{"id":4089,"date":"2022-04-14T18:15:52","date_gmt":"2022-04-14T18:15:52","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-vector-valued-functions-and-space-curves\/"},"modified":"2022-04-25T17:54:14","modified_gmt":"2022-04-25T17:54:14","slug":"skills-review-for-vector-valued-functions-and-space-curves","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-vector-valued-functions-and-space-curves\/","title":{"raw":"Skills Review for Vector-Valued Functions and Space Curves","rendered":"Skills Review for Vector-Valued Functions and Space Curves"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Evaluate trigonometric functions using the unit circle<\/li>\r\n \t<li>Recognize the basic limit laws<\/li>\r\n \t<li>Use the limit laws to evaluate the limit of a function<\/li>\r\n \t<li>Recognize when to apply L\u2019H\u00f4pital\u2019s rule<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Vector-Valued Functions and Space Curves section, we will define and develop skills needed to understand and work with vector-valued functions. Here we will review how to evaluate trigonometric functions at specific angle measures, apply limit laws, and apply L\u2019H\u00f4pital\u2019s rule.\r\n<h2>Evaluate Trigonometric Functions Using the Unit Circle<\/h2>\r\n\r\n<strong><em>(also in Module 1, Skills Review for Polar Coordinates)<\/em><\/strong>\r\nThe unit circle tells us the value of cosine and sine at any of the given angle measures seen below. The first coordinate in each ordered pair is the value of cosine at the given angle measure, while the second coordinate in each ordered pair is the value of sine at the given angle measure. You will learn that all trigonometric functions can be written in terms of sine and cosine. Thus, if you can evaluate sine and cosine at various angle values, you can also evaluate the other trigonometric functions at various angle values. Take time to learn the [latex]\\left(x,y\\right)[\/latex] coordinates of all of the major angles in the first quadrant of the unit circle.\r\n\r\nRemember, every angle in quadrant two, three, or four has a reference angle that lies in quadrant one. The quadrant of the original angle only affects the sign (positive or negative) of a trigonometric function's value at a given angle.\r\n\r\n<img class=\"aligncenter wp-image-12625 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003609\/f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3-IMAGE-IMAGE.png\" alt=\"f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3+IMAGE+IMAGE\" width=\"800\" height=\"728\">\r\n<div class=\"textbox\">\r\n<h3>A General Note: Evaluating Tangent, Secant, Cosecant, and Cotangent Functions<\/h3>\r\nIf [latex] \\theta[\/latex] is an angle measure, then, using the unit circle,\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\tan \\theta=\\frac{sin \\theta}{cos \\theta}\\\\ \\sec \\theta=\\frac{1}{cos \\theta}\\\\ \\csc t=\\frac{1}{sin \\theta}\\\\ \\cot \\theta=\\frac{cos \\theta}{sin \\theta}\\end{gathered}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Unit Circle to Find the Value of Trigonometric Functions<\/h3>\r\nFind [latex]\\sin \\theta,\\cos \\theta,\\tan \\theta,\\sec \\theta,\\csc \\theta[\/latex], and [latex]\\cot \\theta[\/latex] when [latex] \\theta=\\frac{\\pi }{3}[\/latex].\r\n\r\n[reveal-answer q=\"151548\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"151548\"]\r\n\r\n[latex]\\begin{align}&amp;\\sin \\frac{\\pi }{3}=\\frac{\\sqrt{3}}{2}\\\\ &amp;\\cos \\frac{\\pi }{3}=\\frac{1}{2}\\\\ &amp;\\tan \\frac{\\pi }{3}=\\sqrt{3} \\text{ (From}\\frac{sin \\theta}{cos \\theta})\\\\ &amp;\\sec \\frac{\\pi }{3}=2 \\text{ (From}\\frac{1}{cos \\theta})\\\\ &amp;\\csc \\frac{\\pi }{3}=\\frac{2\\sqrt{3}}{3}\\text{ (From}\\frac{1}{sin \\theta} \\text{ - do not forget to rationalize})\\\\ &amp;\\cot \\frac{\\pi }{3}=\\frac{\\sqrt{3}}{3}(\\text{ (From}\\frac{cos \\theta}{sin \\theta} \\text{ - do not forget to rationalize})\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173354[\/ohm_question]\r\n\r\n<\/div>\r\nBelow are the values of all six trigonometric functions evaluated at common angle measures from Quadrant I.\r\n<table id=\"Table_05_03_01\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td><strong>Angle<\/strong><\/td>\r\n<td><strong> [latex]0[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{6},\\text{ or }{30}^{\\circ}[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{4},\\text{ or } {45}^{\\circ }[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{3},\\text{ or }{60}^{\\circ }[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{2},\\text{ or }{90}^{\\circ }[\/latex] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Cosine<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Sine<\/strong><\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Tangent<\/strong><\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\r\n<td>Undefined<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Secant<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\r\n<td>2<\/td>\r\n<td>Undefined<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Cosecant<\/strong><\/td>\r\n<td>Undefined<\/td>\r\n<td>2<\/td>\r\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Cotangent<\/strong><\/td>\r\n<td>Undefined<\/td>\r\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Limit Laws<\/h2>\r\n<p id=\"fs-id1170571680609\"> The basic limit laws stated below, together with the other limit laws, allow us to evaluate limits of many algebraic functions.<\/p>\r\n\r\n<div id=\"fs-id1170572451153\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Basic Limit Results<\/h3>\r\n\r\n<hr>\r\n<p id=\"fs-id1170572205248\">For any real number [latex]a[\/latex] and any constant [latex]c[\/latex],<\/p>\r\n\r\n<ol id=\"fs-id1170572286963\">\r\n \t<li>\r\n<div id=\"fs-id1170572624896\" class=\"equation\">[latex]\\underset{x\\to a}{\\lim}x=a[\/latex]<\/div><\/li>\r\n \t<li>\r\n<div id=\"fs-id1170572209025\" class=\"equation\">[latex]\\underset{x\\to a}{\\lim}c=c[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1170572111463\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating a Basic Limit<\/h3>\r\n<p id=\"fs-id1170571569246\">Evaluate each of the following limits using the basic limit results above.<\/p>\r\n\r\n<ol id=\"fs-id1170572176731\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\underset{x\\to 2}{\\lim}x[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to 2}{\\lim}5[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170572101621\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572101621\"]\r\n<ol id=\"fs-id1170572101621\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>The limit of [latex]x[\/latex] as [latex]x[\/latex] approaches [latex]a[\/latex] is [latex]a[\/latex]: [latex]\\underset{x\\to 2}{\\lim}x=2[\/latex].<\/li>\r\n \t<li>The limit of a constant is that constant: [latex]\\underset{x\\to 2}{\\lim}5=5[\/latex].<\/li>\r\n<\/ol>\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]4886[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572570027\">We now take a look at the limit laws, the individual properties of limits. The proofs that these laws hold are omitted here.<\/p>\r\n\r\n<div id=\"fs-id1170572508800\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Limit Laws<\/h3>\r\n\r\n<hr>\r\n<p id=\"fs-id1170572086164\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be defined for all [latex]x\\ne a[\/latex] over some open interval containing [latex]a[\/latex]. Assume that [latex]L[\/latex] and [latex]M[\/latex] are real numbers such that [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=M[\/latex]. Let [latex]c[\/latex] be a constant. Then, each of the following statements holds:<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1170572204187\"><strong>Sum law for limits<\/strong>: [latex]\\underset{x\\to a}{\\lim}(f(x)+g(x))=\\underset{x\\to a}{\\lim}f(x)+\\underset{x\\to a}{\\lim}g(x)=L+M[\/latex]<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1170572627273\"><strong>Difference law for limits<\/strong>: [latex]\\underset{x\\to a}{\\lim}(f(x)-g(x))=\\underset{x\\to a}{\\lim}f(x)-\\underset{x\\to a}{\\lim}g(x)=L-M[\/latex]<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1170572450574\"><strong>Constant multiple law for limits<\/strong>: [latex]\\underset{x\\to a}{\\lim}cf(x)=c \\cdot \\underset{x\\to a}{\\lim}f(x)=cL[\/latex]<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1170572104032\"><strong>Product law for limits<\/strong>: [latex]\\underset{x\\to a}{\\lim}(f(x) \\cdot g(x))=\\underset{x\\to a}{\\lim}f(x) \\cdot \\underset{x\\to a}{\\lim}g(x)=L \\cdot M[\/latex]<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1170572347458\"><strong>Quotient law for limits<\/strong>: [latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\dfrac{\\underset{x\\to a}{\\lim}f(x)}{\\underset{x\\to a}{\\lim}g(x)}=\\frac{L}{M}[\/latex] for [latex]M\\ne 0[\/latex]<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1170572246193\"><strong>Power law for limits<\/strong>: [latex]\\underset{x\\to a}{\\lim}(f(x))^n=(\\underset{x\\to a}{\\lim}f(x))^n=L^n[\/latex] for every positive integer [latex]n[\/latex].<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1170572232633\"><strong>Root law for limits<\/strong>: [latex]\\underset{x\\to a}{\\lim}\\sqrt[n]{f(x)}=\\sqrt[n]{\\underset{x\\to a}{\\lim}f(x)}=\\sqrt[n]{L}[\/latex] for all [latex]L[\/latex] if [latex]n[\/latex] is odd and for [latex]L\\ge 0[\/latex] if [latex]n[\/latex] is even<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572479215\">We now practice applying these limit laws to evaluate a limit.<\/p>\r\n\r\n<div id=\"fs-id1170572451489\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating a Limit Using Limit Laws<\/h3>\r\n<p id=\"fs-id1170572109838\">Use the limit laws to evaluate [latex]\\underset{x\\to -3}{\\lim}(4x+2)[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170572169042\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572169042\"]\r\n<p id=\"fs-id1170572169042\">Let\u2019s apply the limit laws one step at a time to be sure we understand how they work. We need to keep in mind the requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied.<\/p>\r\n<p id=\"fs-id1170571565987\">[latex]\\begin{array}{ccccc}\\underset{x\\to -3}{\\lim}(4x+2)\\hfill &amp; =\\underset{x\\to -3}{\\lim}4x+\\underset{x\\to -3}{\\lim}2\\hfill &amp; &amp; &amp; \\text{Apply the sum law.}\\hfill \\\\ &amp; =4 \\cdot \\underset{x\\to -3}{\\lim}x+\\underset{x\\to -3}{\\lim}2\\hfill &amp; &amp; &amp; \\text{Apply the constant multiple law.}\\hfill \\\\ &amp; =4 \\cdot (-3)+2=-10\\hfill &amp; &amp; &amp; \\text{Apply the basic limit results and simplify.}\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572509954\" class=\"textbook exercises\">\r\n<h3>Example: Using Limit Laws Repeatedly<\/h3>\r\nUse the limit laws to evaluate [latex]\\underset{x\\to 2}{\\lim}\\dfrac{2x^2-3x+1}{x^3+4}[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1170572506406\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572506406\"]\r\n<p id=\"fs-id1170572506406\">To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccccc}\\\\ \\\\ \\underset{x\\to 2}{\\lim}\\large \\frac{2x^2-3x+1}{x^3+4} &amp; = \\large \\frac{\\underset{x\\to 2}{\\lim}(2x^2-3x+1)}{\\underset{x\\to 2}{\\lim}(x^3+4)} &amp; &amp; &amp; \\text{Apply the quotient law, making sure that} \\, 2^3+4\\ne 0 \\\\ &amp; = \\large \\frac{2 \\cdot \\underset{x\\to 2}{\\lim}x^2-3 \\cdot \\underset{x\\to 2}{\\lim}x+\\underset{x\\to 2}{\\lim}1}{\\underset{x\\to 2}{\\lim}x^3+\\underset{x\\to 2}{\\lim}4} &amp; &amp; &amp; \\text{Apply the sum law and constant multiple law.} \\\\ &amp; = \\large \\frac{2 \\cdot (\\underset{x\\to 2}{\\lim}x)^2-3 \\cdot \\underset{x\\to 2}{\\lim}x+\\underset{x\\to 2}{\\lim}1}{(\\underset{x\\to 2}{\\lim}x)^3+\\underset{x\\to 2}{\\lim}4} &amp; &amp; &amp; \\text{Apply the power law.} \\\\ &amp; = \\large \\frac{2(4)-3(2)+1}{2^3+4}=\\frac{1}{4} &amp; &amp; &amp; \\text{Apply the basic limit laws and simplify.} \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170571655486\">Use the limit laws to evaluate [latex]\\underset{x\\to 6}{\\lim}(2x-1)\\sqrt{x+4}[\/latex]. In each step, indicate the limit law applied.<\/p>\r\n[reveal-answer q=\"6635113\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"6635113\"]\r\n<p id=\"fs-id1170572209920\">Begin by applying the product law.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572094142\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572094142\"]\r\n<p id=\"fs-id1170572094142\">[latex]11\\sqrt{10}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Limits of Polynomial and Rational Functions<\/h2>\r\n<p id=\"fs-id1170572133214\">By now you have probably noticed that, in each of the previous examples, it has been the case that [latex]\\underset{x\\to a}{\\lim}f(x)=f(a)[\/latex]. This is not always true, but it does hold for all polynomials for any choice of [latex]a[\/latex] and for all rational functions at all values of [latex]a[\/latex] for which the rational function is defined.<\/p>\r\n\r\n<div id=\"fs-id1170572557796\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Limits of Polynomial and Rational Functions<\/h3>\r\n\r\n<hr>\r\n<p id=\"fs-id1170572557802\">Let [latex]p(x)[\/latex] and [latex]q(x)[\/latex] be polynomial functions. Let [latex]a[\/latex] be a real number. Then,<\/p>\r\n\r\n<div id=\"fs-id1170572347161\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}p(x)=p(a)[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1170571656084\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{p(x)}{q(x)}=\\dfrac{p(a)}{q(a)} \\, \\text{when} \\, q(a)\\ne 0[\/latex]<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571650163\">To see that this theorem holds, consider the polynomial [latex]p(x)=c_nx^n+c_{n-1}x^{n-1}+\\cdots +c_1x+c_0[\/latex]. By applying the sum, constant multiple, and power laws, we end up with<\/p>\r\n\r\n<div id=\"fs-id1170571648575\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\underset{x\\to a}{\\lim}p(x)&amp; =\\underset{x\\to a}{\\lim}(c_nx^n+c_{n-1}x^{n-1}+\\cdots +c_1x+c_0)\\hfill \\\\ &amp; =c_n(\\underset{x\\to a}{\\lim}x)^n+c_{n-1}(\\underset{x\\to a}{\\lim}x)^{n-1}+\\cdots +c_1(\\underset{x\\to a}{\\lim}x)+\\underset{x\\to a}{\\lim}c_0\\hfill \\\\ &amp; =c_na^n+c_{n-1}a^{n-1}+\\cdots +c_1a+c_0\\hfill \\\\ &amp; =p(a)\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572628443\">It now follows from the quotient law that if [latex]p(x)[\/latex] and [latex]q(x)[\/latex] are polynomials for which [latex]q(a)\\ne 0[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1170571672249\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{p(x)}{q(x)}=\\dfrac{p(a)}{q(a)}[\/latex]<\/div>\r\n<p id=\"fs-id1170572305824\">The example below applies this result.<\/p>\r\n\r\n<div id=\"fs-id1170572305829\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating a Limit of a Rational Function<\/h3>\r\n<p id=\"fs-id1170572305839\">Evaluate the [latex]\\underset{x\\to 3}{\\lim}\\dfrac{2x^2-3x+1}{5x+4}[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170572305892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572305892\"]\r\n<p id=\"fs-id1170572305892\">Since 3 is in the domain of the rational function [latex]f(x)=\\frac{2x^2-3x+1}{5x+4}[\/latex], we can calculate the limit by substituting 3 for [latex]x[\/latex] into the function. Thus,<\/p>\r\n\r\n<div id=\"fs-id1170571686198\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 3}{\\lim}\\dfrac{2x^2-3x+1}{5x+4}=\\dfrac{10}{19}[\/latex]\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571675270\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170571675277\">Evaluate [latex]\\underset{x\\to -2}{\\lim}(3x^3-2x+7)[\/latex].<\/p>\r\n[reveal-answer q=\"4482011\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"4482011\"]\r\n<p id=\"fs-id1170571688063\">Use&nbsp;limits of polynomial and rational functions<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170571688072\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571688072\"]\r\n<p id=\"fs-id1170571688072\">[latex]\u221213[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Apply L'Hopital's Rule<\/h2>\r\n<h3>Indeterminate Form of Type [latex]\\frac{0}{0}[\/latex]<\/h3>\r\n<div id=\"fs-id1165043085155\" class=\"bc-section section\">\r\n<p id=\"fs-id1165042941863\">L\u2019H\u00f4pital\u2019s rule can be used to evaluate limits involving the quotient of two functions. Consider<\/p>\r\n\r\n<div id=\"fs-id1165042458008\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042613148\">If [latex]\\underset{x\\to a}{\\lim}f(x)=L_1[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=L_2 \\ne 0[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1165043088102\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\dfrac{L_1}{L_2}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042330308\">However, what happens if [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex]? We call this one of the <strong>indeterminate forms<\/strong>, of type [latex]\\frac{0}{0}[\/latex]. This is considered an indeterminate form because we cannot determine the exact behavior of [latex]\\frac{f(x)}{g(x)}[\/latex] as [latex]x\\to a[\/latex] without further analysis.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043085155\" class=\"bc-section section\">\r\n<div id=\"fs-id1165043352593\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">L\u2019H\u00f4pital\u2019s Rule (0\/0 Case)<\/h3>\r\n\r\n<hr>\r\n<p id=\"fs-id1165043276140\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval containing [latex]a[\/latex], except possibly at [latex]a[\/latex]. If [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1165043020148\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex],<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<p id=\"fs-id1165043035673\">assuming the limit on the right exists or is [latex]\\infty [\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if we are considering one-sided limits, or if [latex]a=\\infty[\/latex] or [latex]-\\infty[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042712928\" class=\"bc-section section\">\r\n<div id=\"fs-id1165043397578\" class=\"textbook exercises\">\r\n<h3>Example: Applying L\u2019H\u00f4pital\u2019s Rule (0\/0 Case)<\/h3>\r\n<p id=\"fs-id1165042456913\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\r\n\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{1- \\cos x}{x}[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to 1}{\\lim}\\dfrac{\\sin (\\pi x)}{\\ln x}[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{e^{\\frac{1}{x}}-1}{\\frac{1}{x}}[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{\\sin x-x}{x^2}[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1165043104016\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"fs-id1165042535045\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042535045\"]\r\n<ol id=\"fs-id1165042535045\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Since the numerator [latex]1- \\cos x\\to 0[\/latex] and the denominator [latex]x\\to 0[\/latex], we can apply L\u2019H\u00f4pital\u2019s rule to evaluate this limit. We have\r\n<div id=\"fs-id1165042328696\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to 0}{\\lim}\\frac{1- \\cos x}{x} &amp; =\\underset{x\\to 0}{\\lim}\\frac{\\frac{d}{dx}(1- \\cos x)}{\\frac{d}{dx}(x)} \\\\ &amp; =\\underset{x\\to 0}{\\lim}\\frac{\\sin x}{1} \\\\ &amp; =\\frac{\\underset{x\\to 0}{\\lim}(\\sin x)}{\\underset{x\\to 0}{\\lim}(1)} \\\\ &amp; =\\frac{0}{1}=0 \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>As [latex]x\\to 1[\/latex], the numerator [latex]\\sin (\\pi x)\\to 0[\/latex] and the denominator [latex]\\ln x \\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\r\n<div id=\"fs-id1165043116372\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to 1}{\\lim}\\frac{\\sin (\\pi x)}{\\ln x} &amp; =\\underset{x\\to 1}{\\lim}\\frac{\\pi \\cos (\\pi x)}{1\/x} \\\\ &amp; =\\underset{x\\to 1}{\\lim}(\\pi x) \\cos (\\pi x) \\\\ &amp; =(\\pi \\cdot 1)(-1)=\u2212\\pi \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>As [latex]x\\to \\infty[\/latex], the numerator [latex]e^{1\/x}-1\\to 0[\/latex] and the denominator [latex](\\frac{1}{x})\\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\r\n<div id=\"fs-id1165042327460\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\frac{e^{1\/x}-1}{\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\frac{e^{1\/x}(\\frac{-1}{x^2})}{(\\frac{-1}{x^2})}=\\underset{x\\to \\infty}{\\lim} e^{1\/x}=e^0=1[\/latex]<\/div><\/li>\r\n \t<li>As [latex]x\\to 0[\/latex], both the numerator and denominator approach zero. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\r\n<div id=\"fs-id1165042562974\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x-x}{x^2}=\\underset{x\\to 0}{\\lim}\\frac{\\cos x-1}{2x}[\/latex].<\/div>\r\nSince the numerator and denominator of this new quotient both approach zero as [latex]x\\to 0[\/latex], we apply L\u2019H\u00f4pital\u2019s rule again. In doing so, we see that\r\n<div id=\"fs-id1165043281524\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\cos x-1}{2x}=\\underset{x\\to 0}{\\lim}\\frac{\u2212\\sin x}{2}=0[\/latex].<\/div>\r\nTherefore, we conclude that\r\n<div id=\"fs-id1165043284142\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x-x}{x^2}=0[\/latex].<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\underset{x\\to 0}{\\lim}\\dfrac{x}{\\tan x}[\/latex].\r\n\r\n[reveal-answer q=\"37002811\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"37002811\"]\r\n\r\n[latex]\\frac{d}{dx} \\tan x= \\sec ^2 x[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042377480\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042377480\"]\r\n\r\n[latex]1[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<h3>Indeterminate Form of Type [latex]\\frac{\\infty}{\\infty}[\/latex]<\/h3>\r\n<p id=\"fs-id1165042318672\">We can also use L\u2019H\u00f4pital\u2019s rule to evaluate limits of quotients [latex]\\frac{f(x)}{g(x)}[\/latex] in which [latex]f(x)\\to \\pm \\infty [\/latex] and [latex]g(x)\\to \\pm \\infty[\/latex]. Limits of this form are classified as <em>indeterminate forms of type<\/em> [latex]\\infty \/ \\infty[\/latex]. Again, note that we are not actually dividing [latex]\\infty[\/latex] by [latex]\\infty[\/latex]. Since [latex]\\infty[\/latex] is not a real number, that is impossible; rather, [latex]\\infty \/ \\infty[\/latex] is used to represent a quotient of limits, each of which is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165042330826\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/ \\infty[\/latex] Case)<\/h3>\r\n\r\n<hr>\r\n<p id=\"fs-id1165043426174\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval containing [latex]a[\/latex], except possibly at [latex]a[\/latex]. Suppose [latex]\\underset{x\\to a}{\\lim}f(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]) and [latex]\\underset{x\\to a}{\\lim}g(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]). Then,<\/p>\r\n\r\n<div id=\"fs-id1165042373703\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex],<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<p id=\"fs-id1165042707280\">assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if the limit is infinite, if [latex]a=\\infty[\/latex] or [latex]\u2212\\infty[\/latex], or the limit is one-sided.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043427623\" class=\"textbook exercises\">\r\n<h3>Example: Applying L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/\\infty[\/latex] Case)<\/h3>\r\n<p id=\"fs-id1165042709794\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\r\n\r\n<ol id=\"fs-id1165042709798\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1165043427625\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"fs-id1165042376758\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042376758\"]\r\n<ol id=\"fs-id1165042376758\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Since [latex]3x+5[\/latex] and [latex]2x+1[\/latex] are first-degree polynomials with positive leading coefficients, [latex]\\underset{x\\to \\infty }{\\lim}(3x+5)=\\infty [\/latex] and [latex]\\underset{x\\to \\infty }{\\lim}(2x+1)=\\infty[\/latex]. Therefore, we apply L\u2019H\u00f4pital\u2019s rule and obtain\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim}\\dfrac{3x+5}{2x+\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3}{2}=\\dfrac{3}{2}[\/latex].<\/div>\r\nNote that this limit can also be calculated without invoking L\u2019H\u00f4pital\u2019s rule. Earlier in the module we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of [latex]x[\/latex] in the denominator. In doing so, we saw that\r\n<div id=\"fs-id1165042319986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3+\\frac{5}{x}}{2+\\frac{1}{x}}=\\dfrac{3}{2}[\/latex].<\/div>\r\nL\u2019H\u00f4pital\u2019s rule provides us with an alternative means of evaluating this type of limit.<\/li>\r\n \t<li>Here, [latex]\\underset{x\\to 0^+}{\\lim} \\ln x=\u2212\\infty [\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} \\cot x=\\infty[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule and obtain\r\n<div id=\"fs-id1165042374803\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\frac{1}{x}}{\u2212\\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}[\/latex].<\/div>\r\nNow as [latex]x\\to 0^+[\/latex], [latex]\\csc^2 x\\to \\infty[\/latex]. Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of [latex]\\csc x[\/latex] to write\r\n<div id=\"fs-id1165042376466\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}[\/latex].<\/div>\r\nNow [latex]\\underset{x\\to 0^+}{\\lim} \\sin^2 x=0[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} x=0[\/latex], so we apply L\u2019H\u00f4pital\u2019s rule again. We find\r\n<div id=\"fs-id1165043249678\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{2 \\sin x \\cos x}{-1}=\\dfrac{0}{-1}=0[\/latex].<\/div>\r\nWe conclude that\r\n<div id=\"fs-id1165042315721\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=0[\/latex].<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042333392\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{\\ln x}{5x}[\/latex]\r\n\r\n[reveal-answer q=\"30011179\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"30011179\"]\r\n\r\n[latex]\\frac{d}{dx}\\ln x=\\frac{1}{x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042367881\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042367881\"]\r\n\r\n0\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165042331798\">L\u2019H\u00f4pital\u2019s rule is very useful for evaluating limits involving the indeterminate forms [latex]\\frac{0}{0}[\/latex] and [latex]\\frac{\\infty}{\\infty}[\/latex]. However, we can also use L\u2019H\u00f4pital\u2019s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits. The expressions [latex]0 \\cdot \\infty[\/latex], [latex]\\infty - \\infty[\/latex], [latex]1^{\\infty}[\/latex], [latex]\\infty^0[\/latex], and [latex]0^0[\/latex] are all considered indeterminate forms. These expressions are not real numbers. Rather, they represent forms that arise when trying to evaluate certain limits. Next we realize why these are indeterminate forms and then understand how to use L\u2019H\u00f4pital\u2019s rule in these cases. The key idea is that we must rewrite the indeterminate forms in such a way that we arrive at the indeterminate form [latex]\\frac{0}{0}[\/latex] or [latex]\\frac{\\infty}{\\infty}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165042320288\" class=\"bc-section section\">\r\n<h3>Indeterminate Form of Type [latex]0 \\cdot \\infty[\/latex]<\/h3>\r\n<p id=\"fs-id1165042320301\">Suppose we want to evaluate [latex]\\underset{x\\to a}{\\lim}(f(x) \\cdot g(x))[\/latex], where [latex]f(x)\\to 0[\/latex] and [latex]g(x)\\to \\infty[\/latex] (or [latex]\u2212\\infty[\/latex]) as [latex]x\\to a[\/latex]. Since one term in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen to the product. We use the notation [latex]0 \\cdot \\infty[\/latex] to denote the form that arises in this situation. The expression [latex]0 \\cdot \\infty[\/latex] is considered indeterminate because we cannot determine without further analysis the exact behavior of the product [latex]f(x)g(x)[\/latex] as [latex]x\\to {a}[\/latex]. For example, let [latex]n[\/latex] be a positive integer and consider<\/p>\r\n\r\n<div id=\"fs-id1165043259749\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\dfrac{1}{(x^n+1)}[\/latex] and [latex]g(x)=3x^2[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042323522\">As [latex]x\\to \\infty[\/latex], [latex]f(x)\\to 0[\/latex] and [latex]g(x)\\to \\infty[\/latex]. However, the limit as [latex]x\\to \\infty [\/latex] of [latex]f(x)g(x)=\\frac{3x^2}{(x^n+1)}[\/latex] varies, depending on [latex]n[\/latex]. If [latex]n=2[\/latex], then [latex]\\underset{x\\to \\infty }{\\lim}f(x)g(x)=3[\/latex]. If [latex]n=1[\/latex], then [latex]\\underset{x\\to \\infty }{\\lim}f(x)g(x)=\\infty[\/latex]. If [latex]n=3[\/latex], then [latex]\\underset{x\\to \\infty }{\\lim}f(x)g(x)=0[\/latex]. Here we consider another limit involving the indeterminate form [latex]0 \\cdot \\infty[\/latex] and show how to rewrite the function as a quotient to use L\u2019H\u00f4pital\u2019s rule.<\/p>\r\n\r\n<div id=\"fs-id1165042368495\" class=\"textbook exercises\">\r\n<h3>Example: Indeterminate Form of Type [latex]0\u00b7\\infty [\/latex]<\/h3>\r\nEvaluate [latex]\\underset{x\\to 0^+}{\\lim}x \\ln x[\/latex]\r\n<div id=\"fs-id1165042368497\" class=\"exercise\">[reveal-answer q=\"fs-id1165042545829\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042545829\"]\r\n<p id=\"fs-id1165042545829\">First, rewrite the function [latex]x \\ln x[\/latex] as a quotient to apply L\u2019H\u00f4pital\u2019s rule. If we write<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x \\ln x=\\frac{\\ln x}{1\/x}[\/latex],<\/div>\r\n<p id=\"fs-id1165042383922\">we see that [latex]\\ln x\\to \u2212\\infty [\/latex] as [latex]x\\to 0^+[\/latex] and [latex]\\frac{1}{x}\\to \\infty [\/latex] as [latex]x\\to 0^+[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule and obtain<\/p>\r\n\r\n<div id=\"fs-id1165042705715\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{\\ln x}{1\/x}=\\underset{x\\to 0^+}{\\lim}\\frac{\\frac{d}{dx}(\\ln x)}{\\frac{d}{dx}(1\/x)}=\\underset{x\\to 0^+}{\\lim}\\frac{1\/x}{-1\/x^2}=\\underset{x\\to 0^+}{\\lim}(\u2212x)=0[\/latex].<\/div>\r\n<p id=\"fs-id1165042318647\">We conclude that<\/p>\r\n\r\n<div id=\"fs-id1165042318650\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}x \\ln x=0[\/latex].<\/div>\r\n<div><\/div>\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"358\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211307\/CNX_Calc_Figure_04_08_004.jpg\" alt=\"The function y = x ln(x) is graphed for values x \u2265 0. At x = 0, the value of the function is 0.\" width=\"358\" height=\"347\"> Finding the limit at [latex]x=0[\/latex] of the function [latex]f(x)=x \\ln x[\/latex].[\/caption]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165043430905\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\underset{x\\to 0}{\\lim}x \\cot x[\/latex]\r\n\r\n[reveal-answer q=\"404616\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"404616\"]\r\n\r\nWrite [latex]x \\cot x=\\frac{x \\cos x}{\\sin x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043286669\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043286669\"]\r\n\r\n1\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042318802\" class=\"bc-section section\">\r\n<h3>Indeterminate Form of Type [latex]\\infty -\\infty[\/latex]<\/h3>\r\n<p id=\"fs-id1165042318816\">Another type of indeterminate form is [latex]\\infty -\\infty[\/latex]. Consider the following example. Let [latex]n[\/latex] be a positive integer and let [latex]f(x)=3x^n[\/latex] and [latex]g(x)=3x^2+5[\/latex]. As [latex]x\\to \\infty[\/latex], [latex]f(x)\\to \\infty [\/latex] and [latex]g(x)\\to \\infty [\/latex]. We are interested in [latex]\\underset{x\\to \\infty}{\\lim}(f(x)-g(x))[\/latex]. Depending on whether [latex]f(x)[\/latex] grows faster, [latex]g(x)[\/latex] grows faster, or they grow at the same rate, as we see next, anything can happen in this limit. Since [latex]f(x)\\to \\infty [\/latex] and [latex]g(x)\\to \\infty[\/latex], we write [latex]\\infty -\\infty [\/latex] to denote the form of this limit. As with our other indeterminate forms, [latex]\\infty -\\infty [\/latex] has no meaning on its own and we must do more analysis to determine the value of the limit. For example, suppose the exponent [latex]n[\/latex] in the function [latex]f(x)=3x^n[\/latex] is [latex]n=3[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1165043430807\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x^3-3x^2-5)=\\infty[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042333220\">On the other hand, if [latex]n=2[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1165042333235\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x^2-3x^2-5)=-5[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043395183\">However, if [latex]n=1[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1165043254251\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x-3x^2-5)=\u2212\\infty[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043323851\">Therefore, the limit cannot be determined by considering only [latex]\\infty -\\infty[\/latex]. Next we see how to rewrite an expression involving the indeterminate form [latex]\\infty -\\infty [\/latex] as a fraction to apply L\u2019H\u00f4pital\u2019s rule.<\/p>\r\n\r\n<div id=\"fs-id1165043323875\" class=\"textbook exercises\">\r\n<h3>Example: Indeterminate Form of Type [latex]\\infty -\\infty[\/latex]<\/h3>\r\nEvaluate [latex]\\underset{x\\to 0^+}{\\lim}\\left(\\dfrac{1}{x^2}-\\dfrac{1}{\\tan x}\\right)[\/latex].\r\n<div id=\"fs-id1165043323877\" class=\"exercise\">[reveal-answer q=\"fs-id1165043281296\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043281296\"]\r\n<p id=\"fs-id1165043281296\">By combining the fractions, we can write the function as a quotient. Since the least common denominator is [latex]x^2 \\tan x[\/latex], we have<\/p>\r\n\r\n<div id=\"fs-id1165043281316\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{x^2}-\\frac{1}{\\tan x}=\\frac{(\\tan x)-x^2}{x^2 \\tan x}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043259808\">As [latex]x\\to 0^+[\/latex], the numerator [latex]\\tan x-x^2 \\to 0[\/latex] and the denominator [latex]x^2 \\tan x \\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. Taking the derivatives of the numerator and the denominator, we have<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{(\\tan x)-x^2}{x^2 \\tan x}=\\underset{x\\to 0^+}{\\lim}\\frac{(\\sec^2 x)-2x}{x^2 \\sec^2 x+2x \\tan x}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043327626\">As [latex]x\\to 0^+[\/latex], [latex](\\sec^2 x)-2x \\to 1[\/latex] and [latex]x^2 \\sec^2 x+2x \\tan x \\to 0[\/latex]. Since the denominator is positive as [latex]x[\/latex] approaches zero from the right, we conclude that<\/p>\r\n\r\n<div id=\"fs-id1165042710940\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{(\\sec^2 x)-2x}{x^2 \\sec^2 x+2x \\tan x}=\\infty[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043396304\">Therefore,<\/p>\r\n\r\n<div id=\"fs-id1165043396307\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}(\\frac{1}{x^2}-\\frac{1}{ tan x})=\\infty[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165043348549\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\underset{x\\to 0^+}{\\lim}\\left(\\dfrac{1}{x}-\\dfrac{1}{\\sin x}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"1238807\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"1238807\"]\r\n\r\nRewrite the difference of fractions as a single fraction.\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043317356\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043317356\"]\r\n\r\n0\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Other Types of Indeterminate Form<\/h3>\r\n<p id=\"fs-id1165042364139\">Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions [latex]0^0[\/latex], [latex]\\infty^0[\/latex], and [latex]1^{\\infty}[\/latex] are all indeterminate forms. On their own, these expressions are meaningless because we cannot actually evaluate these expressions as we would evaluate an expression involving real numbers. Rather, these expressions represent forms that arise when finding limits. Now we examine how L\u2019H\u00f4pital\u2019s rule can be used to evaluate limits involving these indeterminate forms.<\/p>\r\n<p id=\"fs-id1165042364178\">Since L\u2019H\u00f4pital\u2019s rule applies to quotients, we use the natural logarithm function and its properties to reduce a problem evaluating a limit involving exponents to a related problem involving a limit of a quotient. For example, suppose we want to evaluate [latex]\\underset{x\\to a}{\\lim}f(x)^{g(x)}[\/latex] and we arrive at the indeterminate form [latex]\\infty^0[\/latex]. (The indeterminate forms [latex]0^0[\/latex] and [latex]1^{\\infty}[\/latex] can be handled similarly.)<\/p>\r\n\r\n<div id=\"fs-id1165043390815\" class=\"textbook exercises\">\r\n<h3>Example: Indeterminate Form of Type [latex]\\infty^0[\/latex]<\/h3>\r\nEvaluate [latex]\\underset{x\\to \\infty }{\\lim} x^{\\frac{1}{x}}[\/latex]\r\n<div id=\"fs-id1165043390817\" class=\"exercise\">[reveal-answer q=\"fs-id1165043390866\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043390866\"]\r\n<p id=\"fs-id1165043390866\">Let [latex]y=x^{1\/x}[\/latex]. Then,<\/p>\r\n\r\n<div id=\"fs-id1165043281565\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln (x^{1\/x})=\\frac{1}{x} \\ln x=\\frac{\\ln x}{x}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043281615\">We need to evaluate [latex]\\underset{x\\to \\infty }{\\lim}\\frac{\\ln x}{x}[\/latex]. Applying L\u2019H\u00f4pital\u2019s rule, we obtain<\/p>\r\n\r\n<div id=\"fs-id1165043281645\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim} \\ln y=\\underset{x\\to \\infty}{\\lim}\\frac{\\ln x}{x}=\\underset{x\\to \\infty}{\\lim}\\frac{1\/x}{1}=0[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043173746\">Therefore, [latex]\\underset{x\\to \\infty }{\\lim}\\ln y=0[\/latex]. Since the natural logarithm function is continuous, we conclude that<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln (\\underset{x\\to \\infty}{\\lim} y)=0[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043427387\">which leads to<\/p>\r\n\r\n<div id=\"fs-id1165043427390\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim} y=\\underset{x\\to \\infty}{\\lim}\\frac{\\ln x}{x}=e^0=1[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042407320\">Hence,<\/p>\r\n\r\n<div id=\"fs-id1165042407323\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim} x^{1\/x}=1[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042407362\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\underset{x\\to \\infty}{\\lim} x^{\\frac{1}{\\ln x}}[\/latex]\r\n\r\n[reveal-answer q=\"3762844\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3762844\"]\r\n\r\nLet [latex]y=x^{1\/ \\ln x}[\/latex] and apply the natural logarithm to both sides of the equation.\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043108248\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043108248\"]\r\n\r\n[latex]e[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043108292\" class=\"textbook exercises\">\r\n<h3>Example: Indeterminate Form of Type [latex]0^0[\/latex]<\/h3>\r\nEvaluate [latex]\\underset{x\\to 0^+}{\\lim} x^{\\sin x}[\/latex]\r\n<div id=\"fs-id1165043108295\" class=\"exercise\">[reveal-answer q=\"fs-id1165042657755\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042657755\"]\r\n<p id=\"fs-id1165042657755\">Let<\/p>\r\n\r\n<div id=\"fs-id1165042657759\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=x^{\\sin x}[\/latex]<\/div>\r\n<p id=\"fs-id1165042657780\">Therefore,<\/p>\r\n\r\n<div id=\"fs-id1165042657783\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln y=\\ln (x^{\\sin x})= \\sin x \\ln x[\/latex]<\/div>\r\n<p id=\"fs-id1165042707171\">We now evaluate [latex]\\underset{x\\to 0^+}{\\lim} \\sin x \\ln x[\/latex]. Since [latex]\\underset{x\\to 0^+}{\\lim} \\sin x=0[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} \\ln x=\u2212\\infty[\/latex], we have the indeterminate form [latex]0 \\cdot \\infty[\/latex]. To apply L\u2019H\u00f4pital\u2019s rule, we need to rewrite [latex] \\sin x \\ln x[\/latex] as a fraction. We could write<\/p>\r\n\r\n<div id=\"fs-id1165043173832\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex] \\sin x \\ln x=\\frac{\\sin x}{1\/ \\ln x}[\/latex]<\/div>\r\n<p id=\"fs-id1165043173865\">or<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex] \\sin x \\ln x=\\frac{\\ln x}{1\/ \\sin x}=\\frac{\\ln x}{\\csc x}[\/latex]<\/div>\r\n<p id=\"fs-id1165042364489\">Let\u2019s consider the first option. In this case, applying L\u2019H\u00f4pital\u2019s rule, we would obtain<\/p>\r\n\r\n<div id=\"fs-id1165042364494\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} \\sin x \\ln x=\\underset{x\\to 0^+}{\\lim}\\frac{\\sin x}{1\/ \\ln x}=\\underset{x\\to 0^+}{\\lim}\\frac{\\cos x}{-1\/(x(\\ln x)^2)}=\\underset{x\\to 0^+}{\\lim}(\u2212x(\\ln x)^2 \\cos x)[\/latex]<\/div>\r\n<p id=\"fs-id1165043317274\">Unfortunately, we not only have another expression involving the indeterminate form [latex]0 \\cdot \\infty[\/latex], but the new limit is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By writing<\/p>\r\n\r\n<div id=\"fs-id1165043131555\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex] \\sin x \\ln x=\\frac{\\ln x}{1\/ \\sin x}=\\frac{\\ln x}{\\csc x}[\/latex]<\/div>\r\n<p id=\"fs-id1165043131604\">and applying L\u2019H\u00f4pital\u2019s rule, we obtain<\/p>\r\n\r\n<div id=\"fs-id1165043131609\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} \\sin x \\ln x=\\underset{x\\to 0^+}{\\lim}\\frac{\\ln x}{\\csc x}=\\underset{x\\to 0^+}{\\lim}\\frac{1\/x}{\u2212 \\csc x \\cot x}=\\underset{x\\to 0^+}{\\lim}\\frac{-1}{x \\csc x \\cot x}[\/latex]<\/div>\r\n<p id=\"fs-id1165042651533\">Using the fact that [latex]\\csc x=\\frac{1}{\\sin x}[\/latex] and [latex]\\cot x=\\frac{\\cos x}{\\sin x}[\/latex], we can rewrite the expression on the right-hand side as<\/p>\r\n\r\n<div id=\"fs-id1165043251999\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{\u2212\\sin^2 x}{x \\cos x}=\\underset{x\\to 0^+}{\\lim}[\\frac{\\sin x}{x} \\cdot (\u2212\\tan x)]=(\\underset{x\\to 0^+}{\\lim}\\frac{\\sin x}{x}) \\cdot (\\underset{x\\to 0^+}{\\lim}(\u2212\\tan x))=1 \\cdot 0=0[\/latex]<\/div>\r\n<p id=\"fs-id1165042676314\">We conclude that [latex]\\underset{x\\to 0^+}{\\lim} \\ln y=0[\/latex]. Therefore, [latex]\\ln (\\underset{x\\to 0^+}{\\lim} y)=0[\/latex] and we have<\/p>\r\n\r\n<div id=\"fs-id1165042327527\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} y=\\underset{x\\to 0^+}{\\lim} x^{\\sin x}=e^0=1[\/latex]<\/div>\r\n<p id=\"fs-id1165042327592\">Hence,<\/p>\r\n\r\n<div id=\"fs-id1165042327595\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} x^{\\sin x}=1[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042660254\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\underset{x\\to 0^+}{\\lim} x^x[\/latex]\r\n\r\n[reveal-answer q=\"929037\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"929037\"]\r\n\r\nLet [latex]y=x^x[\/latex] and take the natural logarithm of both sides of the equation.\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042660293\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042660293\"]\r\n\r\n1\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Evaluate trigonometric functions using the unit circle<\/li>\n<li>Recognize the basic limit laws<\/li>\n<li>Use the limit laws to evaluate the limit of a function<\/li>\n<li>Recognize when to apply L\u2019H\u00f4pital\u2019s rule<\/li>\n<\/ul>\n<\/div>\n<p>In the Vector-Valued Functions and Space Curves section, we will define and develop skills needed to understand and work with vector-valued functions. Here we will review how to evaluate trigonometric functions at specific angle measures, apply limit laws, and apply L\u2019H\u00f4pital\u2019s rule.<\/p>\n<h2>Evaluate Trigonometric Functions Using the Unit Circle<\/h2>\n<p><strong><em>(also in Module 1, Skills Review for Polar Coordinates)<\/em><\/strong><br \/>\nThe unit circle tells us the value of cosine and sine at any of the given angle measures seen below. The first coordinate in each ordered pair is the value of cosine at the given angle measure, while the second coordinate in each ordered pair is the value of sine at the given angle measure. You will learn that all trigonometric functions can be written in terms of sine and cosine. Thus, if you can evaluate sine and cosine at various angle values, you can also evaluate the other trigonometric functions at various angle values. Take time to learn the [latex]\\left(x,y\\right)[\/latex] coordinates of all of the major angles in the first quadrant of the unit circle.<\/p>\n<p>Remember, every angle in quadrant two, three, or four has a reference angle that lies in quadrant one. The quadrant of the original angle only affects the sign (positive or negative) of a trigonometric function&#8217;s value at a given angle.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-12625 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003609\/f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3-IMAGE-IMAGE.png\" alt=\"f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3+IMAGE+IMAGE\" width=\"800\" height=\"728\" \/><\/p>\n<div class=\"textbox\">\n<h3>A General Note: Evaluating Tangent, Secant, Cosecant, and Cotangent Functions<\/h3>\n<p>If [latex]\\theta[\/latex] is an angle measure, then, using the unit circle,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\tan \\theta=\\frac{sin \\theta}{cos \\theta}\\\\ \\sec \\theta=\\frac{1}{cos \\theta}\\\\ \\csc t=\\frac{1}{sin \\theta}\\\\ \\cot \\theta=\\frac{cos \\theta}{sin \\theta}\\end{gathered}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Unit Circle to Find the Value of Trigonometric Functions<\/h3>\n<p>Find [latex]\\sin \\theta,\\cos \\theta,\\tan \\theta,\\sec \\theta,\\csc \\theta[\/latex], and [latex]\\cot \\theta[\/latex] when [latex]\\theta=\\frac{\\pi }{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q151548\">Show Solution<\/span><\/p>\n<div id=\"q151548\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{align}&\\sin \\frac{\\pi }{3}=\\frac{\\sqrt{3}}{2}\\\\ &\\cos \\frac{\\pi }{3}=\\frac{1}{2}\\\\ &\\tan \\frac{\\pi }{3}=\\sqrt{3} \\text{ (From}\\frac{sin \\theta}{cos \\theta})\\\\ &\\sec \\frac{\\pi }{3}=2 \\text{ (From}\\frac{1}{cos \\theta})\\\\ &\\csc \\frac{\\pi }{3}=\\frac{2\\sqrt{3}}{3}\\text{ (From}\\frac{1}{sin \\theta} \\text{ - do not forget to rationalize})\\\\ &\\cot \\frac{\\pi }{3}=\\frac{\\sqrt{3}}{3}(\\text{ (From}\\frac{cos \\theta}{sin \\theta} \\text{ - do not forget to rationalize})\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173354\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173354&theme=oea&iframe_resize_id=ohm173354\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Below are the values of all six trigonometric functions evaluated at common angle measures from Quadrant I.<\/p>\n<table id=\"Table_05_03_01\" summary=\"..\">\n<tbody>\n<tr>\n<td><strong>Angle<\/strong><\/td>\n<td><strong> [latex]0[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{6},\\text{ or }{30}^{\\circ}[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{4},\\text{ or } {45}^{\\circ }[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{3},\\text{ or }{60}^{\\circ }[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{2},\\text{ or }{90}^{\\circ }[\/latex] <\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>Cosine<\/strong><\/td>\n<td>1<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td><strong>Sine<\/strong><\/td>\n<td>0<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n<td>1<\/td>\n<\/tr>\n<tr>\n<td><strong>Tangent<\/strong><\/td>\n<td>0<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\n<td>1<\/td>\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\n<td>Undefined<\/td>\n<\/tr>\n<tr>\n<td><strong>Secant<\/strong><\/td>\n<td>1<\/td>\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\n<td>2<\/td>\n<td>Undefined<\/td>\n<\/tr>\n<tr>\n<td><strong>Cosecant<\/strong><\/td>\n<td>Undefined<\/td>\n<td>2<\/td>\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\n<td>1<\/td>\n<\/tr>\n<tr>\n<td><strong>Cotangent<\/strong><\/td>\n<td>Undefined<\/td>\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\n<td>1<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\n<td>0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Limit Laws<\/h2>\n<p id=\"fs-id1170571680609\"> The basic limit laws stated below, together with the other limit laws, allow us to evaluate limits of many algebraic functions.<\/p>\n<div id=\"fs-id1170572451153\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Basic Limit Results<\/h3>\n<hr \/>\n<p id=\"fs-id1170572205248\">For any real number [latex]a[\/latex] and any constant [latex]c[\/latex],<\/p>\n<ol id=\"fs-id1170572286963\">\n<li>\n<div id=\"fs-id1170572624896\" class=\"equation\">[latex]\\underset{x\\to a}{\\lim}x=a[\/latex]<\/div>\n<\/li>\n<li>\n<div id=\"fs-id1170572209025\" class=\"equation\">[latex]\\underset{x\\to a}{\\lim}c=c[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1170572111463\" class=\"textbook exercises\">\n<h3>Example: Evaluating a Basic Limit<\/h3>\n<p id=\"fs-id1170571569246\">Evaluate each of the following limits using the basic limit results above.<\/p>\n<ol id=\"fs-id1170572176731\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\underset{x\\to 2}{\\lim}x[\/latex]<\/li>\n<li>[latex]\\underset{x\\to 2}{\\lim}5[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572101621\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572101621\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572101621\" style=\"list-style-type: lower-alpha;\">\n<li>The limit of [latex]x[\/latex] as [latex]x[\/latex] approaches [latex]a[\/latex] is [latex]a[\/latex]: [latex]\\underset{x\\to 2}{\\lim}x=2[\/latex].<\/li>\n<li>The limit of a constant is that constant: [latex]\\underset{x\\to 2}{\\lim}5=5[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm4886\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=4886&theme=oea&iframe_resize_id=ohm4886&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1170572570027\">We now take a look at the limit laws, the individual properties of limits. The proofs that these laws hold are omitted here.<\/p>\n<div id=\"fs-id1170572508800\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Limit Laws<\/h3>\n<hr \/>\n<p id=\"fs-id1170572086164\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be defined for all [latex]x\\ne a[\/latex] over some open interval containing [latex]a[\/latex]. Assume that [latex]L[\/latex] and [latex]M[\/latex] are real numbers such that [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=M[\/latex]. Let [latex]c[\/latex] be a constant. Then, each of the following statements holds:<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572204187\"><strong>Sum law for limits<\/strong>: [latex]\\underset{x\\to a}{\\lim}(f(x)+g(x))=\\underset{x\\to a}{\\lim}f(x)+\\underset{x\\to a}{\\lim}g(x)=L+M[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572627273\"><strong>Difference law for limits<\/strong>: [latex]\\underset{x\\to a}{\\lim}(f(x)-g(x))=\\underset{x\\to a}{\\lim}f(x)-\\underset{x\\to a}{\\lim}g(x)=L-M[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572450574\"><strong>Constant multiple law for limits<\/strong>: [latex]\\underset{x\\to a}{\\lim}cf(x)=c \\cdot \\underset{x\\to a}{\\lim}f(x)=cL[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572104032\"><strong>Product law for limits<\/strong>: [latex]\\underset{x\\to a}{\\lim}(f(x) \\cdot g(x))=\\underset{x\\to a}{\\lim}f(x) \\cdot \\underset{x\\to a}{\\lim}g(x)=L \\cdot M[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572347458\"><strong>Quotient law for limits<\/strong>: [latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\dfrac{\\underset{x\\to a}{\\lim}f(x)}{\\underset{x\\to a}{\\lim}g(x)}=\\frac{L}{M}[\/latex] for [latex]M\\ne 0[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572246193\"><strong>Power law for limits<\/strong>: [latex]\\underset{x\\to a}{\\lim}(f(x))^n=(\\underset{x\\to a}{\\lim}f(x))^n=L^n[\/latex] for every positive integer [latex]n[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572232633\"><strong>Root law for limits<\/strong>: [latex]\\underset{x\\to a}{\\lim}\\sqrt[n]{f(x)}=\\sqrt[n]{\\underset{x\\to a}{\\lim}f(x)}=\\sqrt[n]{L}[\/latex] for all [latex]L[\/latex] if [latex]n[\/latex] is odd and for [latex]L\\ge 0[\/latex] if [latex]n[\/latex] is even<\/p>\n<\/div>\n<p id=\"fs-id1170572479215\">We now practice applying these limit laws to evaluate a limit.<\/p>\n<div id=\"fs-id1170572451489\" class=\"textbook exercises\">\n<h3>Example: Evaluating a Limit Using Limit Laws<\/h3>\n<p id=\"fs-id1170572109838\">Use the limit laws to evaluate [latex]\\underset{x\\to -3}{\\lim}(4x+2)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572169042\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572169042\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572169042\">Let\u2019s apply the limit laws one step at a time to be sure we understand how they work. We need to keep in mind the requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied.<\/p>\n<p id=\"fs-id1170571565987\">[latex]\\begin{array}{ccccc}\\underset{x\\to -3}{\\lim}(4x+2)\\hfill & =\\underset{x\\to -3}{\\lim}4x+\\underset{x\\to -3}{\\lim}2\\hfill & & & \\text{Apply the sum law.}\\hfill \\\\ & =4 \\cdot \\underset{x\\to -3}{\\lim}x+\\underset{x\\to -3}{\\lim}2\\hfill & & & \\text{Apply the constant multiple law.}\\hfill \\\\ & =4 \\cdot (-3)+2=-10\\hfill & & & \\text{Apply the basic limit results and simplify.}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572509954\" class=\"textbook exercises\">\n<h3>Example: Using Limit Laws Repeatedly<\/h3>\n<p>Use the limit laws to evaluate [latex]\\underset{x\\to 2}{\\lim}\\dfrac{2x^2-3x+1}{x^3+4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572506406\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572506406\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572506406\">To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccccc}\\\\ \\\\ \\underset{x\\to 2}{\\lim}\\large \\frac{2x^2-3x+1}{x^3+4} & = \\large \\frac{\\underset{x\\to 2}{\\lim}(2x^2-3x+1)}{\\underset{x\\to 2}{\\lim}(x^3+4)} & & & \\text{Apply the quotient law, making sure that} \\, 2^3+4\\ne 0 \\\\ & = \\large \\frac{2 \\cdot \\underset{x\\to 2}{\\lim}x^2-3 \\cdot \\underset{x\\to 2}{\\lim}x+\\underset{x\\to 2}{\\lim}1}{\\underset{x\\to 2}{\\lim}x^3+\\underset{x\\to 2}{\\lim}4} & & & \\text{Apply the sum law and constant multiple law.} \\\\ & = \\large \\frac{2 \\cdot (\\underset{x\\to 2}{\\lim}x)^2-3 \\cdot \\underset{x\\to 2}{\\lim}x+\\underset{x\\to 2}{\\lim}1}{(\\underset{x\\to 2}{\\lim}x)^3+\\underset{x\\to 2}{\\lim}4} & & & \\text{Apply the power law.} \\\\ & = \\large \\frac{2(4)-3(2)+1}{2^3+4}=\\frac{1}{4} & & & \\text{Apply the basic limit laws and simplify.} \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170571655486\">Use the limit laws to evaluate [latex]\\underset{x\\to 6}{\\lim}(2x-1)\\sqrt{x+4}[\/latex]. In each step, indicate the limit law applied.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q6635113\">Hint<\/span><\/p>\n<div id=\"q6635113\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572209920\">Begin by applying the product law.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572094142\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572094142\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572094142\">[latex]11\\sqrt{10}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Limits of Polynomial and Rational Functions<\/h2>\n<p id=\"fs-id1170572133214\">By now you have probably noticed that, in each of the previous examples, it has been the case that [latex]\\underset{x\\to a}{\\lim}f(x)=f(a)[\/latex]. This is not always true, but it does hold for all polynomials for any choice of [latex]a[\/latex] and for all rational functions at all values of [latex]a[\/latex] for which the rational function is defined.<\/p>\n<div id=\"fs-id1170572557796\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Limits of Polynomial and Rational Functions<\/h3>\n<hr \/>\n<p id=\"fs-id1170572557802\">Let [latex]p(x)[\/latex] and [latex]q(x)[\/latex] be polynomial functions. Let [latex]a[\/latex] be a real number. Then,<\/p>\n<div id=\"fs-id1170572347161\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}p(x)=p(a)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<div><\/div>\n<div id=\"fs-id1170571656084\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{p(x)}{q(x)}=\\dfrac{p(a)}{q(a)} \\, \\text{when} \\, q(a)\\ne 0[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<p id=\"fs-id1170571650163\">To see that this theorem holds, consider the polynomial [latex]p(x)=c_nx^n+c_{n-1}x^{n-1}+\\cdots +c_1x+c_0[\/latex]. By applying the sum, constant multiple, and power laws, we end up with<\/p>\n<div id=\"fs-id1170571648575\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\underset{x\\to a}{\\lim}p(x)& =\\underset{x\\to a}{\\lim}(c_nx^n+c_{n-1}x^{n-1}+\\cdots +c_1x+c_0)\\hfill \\\\ & =c_n(\\underset{x\\to a}{\\lim}x)^n+c_{n-1}(\\underset{x\\to a}{\\lim}x)^{n-1}+\\cdots +c_1(\\underset{x\\to a}{\\lim}x)+\\underset{x\\to a}{\\lim}c_0\\hfill \\\\ & =c_na^n+c_{n-1}a^{n-1}+\\cdots +c_1a+c_0\\hfill \\\\ & =p(a)\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572628443\">It now follows from the quotient law that if [latex]p(x)[\/latex] and [latex]q(x)[\/latex] are polynomials for which [latex]q(a)\\ne 0[\/latex], then<\/p>\n<div id=\"fs-id1170571672249\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{p(x)}{q(x)}=\\dfrac{p(a)}{q(a)}[\/latex]<\/div>\n<p id=\"fs-id1170572305824\">The example below applies this result.<\/p>\n<div id=\"fs-id1170572305829\" class=\"textbook exercises\">\n<h3>Example: Evaluating a Limit of a Rational Function<\/h3>\n<p id=\"fs-id1170572305839\">Evaluate the [latex]\\underset{x\\to 3}{\\lim}\\dfrac{2x^2-3x+1}{5x+4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572305892\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572305892\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572305892\">Since 3 is in the domain of the rational function [latex]f(x)=\\frac{2x^2-3x+1}{5x+4}[\/latex], we can calculate the limit by substituting 3 for [latex]x[\/latex] into the function. Thus,<\/p>\n<div id=\"fs-id1170571686198\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 3}{\\lim}\\dfrac{2x^2-3x+1}{5x+4}=\\dfrac{10}{19}[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571675270\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170571675277\">Evaluate [latex]\\underset{x\\to -2}{\\lim}(3x^3-2x+7)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q4482011\">Hint<\/span><\/p>\n<div id=\"q4482011\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571688063\">Use&nbsp;limits of polynomial and rational functions<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571688072\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571688072\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571688072\">[latex]\u221213[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Apply L&#8217;Hopital&#8217;s Rule<\/h2>\n<h3>Indeterminate Form of Type [latex]\\frac{0}{0}[\/latex]<\/h3>\n<div id=\"fs-id1165043085155\" class=\"bc-section section\">\n<p id=\"fs-id1165042941863\">L\u2019H\u00f4pital\u2019s rule can be used to evaluate limits involving the quotient of two functions. Consider<\/p>\n<div id=\"fs-id1165042458008\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042613148\">If [latex]\\underset{x\\to a}{\\lim}f(x)=L_1[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=L_2 \\ne 0[\/latex], then<\/p>\n<div id=\"fs-id1165043088102\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\dfrac{L_1}{L_2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042330308\">However, what happens if [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex]? We call this one of the <strong>indeterminate forms<\/strong>, of type [latex]\\frac{0}{0}[\/latex]. This is considered an indeterminate form because we cannot determine the exact behavior of [latex]\\frac{f(x)}{g(x)}[\/latex] as [latex]x\\to a[\/latex] without further analysis.<\/p>\n<\/div>\n<div id=\"fs-id1165043085155\" class=\"bc-section section\">\n<div id=\"fs-id1165043352593\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">L\u2019H\u00f4pital\u2019s Rule (0\/0 Case)<\/h3>\n<hr \/>\n<p id=\"fs-id1165043276140\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval containing [latex]a[\/latex], except possibly at [latex]a[\/latex]. If [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex], then<\/p>\n<div id=\"fs-id1165043020148\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<p id=\"fs-id1165043035673\">assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if we are considering one-sided limits, or if [latex]a=\\infty[\/latex] or [latex]-\\infty[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165042712928\" class=\"bc-section section\">\n<div id=\"fs-id1165043397578\" class=\"textbook exercises\">\n<h3>Example: Applying L\u2019H\u00f4pital\u2019s Rule (0\/0 Case)<\/h3>\n<p id=\"fs-id1165042456913\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{1- \\cos x}{x}[\/latex]<\/li>\n<li>[latex]\\underset{x\\to 1}{\\lim}\\dfrac{\\sin (\\pi x)}{\\ln x}[\/latex]<\/li>\n<li>[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{e^{\\frac{1}{x}}-1}{\\frac{1}{x}}[\/latex]<\/li>\n<li>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{\\sin x-x}{x^2}[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1165043104016\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042535045\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042535045\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165042535045\" style=\"list-style-type: lower-alpha;\">\n<li>Since the numerator [latex]1- \\cos x\\to 0[\/latex] and the denominator [latex]x\\to 0[\/latex], we can apply L\u2019H\u00f4pital\u2019s rule to evaluate this limit. We have\n<div id=\"fs-id1165042328696\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to 0}{\\lim}\\frac{1- \\cos x}{x} & =\\underset{x\\to 0}{\\lim}\\frac{\\frac{d}{dx}(1- \\cos x)}{\\frac{d}{dx}(x)} \\\\ & =\\underset{x\\to 0}{\\lim}\\frac{\\sin x}{1} \\\\ & =\\frac{\\underset{x\\to 0}{\\lim}(\\sin x)}{\\underset{x\\to 0}{\\lim}(1)} \\\\ & =\\frac{0}{1}=0 \\end{array}[\/latex]<\/div>\n<\/li>\n<li>As [latex]x\\to 1[\/latex], the numerator [latex]\\sin (\\pi x)\\to 0[\/latex] and the denominator [latex]\\ln x \\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\n<div id=\"fs-id1165043116372\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to 1}{\\lim}\\frac{\\sin (\\pi x)}{\\ln x} & =\\underset{x\\to 1}{\\lim}\\frac{\\pi \\cos (\\pi x)}{1\/x} \\\\ & =\\underset{x\\to 1}{\\lim}(\\pi x) \\cos (\\pi x) \\\\ & =(\\pi \\cdot 1)(-1)=\u2212\\pi \\end{array}[\/latex]<\/div>\n<\/li>\n<li>As [latex]x\\to \\infty[\/latex], the numerator [latex]e^{1\/x}-1\\to 0[\/latex] and the denominator [latex](\\frac{1}{x})\\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\n<div id=\"fs-id1165042327460\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\frac{e^{1\/x}-1}{\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\frac{e^{1\/x}(\\frac{-1}{x^2})}{(\\frac{-1}{x^2})}=\\underset{x\\to \\infty}{\\lim} e^{1\/x}=e^0=1[\/latex]<\/div>\n<\/li>\n<li>As [latex]x\\to 0[\/latex], both the numerator and denominator approach zero. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. We obtain\n<div id=\"fs-id1165042562974\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x-x}{x^2}=\\underset{x\\to 0}{\\lim}\\frac{\\cos x-1}{2x}[\/latex].<\/div>\n<p>Since the numerator and denominator of this new quotient both approach zero as [latex]x\\to 0[\/latex], we apply L\u2019H\u00f4pital\u2019s rule again. In doing so, we see that<\/p>\n<div id=\"fs-id1165043281524\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\cos x-1}{2x}=\\underset{x\\to 0}{\\lim}\\frac{\u2212\\sin x}{2}=0[\/latex].<\/div>\n<p>Therefore, we conclude that<\/p>\n<div id=\"fs-id1165043284142\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}\\frac{\\sin x-x}{x^2}=0[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\underset{x\\to 0}{\\lim}\\dfrac{x}{\\tan x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q37002811\">Hint<\/span><\/p>\n<div id=\"q37002811\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{d}{dx} \\tan x= \\sec ^2 x[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042377480\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042377480\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]1[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<h3>Indeterminate Form of Type [latex]\\frac{\\infty}{\\infty}[\/latex]<\/h3>\n<p id=\"fs-id1165042318672\">We can also use L\u2019H\u00f4pital\u2019s rule to evaluate limits of quotients [latex]\\frac{f(x)}{g(x)}[\/latex] in which [latex]f(x)\\to \\pm \\infty[\/latex] and [latex]g(x)\\to \\pm \\infty[\/latex]. Limits of this form are classified as <em>indeterminate forms of type<\/em> [latex]\\infty \/ \\infty[\/latex]. Again, note that we are not actually dividing [latex]\\infty[\/latex] by [latex]\\infty[\/latex]. Since [latex]\\infty[\/latex] is not a real number, that is impossible; rather, [latex]\\infty \/ \\infty[\/latex] is used to represent a quotient of limits, each of which is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex].<\/p>\n<div id=\"fs-id1165042330826\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/ \\infty[\/latex] Case)<\/h3>\n<hr \/>\n<p id=\"fs-id1165043426174\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval containing [latex]a[\/latex], except possibly at [latex]a[\/latex]. Suppose [latex]\\underset{x\\to a}{\\lim}f(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]) and [latex]\\underset{x\\to a}{\\lim}g(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]). Then,<\/p>\n<div id=\"fs-id1165042373703\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<p id=\"fs-id1165042707280\">assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if the limit is infinite, if [latex]a=\\infty[\/latex] or [latex]\u2212\\infty[\/latex], or the limit is one-sided.<\/p>\n<\/div>\n<div id=\"fs-id1165043427623\" class=\"textbook exercises\">\n<h3>Example: Applying L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/\\infty[\/latex] Case)<\/h3>\n<p id=\"fs-id1165042709794\">Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\n<ol id=\"fs-id1165042709798\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}[\/latex]<\/li>\n<li>[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1165043427625\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042376758\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042376758\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165042376758\" style=\"list-style-type: lower-alpha;\">\n<li>Since [latex]3x+5[\/latex] and [latex]2x+1[\/latex] are first-degree polynomials with positive leading coefficients, [latex]\\underset{x\\to \\infty }{\\lim}(3x+5)=\\infty[\/latex] and [latex]\\underset{x\\to \\infty }{\\lim}(2x+1)=\\infty[\/latex]. Therefore, we apply L\u2019H\u00f4pital\u2019s rule and obtain\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim}\\dfrac{3x+5}{2x+\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3}{2}=\\dfrac{3}{2}[\/latex].<\/div>\n<p>Note that this limit can also be calculated without invoking L\u2019H\u00f4pital\u2019s rule. Earlier in the module we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of [latex]x[\/latex] in the denominator. In doing so, we saw that<\/p>\n<div id=\"fs-id1165042319986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3+\\frac{5}{x}}{2+\\frac{1}{x}}=\\dfrac{3}{2}[\/latex].<\/div>\n<p>L\u2019H\u00f4pital\u2019s rule provides us with an alternative means of evaluating this type of limit.<\/li>\n<li>Here, [latex]\\underset{x\\to 0^+}{\\lim} \\ln x=\u2212\\infty[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} \\cot x=\\infty[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule and obtain\n<div id=\"fs-id1165042374803\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\frac{1}{x}}{\u2212\\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}[\/latex].<\/div>\n<p>Now as [latex]x\\to 0^+[\/latex], [latex]\\csc^2 x\\to \\infty[\/latex]. Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of [latex]\\csc x[\/latex] to write<\/p>\n<div id=\"fs-id1165042376466\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}[\/latex].<\/div>\n<p>Now [latex]\\underset{x\\to 0^+}{\\lim} \\sin^2 x=0[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} x=0[\/latex], so we apply L\u2019H\u00f4pital\u2019s rule again. We find<\/p>\n<div id=\"fs-id1165043249678\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{2 \\sin x \\cos x}{-1}=\\dfrac{0}{-1}=0[\/latex].<\/div>\n<p>We conclude that<\/p>\n<div id=\"fs-id1165042315721\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=0[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042333392\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{\\ln x}{5x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q30011179\">Hint<\/span><\/p>\n<div id=\"q30011179\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{d}{dx}\\ln x=\\frac{1}{x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042367881\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042367881\" class=\"hidden-answer\" style=\"display: none\">\n<p>0<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165042331798\">L\u2019H\u00f4pital\u2019s rule is very useful for evaluating limits involving the indeterminate forms [latex]\\frac{0}{0}[\/latex] and [latex]\\frac{\\infty}{\\infty}[\/latex]. However, we can also use L\u2019H\u00f4pital\u2019s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits. The expressions [latex]0 \\cdot \\infty[\/latex], [latex]\\infty - \\infty[\/latex], [latex]1^{\\infty}[\/latex], [latex]\\infty^0[\/latex], and [latex]0^0[\/latex] are all considered indeterminate forms. These expressions are not real numbers. Rather, they represent forms that arise when trying to evaluate certain limits. Next we realize why these are indeterminate forms and then understand how to use L\u2019H\u00f4pital\u2019s rule in these cases. The key idea is that we must rewrite the indeterminate forms in such a way that we arrive at the indeterminate form [latex]\\frac{0}{0}[\/latex] or [latex]\\frac{\\infty}{\\infty}[\/latex].<\/p>\n<div id=\"fs-id1165042320288\" class=\"bc-section section\">\n<h3>Indeterminate Form of Type [latex]0 \\cdot \\infty[\/latex]<\/h3>\n<p id=\"fs-id1165042320301\">Suppose we want to evaluate [latex]\\underset{x\\to a}{\\lim}(f(x) \\cdot g(x))[\/latex], where [latex]f(x)\\to 0[\/latex] and [latex]g(x)\\to \\infty[\/latex] (or [latex]\u2212\\infty[\/latex]) as [latex]x\\to a[\/latex]. Since one term in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen to the product. We use the notation [latex]0 \\cdot \\infty[\/latex] to denote the form that arises in this situation. The expression [latex]0 \\cdot \\infty[\/latex] is considered indeterminate because we cannot determine without further analysis the exact behavior of the product [latex]f(x)g(x)[\/latex] as [latex]x\\to {a}[\/latex]. For example, let [latex]n[\/latex] be a positive integer and consider<\/p>\n<div id=\"fs-id1165043259749\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\dfrac{1}{(x^n+1)}[\/latex] and [latex]g(x)=3x^2[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042323522\">As [latex]x\\to \\infty[\/latex], [latex]f(x)\\to 0[\/latex] and [latex]g(x)\\to \\infty[\/latex]. However, the limit as [latex]x\\to \\infty[\/latex] of [latex]f(x)g(x)=\\frac{3x^2}{(x^n+1)}[\/latex] varies, depending on [latex]n[\/latex]. If [latex]n=2[\/latex], then [latex]\\underset{x\\to \\infty }{\\lim}f(x)g(x)=3[\/latex]. If [latex]n=1[\/latex], then [latex]\\underset{x\\to \\infty }{\\lim}f(x)g(x)=\\infty[\/latex]. If [latex]n=3[\/latex], then [latex]\\underset{x\\to \\infty }{\\lim}f(x)g(x)=0[\/latex]. Here we consider another limit involving the indeterminate form [latex]0 \\cdot \\infty[\/latex] and show how to rewrite the function as a quotient to use L\u2019H\u00f4pital\u2019s rule.<\/p>\n<div id=\"fs-id1165042368495\" class=\"textbook exercises\">\n<h3>Example: Indeterminate Form of Type [latex]0\u00b7\\infty[\/latex]<\/h3>\n<p>Evaluate [latex]\\underset{x\\to 0^+}{\\lim}x \\ln x[\/latex]<\/p>\n<div id=\"fs-id1165042368497\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042545829\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042545829\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042545829\">First, rewrite the function [latex]x \\ln x[\/latex] as a quotient to apply L\u2019H\u00f4pital\u2019s rule. If we write<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x \\ln x=\\frac{\\ln x}{1\/x}[\/latex],<\/div>\n<p id=\"fs-id1165042383922\">we see that [latex]\\ln x\\to \u2212\\infty[\/latex] as [latex]x\\to 0^+[\/latex] and [latex]\\frac{1}{x}\\to \\infty[\/latex] as [latex]x\\to 0^+[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule and obtain<\/p>\n<div id=\"fs-id1165042705715\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{\\ln x}{1\/x}=\\underset{x\\to 0^+}{\\lim}\\frac{\\frac{d}{dx}(\\ln x)}{\\frac{d}{dx}(1\/x)}=\\underset{x\\to 0^+}{\\lim}\\frac{1\/x}{-1\/x^2}=\\underset{x\\to 0^+}{\\lim}(\u2212x)=0[\/latex].<\/div>\n<p id=\"fs-id1165042318647\">We conclude that<\/p>\n<div id=\"fs-id1165042318650\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}x \\ln x=0[\/latex].<\/div>\n<div><\/div>\n<div>\n<div style=\"width: 368px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211307\/CNX_Calc_Figure_04_08_004.jpg\" alt=\"The function y = x ln(x) is graphed for values x \u2265 0. At x = 0, the value of the function is 0.\" width=\"358\" height=\"347\" \/><\/p>\n<p class=\"wp-caption-text\">Finding the limit at [latex]x=0[\/latex] of the function [latex]f(x)=x \\ln x[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043430905\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\underset{x\\to 0}{\\lim}x \\cot x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q404616\">Hint<\/span><\/p>\n<div id=\"q404616\" class=\"hidden-answer\" style=\"display: none\">\n<p>Write [latex]x \\cot x=\\frac{x \\cos x}{\\sin x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043286669\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043286669\" class=\"hidden-answer\" style=\"display: none\">\n<p>1<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042318802\" class=\"bc-section section\">\n<h3>Indeterminate Form of Type [latex]\\infty -\\infty[\/latex]<\/h3>\n<p id=\"fs-id1165042318816\">Another type of indeterminate form is [latex]\\infty -\\infty[\/latex]. Consider the following example. Let [latex]n[\/latex] be a positive integer and let [latex]f(x)=3x^n[\/latex] and [latex]g(x)=3x^2+5[\/latex]. As [latex]x\\to \\infty[\/latex], [latex]f(x)\\to \\infty[\/latex] and [latex]g(x)\\to \\infty[\/latex]. We are interested in [latex]\\underset{x\\to \\infty}{\\lim}(f(x)-g(x))[\/latex]. Depending on whether [latex]f(x)[\/latex] grows faster, [latex]g(x)[\/latex] grows faster, or they grow at the same rate, as we see next, anything can happen in this limit. Since [latex]f(x)\\to \\infty[\/latex] and [latex]g(x)\\to \\infty[\/latex], we write [latex]\\infty -\\infty[\/latex] to denote the form of this limit. As with our other indeterminate forms, [latex]\\infty -\\infty[\/latex] has no meaning on its own and we must do more analysis to determine the value of the limit. For example, suppose the exponent [latex]n[\/latex] in the function [latex]f(x)=3x^n[\/latex] is [latex]n=3[\/latex], then<\/p>\n<div id=\"fs-id1165043430807\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x^3-3x^2-5)=\\infty[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042333220\">On the other hand, if [latex]n=2[\/latex], then<\/p>\n<div id=\"fs-id1165042333235\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x^2-3x^2-5)=-5[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043395183\">However, if [latex]n=1[\/latex], then<\/p>\n<div id=\"fs-id1165043254251\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x-3x^2-5)=\u2212\\infty[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043323851\">Therefore, the limit cannot be determined by considering only [latex]\\infty -\\infty[\/latex]. Next we see how to rewrite an expression involving the indeterminate form [latex]\\infty -\\infty[\/latex] as a fraction to apply L\u2019H\u00f4pital\u2019s rule.<\/p>\n<div id=\"fs-id1165043323875\" class=\"textbook exercises\">\n<h3>Example: Indeterminate Form of Type [latex]\\infty -\\infty[\/latex]<\/h3>\n<p>Evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\left(\\dfrac{1}{x^2}-\\dfrac{1}{\\tan x}\\right)[\/latex].<\/p>\n<div id=\"fs-id1165043323877\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043281296\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043281296\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043281296\">By combining the fractions, we can write the function as a quotient. Since the least common denominator is [latex]x^2 \\tan x[\/latex], we have<\/p>\n<div id=\"fs-id1165043281316\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{x^2}-\\frac{1}{\\tan x}=\\frac{(\\tan x)-x^2}{x^2 \\tan x}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043259808\">As [latex]x\\to 0^+[\/latex], the numerator [latex]\\tan x-x^2 \\to 0[\/latex] and the denominator [latex]x^2 \\tan x \\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. Taking the derivatives of the numerator and the denominator, we have<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{(\\tan x)-x^2}{x^2 \\tan x}=\\underset{x\\to 0^+}{\\lim}\\frac{(\\sec^2 x)-2x}{x^2 \\sec^2 x+2x \\tan x}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043327626\">As [latex]x\\to 0^+[\/latex], [latex](\\sec^2 x)-2x \\to 1[\/latex] and [latex]x^2 \\sec^2 x+2x \\tan x \\to 0[\/latex]. Since the denominator is positive as [latex]x[\/latex] approaches zero from the right, we conclude that<\/p>\n<div id=\"fs-id1165042710940\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{(\\sec^2 x)-2x}{x^2 \\sec^2 x+2x \\tan x}=\\infty[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043396304\">Therefore,<\/p>\n<div id=\"fs-id1165043396307\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}(\\frac{1}{x^2}-\\frac{1}{ tan x})=\\infty[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043348549\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\left(\\dfrac{1}{x}-\\dfrac{1}{\\sin x}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q1238807\">Hint<\/span><\/p>\n<div id=\"q1238807\" class=\"hidden-answer\" style=\"display: none\">\n<p>Rewrite the difference of fractions as a single fraction.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043317356\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043317356\" class=\"hidden-answer\" style=\"display: none\">\n<p>0<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Other Types of Indeterminate Form<\/h3>\n<p id=\"fs-id1165042364139\">Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions [latex]0^0[\/latex], [latex]\\infty^0[\/latex], and [latex]1^{\\infty}[\/latex] are all indeterminate forms. On their own, these expressions are meaningless because we cannot actually evaluate these expressions as we would evaluate an expression involving real numbers. Rather, these expressions represent forms that arise when finding limits. Now we examine how L\u2019H\u00f4pital\u2019s rule can be used to evaluate limits involving these indeterminate forms.<\/p>\n<p id=\"fs-id1165042364178\">Since L\u2019H\u00f4pital\u2019s rule applies to quotients, we use the natural logarithm function and its properties to reduce a problem evaluating a limit involving exponents to a related problem involving a limit of a quotient. For example, suppose we want to evaluate [latex]\\underset{x\\to a}{\\lim}f(x)^{g(x)}[\/latex] and we arrive at the indeterminate form [latex]\\infty^0[\/latex]. (The indeterminate forms [latex]0^0[\/latex] and [latex]1^{\\infty}[\/latex] can be handled similarly.)<\/p>\n<div id=\"fs-id1165043390815\" class=\"textbook exercises\">\n<h3>Example: Indeterminate Form of Type [latex]\\infty^0[\/latex]<\/h3>\n<p>Evaluate [latex]\\underset{x\\to \\infty }{\\lim} x^{\\frac{1}{x}}[\/latex]<\/p>\n<div id=\"fs-id1165043390817\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043390866\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043390866\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043390866\">Let [latex]y=x^{1\/x}[\/latex]. Then,<\/p>\n<div id=\"fs-id1165043281565\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln (x^{1\/x})=\\frac{1}{x} \\ln x=\\frac{\\ln x}{x}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043281615\">We need to evaluate [latex]\\underset{x\\to \\infty }{\\lim}\\frac{\\ln x}{x}[\/latex]. Applying L\u2019H\u00f4pital\u2019s rule, we obtain<\/p>\n<div id=\"fs-id1165043281645\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim} \\ln y=\\underset{x\\to \\infty}{\\lim}\\frac{\\ln x}{x}=\\underset{x\\to \\infty}{\\lim}\\frac{1\/x}{1}=0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043173746\">Therefore, [latex]\\underset{x\\to \\infty }{\\lim}\\ln y=0[\/latex]. Since the natural logarithm function is continuous, we conclude that<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln (\\underset{x\\to \\infty}{\\lim} y)=0[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043427387\">which leads to<\/p>\n<div id=\"fs-id1165043427390\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim} y=\\underset{x\\to \\infty}{\\lim}\\frac{\\ln x}{x}=e^0=1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042407320\">Hence,<\/p>\n<div id=\"fs-id1165042407323\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim} x^{1\/x}=1[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042407362\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\underset{x\\to \\infty}{\\lim} x^{\\frac{1}{\\ln x}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3762844\">Hint<\/span><\/p>\n<div id=\"q3762844\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]y=x^{1\/ \\ln x}[\/latex] and apply the natural logarithm to both sides of the equation.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043108248\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043108248\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]e[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043108292\" class=\"textbook exercises\">\n<h3>Example: Indeterminate Form of Type [latex]0^0[\/latex]<\/h3>\n<p>Evaluate [latex]\\underset{x\\to 0^+}{\\lim} x^{\\sin x}[\/latex]<\/p>\n<div id=\"fs-id1165043108295\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042657755\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042657755\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042657755\">Let<\/p>\n<div id=\"fs-id1165042657759\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=x^{\\sin x}[\/latex]<\/div>\n<p id=\"fs-id1165042657780\">Therefore,<\/p>\n<div id=\"fs-id1165042657783\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln y=\\ln (x^{\\sin x})= \\sin x \\ln x[\/latex]<\/div>\n<p id=\"fs-id1165042707171\">We now evaluate [latex]\\underset{x\\to 0^+}{\\lim} \\sin x \\ln x[\/latex]. Since [latex]\\underset{x\\to 0^+}{\\lim} \\sin x=0[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} \\ln x=\u2212\\infty[\/latex], we have the indeterminate form [latex]0 \\cdot \\infty[\/latex]. To apply L\u2019H\u00f4pital\u2019s rule, we need to rewrite [latex]\\sin x \\ln x[\/latex] as a fraction. We could write<\/p>\n<div id=\"fs-id1165043173832\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin x \\ln x=\\frac{\\sin x}{1\/ \\ln x}[\/latex]<\/div>\n<p id=\"fs-id1165043173865\">or<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin x \\ln x=\\frac{\\ln x}{1\/ \\sin x}=\\frac{\\ln x}{\\csc x}[\/latex]<\/div>\n<p id=\"fs-id1165042364489\">Let\u2019s consider the first option. In this case, applying L\u2019H\u00f4pital\u2019s rule, we would obtain<\/p>\n<div id=\"fs-id1165042364494\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} \\sin x \\ln x=\\underset{x\\to 0^+}{\\lim}\\frac{\\sin x}{1\/ \\ln x}=\\underset{x\\to 0^+}{\\lim}\\frac{\\cos x}{-1\/(x(\\ln x)^2)}=\\underset{x\\to 0^+}{\\lim}(\u2212x(\\ln x)^2 \\cos x)[\/latex]<\/div>\n<p id=\"fs-id1165043317274\">Unfortunately, we not only have another expression involving the indeterminate form [latex]0 \\cdot \\infty[\/latex], but the new limit is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By writing<\/p>\n<div id=\"fs-id1165043131555\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin x \\ln x=\\frac{\\ln x}{1\/ \\sin x}=\\frac{\\ln x}{\\csc x}[\/latex]<\/div>\n<p id=\"fs-id1165043131604\">and applying L\u2019H\u00f4pital\u2019s rule, we obtain<\/p>\n<div id=\"fs-id1165043131609\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} \\sin x \\ln x=\\underset{x\\to 0^+}{\\lim}\\frac{\\ln x}{\\csc x}=\\underset{x\\to 0^+}{\\lim}\\frac{1\/x}{\u2212 \\csc x \\cot x}=\\underset{x\\to 0^+}{\\lim}\\frac{-1}{x \\csc x \\cot x}[\/latex]<\/div>\n<p id=\"fs-id1165042651533\">Using the fact that [latex]\\csc x=\\frac{1}{\\sin x}[\/latex] and [latex]\\cot x=\\frac{\\cos x}{\\sin x}[\/latex], we can rewrite the expression on the right-hand side as<\/p>\n<div id=\"fs-id1165043251999\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim}\\frac{\u2212\\sin^2 x}{x \\cos x}=\\underset{x\\to 0^+}{\\lim}[\\frac{\\sin x}{x} \\cdot (\u2212\\tan x)]=(\\underset{x\\to 0^+}{\\lim}\\frac{\\sin x}{x}) \\cdot (\\underset{x\\to 0^+}{\\lim}(\u2212\\tan x))=1 \\cdot 0=0[\/latex]<\/div>\n<p id=\"fs-id1165042676314\">We conclude that [latex]\\underset{x\\to 0^+}{\\lim} \\ln y=0[\/latex]. Therefore, [latex]\\ln (\\underset{x\\to 0^+}{\\lim} y)=0[\/latex] and we have<\/p>\n<div id=\"fs-id1165042327527\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} y=\\underset{x\\to 0^+}{\\lim} x^{\\sin x}=e^0=1[\/latex]<\/div>\n<p id=\"fs-id1165042327592\">Hence,<\/p>\n<div id=\"fs-id1165042327595\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^+}{\\lim} x^{\\sin x}=1[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042660254\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\underset{x\\to 0^+}{\\lim} x^x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q929037\">Hint<\/span><\/p>\n<div id=\"q929037\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]y=x^x[\/latex] and take the natural logarithm of both sides of the equation.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042660293\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042660293\" class=\"hidden-answer\" style=\"display: none\">\n<p>1<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":349141,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4089","chapter","type-chapter","status-publish","hentry"],"part":4118,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4089","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4089\/revisions"}],"predecessor-version":[{"id":4223,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4089\/revisions\/4223"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/4118"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4089\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=4089"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=4089"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=4089"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=4089"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}