (or as [latex]x\\to \u2212\\infty )[\/latex], we write [latex]\\underset{x\\to \\infty }{\\lim}f(x)=\u2212\\infty[\/latex] (or [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=\u2212\\infty )[\/latex].<\/p>\n<p id=\"fs-id1165042606820\">For example, consider the function [latex]f(x)=x^3[\/latex]. The [latex]\\underset{x\\to \\infty }{\\lim}x^3=\\infty[\/latex]. On the other hand, as [latex]x\\to \u2212\\infty[\/latex], the values of [latex]f(x)=x^3[\/latex] are negative but become arbitrarily large in magnitude. Consequently, [latex]\\underset{x\\to \u2212\\infty }{\\lim}x^3=\u2212\\infty[\/latex].<\/p>\n<table id=\"fs-id1165042406634\" class=\"column-header\" summary=\"The table has four rows and six columns. The first column is a header column and it reads x, x3, x, and x3. After the header, the first row reads 10, 20, 50, 100, and 1000. The second row reads 1000, 8000, 125000, 1,000,000, and 1,000,000,000. The third row reads \u221210, \u221220, \u221250, \u2212100, and \u22121000. The forth row reads \u22121000, \u22128000, \u2212125,000, \u22121,000,000, and \u22121,000,000,000.\">\n<caption>Values of a power function as [latex]x\\to \\pm \\infty[\/latex]<\/caption>\n<tbody>\n<tr valign=\"top\">\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>10<\/td>\n<td>20<\/td>\n<td>50<\/td>\n<td>100<\/td>\n<td>1000<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td><strong>[latex]x^3[\/latex]<\/strong><\/td>\n<td>1000<\/td>\n<td>8000<\/td>\n<td>125,000<\/td>\n<td>1,000,000<\/td>\n<td>1,000,000,000<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>-10<\/td>\n<td>-20<\/td>\n<td>-50<\/td>\n<td>-100<\/td>\n<td>-1000<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td><strong>[latex]x^3[\/latex]<\/strong><\/td>\n<td>-1000<\/td>\n<td>-8000<\/td>\n<td>-125,000<\/td>\n<td>-1,000,000<\/td>\n<td>-1,000,000,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div>\n<div style=\"width: 652px\" 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\/11211045\/CNX_Calc_Figure_04_06_022.jpg\" alt=\"The function f(x) = x3 is graphed. It is apparent that this function rapidly approaches infinity as x approaches infinity.\" width=\"642\" height=\"272\" \/><\/p>\n<p class=\"wp-caption-text\">For this function, the functional values approach infinity as [latex]x\\to \\pm \\infty[\/latex].<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043276353\" class=\"textbox shaded\">\n<div class=\"title\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<\/div>\n<p id=\"fs-id1165043276356\">(Informal) We say a function [latex]f[\/latex] has an infinite limit at infinity and write<\/p>\n<div id=\"fs-id1165043276364\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}f(x)=\\infty[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042709557\">if [latex]f(x)[\/latex] becomes arbitrarily large for [latex]x[\/latex] sufficiently large. We say a function has a negative infinite limit at infinity and write<\/p>\n<div id=\"fs-id1165042647077\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}f(x)=\u2212\\infty[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042327355\">if [latex]f(x)<0[\/latex] and [latex]|f(x)|[\/latex] becomes arbitrarily large for [latex]x[\/latex] sufficiently large. Similarly, we can define infinite limits as [latex]x\\to \u2212\\infty[\/latex].<\/p>\n<\/div>\n<h2>Apply L&#8217;Hopital&#8217;s Rule<\/h2>\n<h3>Indeterminate Form of Type [latex]\\frac{0}{0}[\/latex]<\/h3>\n<div id=\"fs-id1165043085155\" class=\"bc-section section\">\n<p id=\"fs-id1165042941863\">L\u2019H\u00f4pital\u2019s rule can be used to evaluate limits involving the quotient of two functions. Consider<\/p>\n<div id=\"fs-id1165042458008\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042613148\">If [latex]\\underset{x\\to a}{\\lim}f(x)=L_1[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=L_2 \\ne 0[\/latex], then<\/p>\n<div id=\"fs-id1165043088102\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\dfrac{L_1}{L_2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042330308\">However, what happens if [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex]? We call this one of the <strong>indeterminate forms<\/strong>, of type [latex]\\frac{0}{0}[\/latex]. This is considered an indeterminate form because we cannot determine the exact behavior of [latex]\\frac{f(x)}{g(x)}[\/latex] as [latex]x\\to a[\/latex] without further analysis.<\/p>\n<\/div>\n<div id=\"fs-id1165043085155\" class=\"bc-section section\">\n<div id=\"fs-id1165043352593\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">L\u2019H\u00f4pital\u2019s Rule (0\/0 Case)<\/h3>\n<hr \/>\n<p id=\"fs-id1165043276140\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval containing [latex]a[\/latex], except possibly at [latex]a[\/latex]. If [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex], then<\/p>\n<div id=\"fs-id1165043020148\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<p id=\"fs-id1165043035673\">assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if we are considering one-sided limits, or if [latex]a=\\infty[\/latex] or [latex]-\\infty[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165042712928\" class=\"bc-section section\">\n<div id=\"fs-id1165043397578\" class=\"textbook exercises\">\n<h3>Example: Applying L\u2019H\u00f4pital\u2019s Rule (0\/0 Case)<\/h3>\n<p id=\"fs-id1165042456913\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{1- \\cos x}{x}[\/latex]<\/li>\n<li>[latex]\\underset{x\\to 1}{\\lim}\\dfrac{\\sin (\\pi x)}{\\ln x}[\/latex]<\/li>\n<li>[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{e^{\\frac{1}{x}}-1}{\\frac{1}{x}}[\/latex]<\/li>\n<li>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{\\sin x-x}{x^2}[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1165043104016\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042535045\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042535045\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165042535045\" style=\"list-style-type: lower-alpha;\">\n<li>Since the numerator [latex]1- \\cos x\\to 0[\/latex] and the denominator [latex]x\\to 0[\/latex], we can apply L\u2019H\u00f4pital\u2019s rule to evaluate this limit. We have\n<div id=\"fs-id1165042328696\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to 0}{\\lim}\\frac{1- \\cos x}{x} & =\\underset{x\\to 0}{\\lim}\\frac{\\frac{d}{dx}(1- \\cos x)}{\\frac{d}{dx}(x)} \\\\ & =\\underset{x\\to 0}{\\lim}\\frac{\\sin x}{1} \\\\ & =\\frac{\\underset{x\\to 0}{\\lim}(\\sin x)}{\\underset{x\\to 0}{\\lim}(1)} \\\\ & =\\frac{0}{1}=0 \\end{array}[\/latex]<\/div>\n<\/li>\n<li>As [latex]x\\to 1[\/latex], the numerator [latex]\\sin (\\pi x)\\to 0[\/latex] and the denominator [latex]\\ln x \\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\n<div id=\"fs-id1165043116372\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to 1}{\\lim}\\frac{\\sin (\\pi x)}{\\ln x} & =\\underset{x\\to 1}{\\lim}\\frac{\\pi \\cos (\\pi x)}{1\/x} \\\\ & =\\underset{x\\to 1}{\\lim}(\\pi x) \\cos (\\pi x) \\\\ & =(\\pi \\cdot 1)(-1)=\u2212\\pi \\end{array}[\/latex]<\/div>\n<\/li>\n<li>As [latex]x\\to \\infty[\/latex], the numerator [latex]e^{1\/x}-1\\to 0[\/latex] and the denominator [latex](\\frac{1}{x})\\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\n<div id=\"fs-id1165042327460\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\frac{e^{1\/x}-1}{\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\frac{e^{1\/x}(\\frac{-1}{x^2})}{(\\frac{-1}{x^2})}=\\underset{x\\to \\infty}{\\lim} e^{1\/x}=e^0=1[\/latex]<\/div>\n<\/li>\n<li>As [latex]x\\to 0[\/latex], both the numerator and denominator approach zero. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\n<div id=\"fs-id1165042562974\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x-x}{x^2}=\\underset{x\\to 0}{\\lim}\\frac{\\cos x-1}{2x}[\/latex].<\/div>\n<p>Since the numerator and denominator of this new quotient both approach zero as [latex]x\\to 0[\/latex], we apply L\u2019H\u00f4pital\u2019s rule again. In doing so, we see that<\/p>\n<div id=\"fs-id1165043281524\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\cos x-1}{2x}=\\underset{x\\to 0}{\\lim}\\frac{\u2212\\sin x}{2}=0[\/latex].<\/div>\n<p>Therefore, we conclude that<\/p>\n<div id=\"fs-id1165043284142\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x-x}{x^2}=0[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\underset{x\\to 0}{\\lim}\\dfrac{x}{\\tan x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q37002811\">Hint<\/span><\/p>\n<div id=\"q37002811\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{d}{dx} \\tan x= \\sec ^2 x[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042377480\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042377480\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]1[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<h3>Indeterminate Form of Type [latex]\\frac{\\infty}{\\infty}[\/latex]<\/h3>\n<p id=\"fs-id1165042318672\">We can also use L\u2019H\u00f4pital\u2019s rule to evaluate limits of quotients [latex]\\frac{f(x)}{g(x)}[\/latex] in which [latex]f(x)\\to \\pm \\infty[\/latex] and [latex]g(x)\\to \\pm \\infty[\/latex]. Limits of this form are classified as <em>indeterminate forms of type<\/em> [latex]\\infty \/ \\infty[\/latex]. Again, note that we are not actually dividing [latex]\\infty[\/latex] by [latex]\\infty[\/latex]. Since [latex]\\infty[\/latex] is not a real number, that is impossible; rather, [latex]\\infty \/ \\infty[\/latex] is used to represent a quotient of limits, each of which is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex].<\/p>\n<div id=\"fs-id1165042330826\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/ \\infty[\/latex] Case)<\/h3>\n<hr \/>\n<p id=\"fs-id1165043426174\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval containing [latex]a[\/latex], except possibly at [latex]a[\/latex]. Suppose [latex]\\underset{x\\to a}{\\lim}f(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]) and [latex]\\underset{x\\to a}{\\lim}g(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]). Then,<\/p>\n<div id=\"fs-id1165042373703\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<p id=\"fs-id1165042707280\">assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if the limit is infinite, if [latex]a=\\infty[\/latex] or [latex]\u2212\\infty[\/latex], or the limit is one-sided.<\/p>\n<\/div>\n<div id=\"fs-id1165043427623\" class=\"textbook exercises\">\n<h3>Example: Applying L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/\\infty[\/latex] Case)<\/h3>\n<p id=\"fs-id1165042709794\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\n<ol id=\"fs-id1165042709798\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}[\/latex]<\/li>\n<li>[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1165043427625\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042376758\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042376758\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165042376758\" style=\"list-style-type: lower-alpha;\">\n<li>Since [latex]3x+5[\/latex] and [latex]2x+1[\/latex] are first-degree polynomials with positive leading coefficients, [latex]\\underset{x\\to \\infty }{\\lim}(3x+5)=\\infty[\/latex] and [latex]\\underset{x\\to \\infty }{\\lim}(2x+1)=\\infty[\/latex]. Therefore, we apply L\u2019H\u00f4pital\u2019s rule and obtain\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim}\\dfrac{3x+5}{2x+\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3}{2}=\\dfrac{3}{2}[\/latex].<\/div>\n<p>Note that this limit can also be calculated without invoking L\u2019H\u00f4pital\u2019s rule. Earlier in the module we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of [latex]x[\/latex] in the denominator. In doing so, we saw that<\/p>\n<div id=\"fs-id1165042319986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3+\\frac{5}{x}}{2+\\frac{1}{x}}=\\dfrac{3}{2}[\/latex].<\/div>\n<p>L\u2019H\u00f4pital\u2019s rule provides us with an alternative means of evaluating this type of limit.<\/li>\n<li>Here, [latex]\\underset{x\\to 0^+}{\\lim} \\ln x=\u2212\\infty[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} \\cot x=\\infty[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule and obtain\n<div id=\"fs-id1165042374803\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\frac{1}{x}}{\u2212\\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}[\/latex].<\/div>\n<p>Now as [latex]x\\to 0^+[\/latex], [latex]\\csc^2 x\\to \\infty[\/latex]. Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of [latex]\\csc x[\/latex] to write<\/p>\n<div id=\"fs-id1165042376466\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}[\/latex].<\/div>\n<p>Now [latex]\\underset{x\\to 0^+}{\\lim} \\sin^2 x=0[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} x=0[\/latex], so we apply L\u2019H\u00f4pital\u2019s rule again. We find<\/p>\n<div id=\"fs-id1165043249678\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{2 \\sin x \\cos x}{-1}=\\dfrac{0}{-1}=0[\/latex].<\/div>\n<p>We conclude that<\/p>\n<div id=\"fs-id1165042315721\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=0[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042333392\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{\\ln x}{5x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q30011179\">Hint<\/span><\/p>\n<div id=\"q30011179\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{d}{dx}\\ln x=\\frac{1}{x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042367881\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042367881\" class=\"hidden-answer\" style=\"display: none\">\n<p>0<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165042331798\">L\u2019H\u00f4pital\u2019s rule is very useful for evaluating limits involving the indeterminate forms [latex]\\frac{0}{0}[\/latex] and [latex]\\frac{\\infty}{\\infty}[\/latex]. However, we can also use L\u2019H\u00f4pital\u2019s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits. The expressions [latex]0 \\cdot \\infty[\/latex], [latex]\\infty - \\infty[\/latex], [latex]1^{\\infty}[\/latex], [latex]\\infty^0[\/latex], and [latex]0^0[\/latex] are all considered indeterminate forms. These expressions are not real numbers. Rather, they represent forms that arise when trying to evaluate certain limits. Next we realize why these are indeterminate forms and then understand how to use L\u2019H\u00f4pital\u2019s rule in these cases. The key idea is that we must rewrite the indeterminate forms in such a way that we arrive at the indeterminate form [latex]\\frac{0}{0}[\/latex] or [latex]\\frac{\\infty}{\\infty}[\/latex].<\/p>\n<div id=\"fs-id1165042320288\" class=\"bc-section 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]. 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. 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>&nbsp;<\/p>\n<p id=\"fs-id1165042323522\">As [latex]x\\to \\infty[\/latex], [latex]f(x)\\to 0[\/latex] and [latex]g(x)\\to \\infty[\/latex]. 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]. 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<div id=\"fs-id1165042368495\" class=\"textbook exercises\">\n<h3>Example: Indeterminate Form of Type [latex]0\u00b7\\infty[\/latex]<\/h3>\n<p>Evaluate [latex]\\underset{x\\to 0^+}{\\lim}x \\ln x[\/latex]<\/p>\n<div id=\"fs-id1165042368497\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042545829\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042545829\" 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<div><\/div>\n<div>\n<div style=\"width: 368px\" 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\" \/><\/p>\n<p class=\"wp-caption-text\">Finding the limit at [latex]x=0[\/latex] of the function [latex]f(x)=x \\ln x[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043430905\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\underset{x\\to 0}{\\lim}x \\cot x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q404616\">Hint<\/span><\/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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043286669\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043286669\" class=\"hidden-answer\" style=\"display: none\">\n<p>1<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042318802\" class=\"bc-section section\">\n<h3>Indeterminate Form of Type [latex]\\infty -\\infty[\/latex]<\/h3>\n<p id=\"fs-id1165042318816\">Another type of indeterminate form is [latex]\\infty -\\infty[\/latex]. Consider the following 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]. 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]. 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. 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. For example, 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>&nbsp;<\/p>\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>&nbsp;<\/p>\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>&nbsp;<\/p>\n<p id=\"fs-id1165043323851\">Therefore, the limit cannot be determined by considering only [latex]\\infty -\\infty[\/latex]. 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.<\/p>\n<div id=\"fs-id1165043323875\" class=\"textbook exercises\">\n<h3>Example: Indeterminate Form of Type [latex]\\infty -\\infty[\/latex]<\/h3>\n<p>Evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\left(\\dfrac{1}{x^2}-\\dfrac{1}{\\tan x}\\right)[\/latex].<\/p>\n<div id=\"fs-id1165043323877\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043281296\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043281296\" 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>&nbsp;<\/p>\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>&nbsp;<\/p>\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>&nbsp;<\/p>\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]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043348549\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q1238807\">Hint<\/span><\/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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043317356\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043317356\" class=\"hidden-answer\" style=\"display: none\">\n<p>0<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Other Types of Indeterminate Form<\/h3>\n<p id=\"fs-id1165042364139\">Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions [latex]0^0[\/latex], [latex]\\infty^0[\/latex], and [latex]1^{\\infty}[\/latex] are all indeterminate forms. On their own, these expressions are meaningless because we cannot actually evaluate these expressions as we would evaluate an expression involving real numbers. Rather, these expressions represent forms that arise when finding limits. Now we examine how L\u2019H\u00f4pital\u2019s rule can be used to evaluate limits involving these indeterminate forms.<\/p>\n<p id=\"fs-id1165042364178\">Since L\u2019H\u00f4pital\u2019s rule applies to quotients, we use the natural logarithm function and its properties to reduce a problem evaluating a limit involving exponents to a related problem involving a limit of a quotient. For example, suppose we want to evaluate [latex]\\underset{x\\to a}{\\lim}f(x)^{g(x)}[\/latex] and we arrive at the indeterminate form [latex]\\infty^0[\/latex]. (The indeterminate forms [latex]0^0[\/latex] and [latex]1^{\\infty}[\/latex] can be handled similarly.)<\/p>\n<div id=\"fs-id1165043390815\" class=\"textbook exercises\">\n<h3>Example: Indeterminate Form of Type [latex]\\infty^0[\/latex]<\/h3>\n<p>Evaluate [latex]\\underset{x\\to \\infty }{\\lim} x^{\\frac{1}{x}}[\/latex]<\/p>\n<div id=\"fs-id1165043390817\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043390866\">Show Solution<\/span><\/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<div class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042407362\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\underset{x\\to \\infty}{\\lim} x^{\\frac{1}{\\ln x}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3762844\">Hint<\/span><\/p>\n<div id=\"q3762844\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]y=x^{1\/ \\ln x}[\/latex] and apply the natural logarithm to both sides of the equation.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043108248\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043108248\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]e[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043108292\" class=\"textbook exercises\">\n<h3>Example: Indeterminate Form of Type [latex]0^0[\/latex]<\/h3>\n<p>Evaluate [latex]\\underset{x\\to 0^+}{\\lim} x^{\\sin x}[\/latex]<\/p>\n<div id=\"fs-id1165043108295\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042657755\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042657755\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042657755\">Let<\/p>\n<div id=\"fs-id1165042657759\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=x^{\\sin x}[\/latex]<\/div>\n<p id=\"fs-id1165042657780\">Therefore,<\/p>\n<div id=\"fs-id1165042657783\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln y=\\ln (x^{\\sin x})= \\sin x \\ln x[\/latex]<\/div>\n<p id=\"fs-id1165042707171\">We now evaluate [latex]\\underset{x\\to 0^+}{\\lim} \\sin x \\ln x[\/latex]. Since [latex]\\underset{x\\to 0^+}{\\lim} \\sin x=0[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} \\ln x=\u2212\\infty[\/latex], we have the indeterminate form [latex]0 \\cdot \\infty[\/latex]. To apply L\u2019H\u00f4pital\u2019s rule, we need to rewrite [latex]\\sin x \\ln x[\/latex] as a fraction. We could write<\/p>\n<div id=\"fs-id1165043173832\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin x \\ln x=\\frac{\\sin x}{1\/ \\ln x}[\/latex]<\/div>\n<p id=\"fs-id1165043173865\">or<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin x \\ln x=\\frac{\\ln x}{1\/ \\sin x}=\\frac{\\ln x}{\\csc x}[\/latex]<\/div>\n<p id=\"fs-id1165042364489\">Let\u2019s consider the first option. In this case, applying L\u2019H\u00f4pital\u2019s rule, we would obtain<\/p>\n<div id=\"fs-id1165042364494\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} \\sin x \\ln x=\\underset{x\\to 0^+}{\\lim}\\frac{\\sin x}{1\/ \\ln x}=\\underset{x\\to 0^+}{\\lim}\\frac{\\cos x}{-1\/(x(\\ln x)^2)}=\\underset{x\\to 0^+}{\\lim}(\u2212x(\\ln x)^2 \\cos x)[\/latex]<\/div>\n<p id=\"fs-id1165043317274\">Unfortunately, we not only have another expression involving the indeterminate form [latex]0 \\cdot \\infty[\/latex], but the new limit is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By writing<\/p>\n<div id=\"fs-id1165043131555\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin x \\ln x=\\frac{\\ln x}{1\/ \\sin x}=\\frac{\\ln x}{\\csc x}[\/latex]<\/div>\n<p id=\"fs-id1165043131604\">and applying L\u2019H\u00f4pital\u2019s rule, we obtain<\/p>\n<div id=\"fs-id1165043131609\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} \\sin x \\ln x=\\underset{x\\to 0^+}{\\lim}\\frac{\\ln x}{\\csc x}=\\underset{x\\to 0^+}{\\lim}\\frac{1\/x}{\u2212 \\csc x \\cot x}=\\underset{x\\to 0^+}{\\lim}\\frac{-1}{x \\csc x \\cot x}[\/latex]<\/div>\n<p id=\"fs-id1165042651533\">Using the fact that [latex]\\csc x=\\frac{1}{\\sin x}[\/latex] and [latex]\\cot x=\\frac{\\cos x}{\\sin x}[\/latex], we can rewrite the expression on the right-hand side as<\/p>\n<div id=\"fs-id1165043251999\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{\u2212\\sin^2 x}{x \\cos x}=\\underset{x\\to 0^+}{\\lim}[\\frac{\\sin x}{x} \\cdot (\u2212\\tan x)]=(\\underset{x\\to 0^+}{\\lim}\\frac{\\sin x}{x}) \\cdot (\\underset{x\\to 0^+}{\\lim}(\u2212\\tan x))=1 \\cdot 0=0[\/latex]<\/div>\n<p id=\"fs-id1165042676314\">We conclude that [latex]\\underset{x\\to 0^+}{\\lim} \\ln y=0[\/latex]. Therefore, [latex]\\ln (\\underset{x\\to 0^+}{\\lim} y)=0[\/latex] and we have<\/p>\n<div id=\"fs-id1165042327527\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} y=\\underset{x\\to 0^+}{\\lim} x^{\\sin x}=e^0=1[\/latex]<\/div>\n<p id=\"fs-id1165042327592\">Hence,<\/p>\n<div id=\"fs-id1165042327595\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} x^{\\sin x}=1[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042660254\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\underset{x\\to 0^+}{\\lim} x^x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q929037\">Hint<\/span><\/p>\n<div id=\"q929037\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]y=x^x[\/latex] and take the natural logarithm of both sides of the equation.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042660293\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042660293\" class=\"hidden-answer\" style=\"display: none\">\n<p>1<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1989\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Modification and Revision. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra Corequisite\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"\",\"organization\":\"Lumen 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