{"id":414,"date":"2021-02-04T02:04:42","date_gmt":"2021-02-04T02:04:42","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=414"},"modified":"2022-03-16T05:50:25","modified_gmt":"2022-03-16T05:50:25","slug":"other-indeterminate-forms","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/other-indeterminate-forms\/","title":{"raw":"Other Indeterminate Forms","rendered":"Other Indeterminate Forms"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\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\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\" \/> Figure 2. 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\nWatch the following video to see the worked solution to Example: Indeterminate Form of Type [latex]\\infty -\\infty[\/latex].\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/e58sGIZe1wU?controls=0&amp;start=671&amp;end=790&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.8LHopitalsRule671to790_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.8 L'Hopital's Rule\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<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<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.) We proceed as follows. Let<\/p>\r\n\r\n<div id=\"fs-id1165042709633\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=f(x)^{g(x)}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043250963\">Then,<\/p>\r\n\r\n<div id=\"fs-id1165043250966\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln y=\\ln (f(x)^{g(x)})=g(x) \\ln (f(x))[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042640888\">Therefore,<\/p>\r\n\r\n<div id=\"fs-id1165042640891\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}[\\ln y]=\\underset{x\\to a}{\\lim}[g(x) \\ln (f(x))][\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043393684\">Since [latex]\\underset{x\\to a}{\\lim}f(x)=\\infty[\/latex], we know that [latex]\\underset{x\\to a}{\\lim}\\ln (f(x))=\\infty[\/latex]. Therefore, [latex]\\underset{x\\to a}{\\lim}g(x) \\ln (f(x))[\/latex] is of the indeterminate form [latex]0 \\cdot \\infty[\/latex], and we can use the techniques discussed earlier to rewrite the expression [latex]g(x) \\ln (f(x))[\/latex] in a form so that we can apply L\u2019H\u00f4pital\u2019s rule. Suppose [latex]\\underset{x\\to a}{\\lim}g(x) \\ln (f(x))=L[\/latex], where [latex]L[\/latex] may be [latex]\\infty [\/latex] or [latex]\u2212\\infty[\/latex]. Then<\/p>\r\n\r\n<div id=\"fs-id1165042638509\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\ln y=L[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042638551\">Since the natural logarithm function is continuous, we conclude that<\/p>\r\n\r\n<div id=\"fs-id1165042638554\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln (\\underset{x\\to a}{\\lim} y)=L[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042708323\">which gives us<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim} y=\\underset{x\\to a}{\\lim}f(x)^{g(x)}=e^L[\/latex]<\/div>\r\n&nbsp;\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\nWatch the following video to see the worked solution to Example: Indeterminate Form of Type [latex]\\infty^0[\/latex].\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/e58sGIZe1wU?controls=0&amp;start=797&amp;end=919&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.8LHopitalsRule797to919_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.8 L'Hopital's Rule\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<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\nWatch the following video to see the worked solution to Example: Indeterminate Form of Type [latex]0^0[\/latex].\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/e58sGIZe1wU?controls=0&amp;start=932&amp;end=1073&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266835\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266835\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.8LHopitalsRule932to1073_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.8 L'Hopital's Rule\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div 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 class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5287[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\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 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\">Figure 2. 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<p>Watch the following video to see the worked solution to Example: Indeterminate Form of Type [latex]\\infty -\\infty[\/latex].<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/e58sGIZe1wU?controls=0&amp;start=671&amp;end=790&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.8LHopitalsRule671to790_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.8 L&#8217;Hopital&#8217;s Rule&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<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<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.) We proceed as follows. Let<\/p>\n<div id=\"fs-id1165042709633\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=f(x)^{g(x)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043250963\">Then,<\/p>\n<div id=\"fs-id1165043250966\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln y=\\ln (f(x)^{g(x)})=g(x) \\ln (f(x))[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042640888\">Therefore,<\/p>\n<div id=\"fs-id1165042640891\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}[\\ln y]=\\underset{x\\to a}{\\lim}[g(x) \\ln (f(x))][\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043393684\">Since [latex]\\underset{x\\to a}{\\lim}f(x)=\\infty[\/latex], we know that [latex]\\underset{x\\to a}{\\lim}\\ln (f(x))=\\infty[\/latex]. Therefore, [latex]\\underset{x\\to a}{\\lim}g(x) \\ln (f(x))[\/latex] is of the indeterminate form [latex]0 \\cdot \\infty[\/latex], and we can use the techniques discussed earlier to rewrite the expression [latex]g(x) \\ln (f(x))[\/latex] in a form so that we can apply L\u2019H\u00f4pital\u2019s rule. Suppose [latex]\\underset{x\\to a}{\\lim}g(x) \\ln (f(x))=L[\/latex], where [latex]L[\/latex] may be [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex]. Then<\/p>\n<div id=\"fs-id1165042638509\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\ln y=L[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042638551\">Since the natural logarithm function is continuous, we conclude that<\/p>\n<div id=\"fs-id1165042638554\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln (\\underset{x\\to a}{\\lim} y)=L[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042708323\">which gives us<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim} y=\\underset{x\\to a}{\\lim}f(x)^{g(x)}=e^L[\/latex]<\/div>\n<p>&nbsp;<\/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<p>Watch the following video to see the worked solution to Example: Indeterminate Form of Type [latex]\\infty^0[\/latex].<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/e58sGIZe1wU?controls=0&amp;start=797&amp;end=919&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.8LHopitalsRule797to919_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.8 L&#8217;Hopital&#8217;s Rule&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div 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<p>Watch the following video to see the worked solution to Example: Indeterminate Form of Type [latex]0^0[\/latex].<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/e58sGIZe1wU?controls=0&amp;start=932&amp;end=1073&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266835\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266835\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.8LHopitalsRule932to1073_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.8 L&#8217;Hopital&#8217;s Rule&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div 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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5287\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5287&theme=oea&iframe_resize_id=ohm5287&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-414\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>4.8 L&#039;Hopital&#039;s Rule. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":29,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.8 L\\'Hopital\\'s Rule\",\"author\":\"Ryan 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