{"id":2282,"date":"2021-09-21T15:11:49","date_gmt":"2021-09-21T15:11:49","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-sequences\/"},"modified":"2021-11-19T03:15:14","modified_gmt":"2021-11-19T03:15:14","slug":"skills-review-for-sequences","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-sequences\/","title":{"raw":"Skills Review for Sequences","rendered":"Skills Review for Sequences"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write the terms of a sequence defined by a recursive formula<\/li>\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<\/ul>\r\n<\/div>\r\nIn the Sequences section, we will look at ordered lists of numbers (sequences) and determine whether they converge or diverge. Here we will review how to evaluate a recursive (recurrence) formula, take limits at infinity, and apply L\u2019H\u00f4pital\u2019s Rule.\r\n<h2>Use a Recursive Formula<\/h2>\r\nA <strong>recursive formula<\/strong> is\u00a0a formula that defines its value at a particular input using the result of the previous input(s).\r\n\r\nA recursive formula always has two parts: the value of an initial input and an equation defining each term in terms of preceding terms. For example, suppose we know the following:\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;{a}_{1}=3 \\\\ &amp;{a}_{n}=2{a}_{n - 1}-1, \\text{ for } n\\ge 2 \\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div style=\"text-align: left;\">We can find the subsequent terms of the recursive formula using the first term.<\/div>\r\n<div><\/div>\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;{a}_{1}=3\\\\ &amp;{a}_{2}=2{a}_{1}-1=2\\left(3\\right)-1=5\\\\ &amp;{a}_{3}=2{a}_{2}-1=2\\left(5\\right)-1=9\\\\ &amp;{a}_{4}=2{a}_{3}-1=2\\left(9\\right)-1=17\\end{align}[\/latex]<\/div>\r\nSo, the first four terms are [latex]3,5,9,\\text{ and},17[\/latex].\r\n<div class=\"textbox\">\r\n<h3>How To: Given a recursive formula with only the first term provided, write the first [latex]n[\/latex] terms of a sequence.<\/h3>\r\n<ol>\r\n \t<li>Identify the initial term, [latex]{a}_{1}[\/latex], which is given as part of the formula. This is the first term.<\/li>\r\n \t<li>To find the second term, [latex]{a}_{2}[\/latex], substitute the initial term into the formula for [latex]{a}_{n - 1}[\/latex]. Solve.<\/li>\r\n \t<li>To find the third term, [latex]{a}_{3}[\/latex], substitute the second term into the formula. Solve.<\/li>\r\n \t<li>Repeat until you have solved for the [latex]n\\text{th}[\/latex] term.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing the Terms of a Sequence Defined by a Recursive Formula<\/h3>\r\nWrite the first five terms of the sequence defined by the recursive formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} {a}_{1}&amp;=9 \\\\ {a}_{n}&amp;=3{a}_{n - 1}-20\\text{, for }n\\ge 2 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"748916\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"748916\"]\r\n\r\nThe first term is given in the formula. For each subsequent term, we replace [latex]{a}_{n - 1}[\/latex] with the value of the preceding term.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;n=1 &amp;&amp; {a}_{1}=9 \\\\ &amp;n=2 &amp;&amp; {a}_{2}=3{a}_{1}-20=3\\left(9\\right)-20=27 - 20=7 \\\\ &amp;n=3 &amp;&amp; {a}_{3}=3{a}_{2}-20=3\\left(7\\right)-20=21 - 20=1 \\\\ &amp;n=4 &amp;&amp; {a}_{4}=3{a}_{3}-20=3\\left(1\\right)-20=3 - 20=-17 \\\\ &amp;n=5 &amp;&amp; {a}_{5}=3{a}_{4}-20=3\\left(-17\\right)-20=-51 - 20=-71 \\end{align}[\/latex]<\/p>\r\nThe first five terms are [latex]\\left\\{9,7,1,-17,-71\\right\\}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWrite the first five terms of the sequence defined by the recursive formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{a}_{1}&amp;=2\\\\ {a}_{n}&amp;=2{a}_{n - 1}+1\\text{, for }n\\ge 2\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"378600\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"378600\"]\r\n\r\n[latex]\\left\\{2, 5, 11, 23, 47\\right\\}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5812&amp;theme=oea&amp;iframe_resize_id=mom21[\/embed]\r\n\r\n<\/div>\r\n<h2>Take 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 in Figure 1 and numerically in the table beneath it, 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\" \/> Figure 1. The function approaches the asymptote [latex]y=2[\/latex] as [latex]x[\/latex] approaches [latex]\\pm \\infty[\/latex].[\/caption]\r\n<table id=\"fs-id1165043428402\" class=\"column-header\" summary=\"The table has four rows and five columns. The first column is a header column and it reads x, 2 + 1\/x, x, and 2 + 1\/x. After the header, the first row reads 10, 100, 1000, and 10000. The second row reads 2.1, 2.01, 2.001, and 2.0001. The third row reads \u221210, \u2212100, \u22121000, and \u221210000. The fourth row reads 1.9, 1.99, 1.999, and 1.9999.\"><caption>Values of a function [latex]f[\/latex] 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>100<\/td>\r\n<td>1,000<\/td>\r\n<td>10,000<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td><strong>[latex]2+\\frac{1}{x}[\/latex]<\/strong><\/td>\r\n<td>2.1<\/td>\r\n<td>2.01<\/td>\r\n<td>2.001<\/td>\r\n<td>2.0001<\/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>-100<\/td>\r\n<td>-1000<\/td>\r\n<td>-10,000<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td><strong>[latex]2+\\frac{1}{x}[\/latex]<\/strong><\/td>\r\n<td>1.9<\/td>\r\n<td>1.99<\/td>\r\n<td>1.999<\/td>\r\n<td>1.9999<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\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<h3 style=\"text-align: center;\">Definition<\/h3>\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<p id=\"fs-id1165043157752\">If the values [latex]f(x)[\/latex] are getting arbitrarily close to some finite value [latex]L[\/latex] as [latex]x\\to \\infty [\/latex] or [latex]x\\to \u2212\\infty[\/latex], the graph of [latex]f[\/latex] approaches the line [latex]y=L[\/latex]. In that case, the line [latex]y=L[\/latex] is a horizontal asymptote of [latex]f[\/latex] (Figure 2). For example, for the function [latex]f(x)=\\frac{1}{x}[\/latex], since [latex]\\underset{x\\to \\infty }{\\lim}f(x)=0[\/latex], the line [latex]y=0[\/latex] is a horizontal asymptote of [latex]f(x)=\\frac{1}{x}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165043262534\" 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-id1165042973921\">If [latex]\\underset{x\\to \\infty }{\\lim}f(x)=L[\/latex] or [latex]\\underset{x \\to \u2212\\infty}{\\lim}f(x)=L[\/latex], we say the line [latex]y=L[\/latex] is a <strong>horizontal asymptote<\/strong> of [latex]f[\/latex].<\/p>\r\n\r\n<\/div>\r\n[caption id=\"\" align=\"aligncenter\" width=\"766\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211028\/CNX_Calc_Figure_04_06_020.jpg\" alt=\"The figure is broken up into two figures labeled a and b. Figure a shows a function f(x) approaching but never touching a horizontal dashed line labeled L from above. Figure b shows a function f(x) approaching but never a horizontal dashed line labeled M from below.\" width=\"766\" height=\"273\" \/> Figure 2. (a) As [latex]x\\to \\infty[\/latex], the values of [latex]f[\/latex] are getting arbitrarily close to [latex]L[\/latex]. The line [latex]y=L[\/latex] is a horizontal asymptote of [latex]f[\/latex]. (b) As [latex]x\\to \u2212\\infty[\/latex], the values of [latex]f[\/latex] are getting arbitrarily close to [latex]M[\/latex]. The line [latex]y=M[\/latex] is a horizontal asymptote of [latex]f[\/latex].[\/caption]\r\n<p id=\"fs-id1165042647732\">A function cannot cross a vertical asymptote because the graph must approach infinity (or negative infinity) from at least one direction as [latex]x[\/latex] approaches the vertical asymptote. However, a function may cross a horizontal asymptote. In fact, a function may cross a horizontal asymptote an unlimited number of times. For example, the function [latex]f(x)=\\frac{ \\cos x}{x}+1[\/latex] shown in Figure 3 intersects the horizontal asymptote [latex]y=1[\/latex] an infinite number of times as it oscillates around the asymptote with ever-decreasing amplitude.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"529\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211031\/CNX_Calc_Figure_04_06_002.jpg\" alt=\"The function f(x) = (cos x)\/x + 1 is shown. It decreases from (0, \u221e) and then proceeds to oscillate around y = 1 with decreasing amplitude.\" width=\"529\" height=\"230\" \/> Figure 3. The graph of [latex]f(x)=\\cos x\/x+1[\/latex] crosses its horizontal asymptote [latex]y=1[\/latex] an infinite number of times.[\/caption]\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 \t<li>[latex]f(x)= \\tan^{-1} (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].<\/li>\r\n \t<li>Since [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>\r\n&nbsp;<\/li>\r\n \t<li>To evaluate [latex]\\underset{x\\to \\infty }{\\lim} \\tan^{-1} (x)[\/latex] and [latex]\\underset{x\\to \u2212\\infty}{\\lim} \\tan^{-1} (x)[\/latex], we first consider the graph of [latex]y= \\tan (x)[\/latex] over the interval [latex](\u2212\\pi \/2,\\pi \/2)[\/latex] as shown in the following graph.[caption id=\"\" align=\"aligncenter\" width=\"492\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211039\/CNX_Calc_Figure_04_06_021.jpg\" alt=\"The function f(x) = tan x is shown. It increases from (\u2212\u03c0\/2, \u2212\u221e), passes through the origin, and then increases toward (\u03c0\/2, \u221e). There are vertical dashed lines marking x = \u00b1\u03c0\/2.\" width=\"492\" height=\"347\" \/> Figure 6. The graph of [latex] \\tan x[\/latex] has vertical asymptotes at [latex]x=\\pm \\frac{\\pi }{2}[\/latex][\/caption]<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1165043092430\">Since<\/p>\r\n\r\n<div id=\"fs-id1165043119614\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to (\\pi\/2)^-}{\\lim} \\tan x=\\infty [\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042514177\">it follows that<\/p>\r\n\r\n<div id=\"fs-id1165042563973\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim} \\tan^{-1} (x)=\\frac{\\pi }{2}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043097156\">Similarly, since<\/p>\r\n\r\n<div id=\"fs-id1165042923290\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to (\\pi\/2)^+}{\\lim} \\tan x=\u2212\\infty[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043131939\">it follows that<\/p>\r\n\r\n<div id=\"fs-id1165043056813\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \u2212\\infty}{\\lim} \\tan^{-1} (x)=-\\frac{\\pi }{2}[\/latex]<\/div>\r\n<span style=\"font-size: 1rem; text-align: initial;\">[\/hidden-answer]<\/span>\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]. Determine the horizontal asymptotes of [latex]f(x)=3+\\frac{4}{x}[\/latex], if any.<\/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<h2>Infinite Limits at Infinity<\/h2>\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]. As seen in the table below and Figure 2, as [latex]x\\to \\infty [\/latex] the values [latex]f(x)[\/latex] become arbitrarily large. Therefore, [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\" \/> Figure 2. For this function, the functional values approach infinity as [latex]x\\to \\pm \\infty[\/latex].[\/caption]<\/div>\r\n<\/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<\/div>\r\n<div id=\"fs-id1165042323710\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165042320226\">Find [latex]\\underset{x\\to \\infty }{\\lim}3x^2[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1165042708272\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042708272\"]\r\n<p id=\"fs-id1165042383154\">[latex]\\infty[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]16109[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Apply L'H\u00f4pital's Rule<\/h2>\r\n<p id=\"fs-id1165042941863\">L\u2019H\u00f4pital\u2019s rule can be used to evaluate limits involving the quotient of two functions. Consider<\/p>\r\n\r\n<div id=\"fs-id1165042458008\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042613148\">If [latex]\\underset{x\\to a}{\\lim}f(x)=L_1[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=L_2 \\ne 0[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1165043088102\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\dfrac{L_1}{L_2}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042330308\">However, what happens if [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex]? We call this one of the <strong>indeterminate forms<\/strong>, of type [latex]\\frac{0}{0}[\/latex]. This is considered an indeterminate form because we cannot determine the exact behavior of [latex]\\frac{f(x)}{g(x)}[\/latex] as [latex]x\\to a[\/latex] without further analysis. We have seen examples of this earlier in the text. For example, consider<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 2}{\\lim}\\dfrac{x^2-4}{x-2}[\/latex] and [latex]\\underset{x\\to 0}{\\lim}\\dfrac{ \\sin x}{x}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043036412\">For the first of these examples, we can evaluate the limit by factoring the numerator and writing<\/p>\r\n\r\n<div id=\"fs-id1165043067705\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 2}{\\lim}\\dfrac{x^2-4}{x-2}=\\underset{x\\to 2}{\\lim}\\dfrac{(x+2)(x-2)}{x-2}=\\underset{x\\to 2}{\\lim}(x+2)=2+2=4[\/latex]<\/div>\r\nFor [latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x}{x}[\/latex] we were able to show, using a geometric argument, that\r\n<div id=\"fs-id1165042954700\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\dfrac{\\sin x}{x}=1[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043001194\">Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could not calculate previously.<\/p>\r\n<p id=\"fs-id1165043062373\">The idea behind L\u2019H\u00f4pital\u2019s rule can be explained using local linear approximations. Consider two differentiable functions [latex]f[\/latex] and [latex]g[\/latex] such that [latex]\\underset{x\\to a}{\\lim}f(x)=0=\\underset{x\\to a}{\\lim}g(x)[\/latex] and such that [latex]g^{\\prime}(a)\\ne 0[\/latex] For [latex]x[\/latex] near [latex]a[\/latex], we can write<\/p>\r\n\r\n<div id=\"fs-id1165043199940\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)\\approx f(a)+f^{\\prime}(a)(x-a)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042355300\">and<\/p>\r\n\r\n<div id=\"fs-id1165042320393\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g(x)\\approx g(a)+g^{\\prime}(a)(x-a)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043178271\">Therefore,<\/p>\r\n\r\n<div id=\"fs-id1165042331440\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(x)}{g(x)}\\approx \\dfrac{f(a)+f^{\\prime}(a)(x-a)}{g(a)+g^{\\prime}(a)(x-a)}[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"618\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211304\/CNX_Calc_Figure_04_08_003.jpg\" alt=\"Two functions y = f(x) and y = g(x) are drawn such that they cross at a point above x = a. The linear approximations of these two functions y = f(a) + f\u2019(a)(x \u2013 a) and y = g(a) + g\u2019(a)(x \u2013 a) are also drawn.\" width=\"618\" height=\"390\" \/> Figure 1. If [latex]\\underset{x\\to a}{\\lim}f(x)=\\underset{x\\to a}{\\lim}g(x)[\/latex], then the ratio [latex]f(x)\/g(x)[\/latex] is approximately equal to the ratio of their linear approximations near [latex]a[\/latex].[\/caption]<\/div>\r\nSince [latex]f[\/latex] is differentiable at [latex]a[\/latex], then [latex]f[\/latex] is continuous at [latex]a[\/latex], and therefore [latex]f(a)=\\underset{x\\to a}{\\lim}f(x)=0[\/latex]. Similarly, [latex]g(a)=\\underset{x\\to a}{\\lim}g(x)=0[\/latex]. If we also assume that [latex]f^{\\prime}[\/latex] and [latex]g^{\\prime}[\/latex] are continuous at [latex]x=a[\/latex], then [latex]f^{\\prime}(a)=\\underset{x\\to a}{\\lim}f^{\\prime}(x)[\/latex] and [latex]g^{\\prime}(a)=\\underset{x\\to a}{\\lim}g^{\\prime}(x)[\/latex]. Using these ideas, we conclude that\r\n<div id=\"fs-id1165042373953\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)(x-a)}{g^{\\prime}(x)(x-a)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1165043085155\" class=\"bc-section section\">\r\n\r\nNote that the assumption that [latex]f^{\\prime}[\/latex] and [latex]g^{\\prime}[\/latex] are continuous at [latex]a[\/latex] and [latex]g^{\\prime}(a)\\ne 0[\/latex] can be loosened. We state L\u2019H\u00f4pital\u2019s rule formally for the indeterminate form [latex]\\frac{0}{0}[\/latex]. Also note that the notation [latex]\\frac{0}{0}[\/latex] does not mean we are actually dividing zero by zero. Rather, we are using the notation [latex]\\frac{0}{0}[\/latex] to represent a quotient of limits, each of which is zero.\r\n<div id=\"fs-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<p id=\"fs-id1165042318672\">We can also use L\u2019H\u00f4pital\u2019s rule to evaluate limits of quotients [latex]\\frac{f(x)}{g(x)}[\/latex] in which [latex]f(x)\\to \\pm \\infty [\/latex] and [latex]g(x)\\to \\pm \\infty[\/latex]. Limits of this form are classified as <em>indeterminate forms of type<\/em> [latex]\\infty \/ \\infty[\/latex]. Again, note that we are not actually dividing [latex]\\infty[\/latex] by [latex]\\infty[\/latex]. Since [latex]\\infty[\/latex] is not a real number, that is impossible; rather, [latex]\\infty \/ \\infty[\/latex] is used to represent a quotient of limits, each of which is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165042330826\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/ \\infty[\/latex] Case)<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165043426174\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval containing [latex]a[\/latex], except possibly at [latex]a[\/latex]. Suppose [latex]\\underset{x\\to a}{\\lim}f(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]) and [latex]\\underset{x\\to a}{\\lim}g(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]). Then,<\/p>\r\n\r\n<div id=\"fs-id1165042373703\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex],<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<p id=\"fs-id1165042707280\">assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if the limit is infinite, if [latex]a=\\infty[\/latex] or [latex]\u2212\\infty[\/latex], or the limit is one-sided.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043427623\" class=\"textbook exercises\">\r\n<h3>Example: Applying L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/\\infty[\/latex] Case)<\/h3>\r\n<p id=\"fs-id1165042709794\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\r\n\r\n<ol id=\"fs-id1165042709798\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1165043427625\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"fs-id1165042376758\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042376758\"]\r\n<ol id=\"fs-id1165042376758\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Since [latex]3x+5[\/latex] and [latex]2x+1[\/latex] are first-degree polynomials with positive leading coefficients, [latex]\\underset{x\\to \\infty }{\\lim}(3x+5)=\\infty [\/latex] and [latex]\\underset{x\\to \\infty }{\\lim}(2x+1)=\\infty[\/latex]. Therefore, we apply L\u2019H\u00f4pital\u2019s rule and obtain\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim}\\dfrac{3x+5}{2x+\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3}{2}=\\dfrac{3}{2}[\/latex].<\/div>\r\nNote that this limit can also be calculated without invoking L\u2019H\u00f4pital\u2019s rule. Earlier in the module we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of [latex]x[\/latex] in the denominator. In doing so, we saw that\r\n<div id=\"fs-id1165042319986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3+\\frac{5}{x}}{2+\\frac{1}{x}}=\\dfrac{3}{2}[\/latex].<\/div>\r\nL\u2019H\u00f4pital\u2019s rule provides us with an alternative means of evaluating this type of limit.<\/li>\r\n \t<li>Here, [latex]\\underset{x\\to 0^+}{\\lim} \\ln x=\u2212\\infty [\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} \\cot x=\\infty[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule and obtain\r\n<div id=\"fs-id1165042374803\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\frac{1}{x}}{\u2212\\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}[\/latex].<\/div>\r\nNow as [latex]x\\to 0^+[\/latex], [latex]\\csc^2 x\\to \\infty[\/latex]. Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of [latex]\\csc x[\/latex] to write\r\n<div id=\"fs-id1165042376466\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}[\/latex].<\/div>\r\nNow [latex]\\underset{x\\to 0^+}{\\lim} \\sin^2 x=0[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} x=0[\/latex], so we apply L\u2019H\u00f4pital\u2019s rule again. We find\r\n<div id=\"fs-id1165043249678\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{2 \\sin x \\cos x}{-1}=\\dfrac{0}{-1}=0[\/latex].<\/div>\r\nWe conclude that\r\n<div id=\"fs-id1165042315721\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=0[\/latex].<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042333392\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{\\ln x}{5x}[\/latex]\r\n\r\n[reveal-answer q=\"30011179\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"30011179\"]\r\n\r\n[latex]\\frac{d}{dx}\\ln x=\\frac{1}{x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042367881\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042367881\"]\r\n\r\n0\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write the terms of a sequence defined by a recursive formula<\/li>\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<\/ul>\n<\/div>\n<p>In the Sequences section, we will look at ordered lists of numbers (sequences) and determine whether they converge or diverge. Here we will review how to evaluate a recursive (recurrence) formula, take limits at infinity, and apply L\u2019H\u00f4pital\u2019s Rule.<\/p>\n<h2>Use a Recursive Formula<\/h2>\n<p>A <strong>recursive formula<\/strong> is\u00a0a formula that defines its value at a particular input using the result of the previous input(s).<\/p>\n<p>A recursive formula always has two parts: the value of an initial input and an equation defining each term in terms of preceding terms. For example, suppose we know the following:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}&{a}_{1}=3 \\\\ &{a}_{n}=2{a}_{n - 1}-1, \\text{ for } n\\ge 2 \\end{align}[\/latex]<\/div>\n<div><\/div>\n<div><\/div>\n<div style=\"text-align: left;\">We can find the subsequent terms of the recursive formula using the first term.<\/div>\n<div><\/div>\n<div style=\"text-align: center;\">[latex]\\begin{align}&{a}_{1}=3\\\\ &{a}_{2}=2{a}_{1}-1=2\\left(3\\right)-1=5\\\\ &{a}_{3}=2{a}_{2}-1=2\\left(5\\right)-1=9\\\\ &{a}_{4}=2{a}_{3}-1=2\\left(9\\right)-1=17\\end{align}[\/latex]<\/div>\n<p>So, the first four terms are [latex]3,5,9,\\text{ and},17[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a recursive formula with only the first term provided, write the first [latex]n[\/latex] terms of a sequence.<\/h3>\n<ol>\n<li>Identify the initial term, [latex]{a}_{1}[\/latex], which is given as part of the formula. This is the first term.<\/li>\n<li>To find the second term, [latex]{a}_{2}[\/latex], substitute the initial term into the formula for [latex]{a}_{n - 1}[\/latex]. Solve.<\/li>\n<li>To find the third term, [latex]{a}_{3}[\/latex], substitute the second term into the formula. Solve.<\/li>\n<li>Repeat until you have solved for the [latex]n\\text{th}[\/latex] term.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing the Terms of a Sequence Defined by a Recursive Formula<\/h3>\n<p>Write the first five terms of the sequence defined by the recursive formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} {a}_{1}&=9 \\\\ {a}_{n}&=3{a}_{n - 1}-20\\text{, for }n\\ge 2 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q748916\">Show Solution<\/span><\/p>\n<div id=\"q748916\" class=\"hidden-answer\" style=\"display: none\">\n<p>The first term is given in the formula. For each subsequent term, we replace [latex]{a}_{n - 1}[\/latex] with the value of the preceding term.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&n=1 && {a}_{1}=9 \\\\ &n=2 && {a}_{2}=3{a}_{1}-20=3\\left(9\\right)-20=27 - 20=7 \\\\ &n=3 && {a}_{3}=3{a}_{2}-20=3\\left(7\\right)-20=21 - 20=1 \\\\ &n=4 && {a}_{4}=3{a}_{3}-20=3\\left(1\\right)-20=3 - 20=-17 \\\\ &n=5 && {a}_{5}=3{a}_{4}-20=3\\left(-17\\right)-20=-51 - 20=-71 \\end{align}[\/latex]<\/p>\n<p>The first five terms are [latex]\\left\\{9,7,1,-17,-71\\right\\}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write the first five terms of the sequence defined by the recursive formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{a}_{1}&=2\\\\ {a}_{n}&=2{a}_{n - 1}+1\\text{, for }n\\ge 2\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q378600\">Show Solution<\/span><\/p>\n<div id=\"q378600\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left\\{2, 5, 11, 23, 47\\right\\}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5812\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5812&#38;theme=oea&#38;iframe_resize_id=ohm5812&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Take 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 in Figure 1 and numerically in the table beneath it, 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\">Figure 1. The function approaches the asymptote [latex]y=2[\/latex] as [latex]x[\/latex] approaches [latex]\\pm \\infty[\/latex].<\/p>\n<\/div>\n<table id=\"fs-id1165043428402\" class=\"column-header\" summary=\"The table has four rows and five columns. The first column is a header column and it reads x, 2 + 1\/x, x, and 2 + 1\/x. After the header, the first row reads 10, 100, 1000, and 10000. The second row reads 2.1, 2.01, 2.001, and 2.0001. The third row reads \u221210, \u2212100, \u22121000, and \u221210000. The fourth row reads 1.9, 1.99, 1.999, and 1.9999.\">\n<caption>Values of a function [latex]f[\/latex] 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>100<\/td>\n<td>1,000<\/td>\n<td>10,000<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td><strong>[latex]2+\\frac{1}{x}[\/latex]<\/strong><\/td>\n<td>2.1<\/td>\n<td>2.01<\/td>\n<td>2.001<\/td>\n<td>2.0001<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>-10<\/td>\n<td>-100<\/td>\n<td>-1000<\/td>\n<td>-10,000<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td><strong>[latex]2+\\frac{1}{x}[\/latex]<\/strong><\/td>\n<td>1.9<\/td>\n<td>1.99<\/td>\n<td>1.999<\/td>\n<td>1.9999<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\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<h3 style=\"text-align: center;\">Definition<\/h3>\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<p id=\"fs-id1165043157752\">If the values [latex]f(x)[\/latex] are getting arbitrarily close to some finite value [latex]L[\/latex] as [latex]x\\to \\infty[\/latex] or [latex]x\\to \u2212\\infty[\/latex], the graph of [latex]f[\/latex] approaches the line [latex]y=L[\/latex]. In that case, the line [latex]y=L[\/latex] is a horizontal asymptote of [latex]f[\/latex] (Figure 2). For example, for the function [latex]f(x)=\\frac{1}{x}[\/latex], since [latex]\\underset{x\\to \\infty }{\\lim}f(x)=0[\/latex], the line [latex]y=0[\/latex] is a horizontal asymptote of [latex]f(x)=\\frac{1}{x}[\/latex].<\/p>\n<div id=\"fs-id1165043262534\" class=\"textbox shaded\">\n<div class=\"title\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<\/div>\n<p id=\"fs-id1165042973921\">If [latex]\\underset{x\\to \\infty }{\\lim}f(x)=L[\/latex] or [latex]\\underset{x \\to \u2212\\infty}{\\lim}f(x)=L[\/latex], we say the line [latex]y=L[\/latex] is a <strong>horizontal asymptote<\/strong> of [latex]f[\/latex].<\/p>\n<\/div>\n<div style=\"width: 776px\" 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\/11211028\/CNX_Calc_Figure_04_06_020.jpg\" alt=\"The figure is broken up into two figures labeled a and b. Figure a shows a function f(x) approaching but never touching a horizontal dashed line labeled L from above. Figure b shows a function f(x) approaching but never a horizontal dashed line labeled M from below.\" width=\"766\" height=\"273\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. (a) As [latex]x\\to \\infty[\/latex], the values of [latex]f[\/latex] are getting arbitrarily close to [latex]L[\/latex]. The line [latex]y=L[\/latex] is a horizontal asymptote of [latex]f[\/latex]. (b) As [latex]x\\to \u2212\\infty[\/latex], the values of [latex]f[\/latex] are getting arbitrarily close to [latex]M[\/latex]. The line [latex]y=M[\/latex] is a horizontal asymptote of [latex]f[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1165042647732\">A function cannot cross a vertical asymptote because the graph must approach infinity (or negative infinity) from at least one direction as [latex]x[\/latex] approaches the vertical asymptote. However, a function may cross a horizontal asymptote. In fact, a function may cross a horizontal asymptote an unlimited number of times. For example, the function [latex]f(x)=\\frac{ \\cos x}{x}+1[\/latex] shown in Figure 3 intersects the horizontal asymptote [latex]y=1[\/latex] an infinite number of times as it oscillates around the asymptote with ever-decreasing amplitude.<\/p>\n<div style=\"width: 539px\" 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\/11211031\/CNX_Calc_Figure_04_06_002.jpg\" alt=\"The function f(x) = (cos x)\/x + 1 is shown. It decreases from (0, \u221e) and then proceeds to oscillate around y = 1 with decreasing amplitude.\" width=\"529\" height=\"230\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. The graph of [latex]f(x)=\\cos x\/x+1[\/latex] crosses its horizontal asymptote [latex]y=1[\/latex] an infinite number of times.<\/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<li>[latex]f(x)= \\tan^{-1} (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].<\/li>\n<li>Since [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<p>&nbsp;<\/li>\n<li>To evaluate [latex]\\underset{x\\to \\infty }{\\lim} \\tan^{-1} (x)[\/latex] and [latex]\\underset{x\\to \u2212\\infty}{\\lim} \\tan^{-1} (x)[\/latex], we first consider the graph of [latex]y= \\tan (x)[\/latex] over the interval [latex](\u2212\\pi \/2,\\pi \/2)[\/latex] as shown in the following graph.\n<div style=\"width: 502px\" 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\/11211039\/CNX_Calc_Figure_04_06_021.jpg\" alt=\"The function f(x) = tan x is shown. It increases from (\u2212\u03c0\/2, \u2212\u221e), passes through the origin, and then increases toward (\u03c0\/2, \u221e). There are vertical dashed lines marking x = \u00b1\u03c0\/2.\" width=\"492\" height=\"347\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. The graph of [latex] \\tan x[\/latex] has vertical asymptotes at [latex]x=\\pm \\frac{\\pi }{2}[\/latex]<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<p id=\"fs-id1165043092430\">Since<\/p>\n<div id=\"fs-id1165043119614\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to (\\pi\/2)^-}{\\lim} \\tan x=\\infty[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042514177\">it follows that<\/p>\n<div id=\"fs-id1165042563973\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim} \\tan^{-1} (x)=\\frac{\\pi }{2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043097156\">Similarly, since<\/p>\n<div id=\"fs-id1165042923290\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to (\\pi\/2)^+}{\\lim} \\tan x=\u2212\\infty[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043131939\">it follows that<\/p>\n<div id=\"fs-id1165043056813\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \u2212\\infty}{\\lim} \\tan^{-1} (x)=-\\frac{\\pi }{2}[\/latex]<\/div>\n<p><span style=\"font-size: 1rem; text-align: initial;\"><\/div>\n<\/div>\n<p><\/span><\/p>\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]. Determine the horizontal asymptotes of [latex]f(x)=3+\\frac{4}{x}[\/latex], if any.<\/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<h2>Infinite Limits at Infinity<\/h2>\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]. As seen in the table below and Figure 2, as [latex]x\\to \\infty[\/latex] the values [latex]f(x)[\/latex] become arbitrarily large. Therefore, [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\">Figure 2. For this function, the functional values approach infinity as [latex]x\\to \\pm \\infty[\/latex].<\/p>\n<\/div>\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<\/div>\n<div id=\"fs-id1165042323710\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165042320226\">Find [latex]\\underset{x\\to \\infty }{\\lim}3x^2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042708272\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042708272\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042383154\">[latex]\\infty[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm16109\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16109&theme=oea&iframe_resize_id=ohm16109&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Apply L&#8217;H\u00f4pital&#8217;s Rule<\/h2>\n<p id=\"fs-id1165042941863\">L\u2019H\u00f4pital\u2019s rule can be used to evaluate limits involving the quotient of two functions. Consider<\/p>\n<div id=\"fs-id1165042458008\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042613148\">If [latex]\\underset{x\\to a}{\\lim}f(x)=L_1[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=L_2 \\ne 0[\/latex], then<\/p>\n<div id=\"fs-id1165043088102\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\dfrac{L_1}{L_2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042330308\">However, what happens if [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex]? We call this one of the <strong>indeterminate forms<\/strong>, of type [latex]\\frac{0}{0}[\/latex]. This is considered an indeterminate form because we cannot determine the exact behavior of [latex]\\frac{f(x)}{g(x)}[\/latex] as [latex]x\\to a[\/latex] without further analysis. We have seen examples of this earlier in the text. For example, consider<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 2}{\\lim}\\dfrac{x^2-4}{x-2}[\/latex] and [latex]\\underset{x\\to 0}{\\lim}\\dfrac{ \\sin x}{x}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043036412\">For the first of these examples, we can evaluate the limit by factoring the numerator and writing<\/p>\n<div id=\"fs-id1165043067705\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 2}{\\lim}\\dfrac{x^2-4}{x-2}=\\underset{x\\to 2}{\\lim}\\dfrac{(x+2)(x-2)}{x-2}=\\underset{x\\to 2}{\\lim}(x+2)=2+2=4[\/latex]<\/div>\n<p>For [latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x}{x}[\/latex] we were able to show, using a geometric argument, that<\/p>\n<div id=\"fs-id1165042954700\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\dfrac{\\sin x}{x}=1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043001194\">Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could not calculate previously.<\/p>\n<p id=\"fs-id1165043062373\">The idea behind L\u2019H\u00f4pital\u2019s rule can be explained using local linear approximations. Consider two differentiable functions [latex]f[\/latex] and [latex]g[\/latex] such that [latex]\\underset{x\\to a}{\\lim}f(x)=0=\\underset{x\\to a}{\\lim}g(x)[\/latex] and such that [latex]g^{\\prime}(a)\\ne 0[\/latex] For [latex]x[\/latex] near [latex]a[\/latex], we can write<\/p>\n<div id=\"fs-id1165043199940\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)\\approx f(a)+f^{\\prime}(a)(x-a)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042355300\">and<\/p>\n<div id=\"fs-id1165042320393\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g(x)\\approx g(a)+g^{\\prime}(a)(x-a)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043178271\">Therefore,<\/p>\n<div id=\"fs-id1165042331440\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(x)}{g(x)}\\approx \\dfrac{f(a)+f^{\\prime}(a)(x-a)}{g(a)+g^{\\prime}(a)(x-a)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<div>\n<div style=\"width: 628px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211304\/CNX_Calc_Figure_04_08_003.jpg\" alt=\"Two functions y = f(x) and y = g(x) are drawn such that they cross at a point above x = a. The linear approximations of these two functions y = f(a) + f\u2019(a)(x \u2013 a) and y = g(a) + g\u2019(a)(x \u2013 a) are also drawn.\" width=\"618\" height=\"390\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. If [latex]\\underset{x\\to a}{\\lim}f(x)=\\underset{x\\to a}{\\lim}g(x)[\/latex], then the ratio [latex]f(x)\/g(x)[\/latex] is approximately equal to the ratio of their linear approximations near [latex]a[\/latex].<\/p>\n<\/div>\n<\/div>\n<p>Since [latex]f[\/latex] is differentiable at [latex]a[\/latex], then [latex]f[\/latex] is continuous at [latex]a[\/latex], and therefore [latex]f(a)=\\underset{x\\to a}{\\lim}f(x)=0[\/latex]. Similarly, [latex]g(a)=\\underset{x\\to a}{\\lim}g(x)=0[\/latex]. If we also assume that [latex]f^{\\prime}[\/latex] and [latex]g^{\\prime}[\/latex] are continuous at [latex]x=a[\/latex], then [latex]f^{\\prime}(a)=\\underset{x\\to a}{\\lim}f^{\\prime}(x)[\/latex] and [latex]g^{\\prime}(a)=\\underset{x\\to a}{\\lim}g^{\\prime}(x)[\/latex]. Using these ideas, we conclude that<\/p>\n<div id=\"fs-id1165042373953\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)(x-a)}{g^{\\prime}(x)(x-a)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex]<\/div>\n<div><\/div>\n<div id=\"fs-id1165043085155\" class=\"bc-section section\">\n<p>Note that the assumption that [latex]f^{\\prime}[\/latex] and [latex]g^{\\prime}[\/latex] are continuous at [latex]a[\/latex] and [latex]g^{\\prime}(a)\\ne 0[\/latex] can be loosened. We state L\u2019H\u00f4pital\u2019s rule formally for the indeterminate form [latex]\\frac{0}{0}[\/latex]. Also note that the notation [latex]\\frac{0}{0}[\/latex] does not mean we are actually dividing zero by zero. Rather, we are using the notation [latex]\\frac{0}{0}[\/latex] to represent a quotient of limits, each of which is zero.<\/p>\n<div id=\"fs-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<p id=\"fs-id1165042318672\">We can also use L\u2019H\u00f4pital\u2019s rule to evaluate limits of quotients [latex]\\frac{f(x)}{g(x)}[\/latex] in which [latex]f(x)\\to \\pm \\infty[\/latex] and [latex]g(x)\\to \\pm \\infty[\/latex]. Limits of this form are classified as <em>indeterminate forms of type<\/em> [latex]\\infty \/ \\infty[\/latex]. Again, note that we are not actually dividing [latex]\\infty[\/latex] by [latex]\\infty[\/latex]. Since [latex]\\infty[\/latex] is not a real number, that is impossible; rather, [latex]\\infty \/ \\infty[\/latex] is used to represent a quotient of limits, each of which is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex].<\/p>\n<div id=\"fs-id1165042330826\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/ \\infty[\/latex] Case)<\/h3>\n<hr \/>\n<p id=\"fs-id1165043426174\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval containing [latex]a[\/latex], except possibly at [latex]a[\/latex]. Suppose [latex]\\underset{x\\to a}{\\lim}f(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]) and [latex]\\underset{x\\to a}{\\lim}g(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]). Then,<\/p>\n<div id=\"fs-id1165042373703\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<p id=\"fs-id1165042707280\">assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if the limit is infinite, if [latex]a=\\infty[\/latex] or [latex]\u2212\\infty[\/latex], or the limit is one-sided.<\/p>\n<\/div>\n<div id=\"fs-id1165043427623\" class=\"textbook exercises\">\n<h3>Example: Applying L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/\\infty[\/latex] Case)<\/h3>\n<p id=\"fs-id1165042709794\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\n<ol id=\"fs-id1165042709798\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}[\/latex]<\/li>\n<li>[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1165043427625\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042376758\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042376758\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165042376758\" style=\"list-style-type: lower-alpha;\">\n<li>Since [latex]3x+5[\/latex] and [latex]2x+1[\/latex] are first-degree polynomials with positive leading coefficients, [latex]\\underset{x\\to \\infty }{\\lim}(3x+5)=\\infty[\/latex] and [latex]\\underset{x\\to \\infty }{\\lim}(2x+1)=\\infty[\/latex]. Therefore, we apply L\u2019H\u00f4pital\u2019s rule and obtain\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim}\\dfrac{3x+5}{2x+\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3}{2}=\\dfrac{3}{2}[\/latex].<\/div>\n<p>Note that this limit can also be calculated without invoking L\u2019H\u00f4pital\u2019s rule. Earlier in the module we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of [latex]x[\/latex] in the denominator. In doing so, we saw that<\/p>\n<div id=\"fs-id1165042319986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3+\\frac{5}{x}}{2+\\frac{1}{x}}=\\dfrac{3}{2}[\/latex].<\/div>\n<p>L\u2019H\u00f4pital\u2019s rule provides us with an alternative means of evaluating this type of limit.<\/li>\n<li>Here, [latex]\\underset{x\\to 0^+}{\\lim} \\ln x=\u2212\\infty[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} \\cot x=\\infty[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule and obtain\n<div id=\"fs-id1165042374803\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\frac{1}{x}}{\u2212\\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}[\/latex].<\/div>\n<p>Now as [latex]x\\to 0^+[\/latex], [latex]\\csc^2 x\\to \\infty[\/latex]. Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of [latex]\\csc x[\/latex] to write<\/p>\n<div id=\"fs-id1165042376466\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}[\/latex].<\/div>\n<p>Now [latex]\\underset{x\\to 0^+}{\\lim} \\sin^2 x=0[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} x=0[\/latex], so we apply L\u2019H\u00f4pital\u2019s rule again. We find<\/p>\n<div id=\"fs-id1165043249678\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{2 \\sin x \\cos x}{-1}=\\dfrac{0}{-1}=0[\/latex].<\/div>\n<p>We conclude that<\/p>\n<div id=\"fs-id1165042315721\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=0[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042333392\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{\\ln x}{5x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q30011179\">Hint<\/span><\/p>\n<div id=\"q30011179\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{d}{dx}\\ln x=\\frac{1}{x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042367881\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042367881\" class=\"hidden-answer\" style=\"display: none\">\n<p>0\n<\/p><\/div>\n<\/div>\n<\/div>\n<\/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-2282\">\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 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