{"id":285,"date":"2021-02-04T00:48:59","date_gmt":"2021-02-04T00:48:59","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=285"},"modified":"2022-03-11T21:55:09","modified_gmt":"2022-03-11T21:55:09","slug":"more-limit-evaluation-techniques","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/more-limit-evaluation-techniques\/","title":{"raw":"More Limit Evaluation Techniques","rendered":"More Limit Evaluation Techniques"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Evaluate the limit of a function by factoring<\/li>\r\n \t<li>Use the limit laws to evaluate the limit of a polynomial or rational function<\/li>\r\n \t<li><span class=\"os-abstract-content\">Evaluate the limit of a function by factoring or by using conjugates<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170571688085\">As we have seen, we may evaluate easily the limits of polynomials and limits of some (but not all) rational functions by direct substitution. However, as we saw in the introductory section on limits, it is certainly possible for [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] to exist when [latex]f(a)[\/latex] is undefined. The following observation allows us to evaluate many limits of this type:<\/p>\r\n<p id=\"fs-id1170571688129\">If for all [latex]x\\ne a, \\, f(x)=g(x)[\/latex] over some open interval containing [latex]a[\/latex], then [latex]\\underset{x\\to a}{\\lim}f(x)=\\underset{x\\to a}{\\lim}g(x)[\/latex].<\/p>\r\n<p id=\"fs-id1170572404968\">To understand this idea better, consider the limit [latex]\\underset{x\\to 1}{\\lim}\\frac{x^2-1}{x-1}[\/latex].<\/p>\r\n<p id=\"fs-id1170572405008\">The function<\/p>\r\n\r\n<div id=\"fs-id1170572405011\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill f(x)&amp; =\\dfrac{x^2-1}{x-1}\\hfill \\\\ &amp; =\\dfrac{(x-1)(x+1)}{x-1}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571689728\">and the function [latex]g(x)=x+1[\/latex] are identical for all values of [latex]x\\ne 1.[\/latex] The graphs of these two functions are shown in Figure 1.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203417\/CNX_Calc_Figure_02_03_001.jpg\" alt=\"Two graphs side by side. The first is a graph of g(x) = x + 1, a linear function with y intercept at (0,1) and x intercept at (-1,0). The second is a graph of f(x) = (x^2 \u2013 1) \/ (x \u2013 1). This graph is identical to the first for all x not equal to 1, as there is an open circle at (1,2) in the second graph.\" width=\"975\" height=\"468\" \/> Figure 1. The graphs of [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are identical for all [latex]x\\ne 1[\/latex]. Their limits at 1 are equal.[\/caption]\r\n<p id=\"fs-id1170571650070\">We see that<\/p>\r\n\r\n<div id=\"fs-id1170571650073\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\underset{x\\to 1}{\\lim}\\dfrac{x^2-1}{x-1}&amp; =\\underset{x\\to 1}{\\lim}\\dfrac{(x-1)(x+1)}{x-1}\\hfill \\\\ &amp; =\\underset{x\\to 1}{\\lim}(x+1)\\hfill \\\\ &amp; =2\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571649704\">The limit has the form [latex]\\underset{x\\to a}{\\lim}\\frac{f(x)}{g(x)}[\/latex], where [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex]. (In this case, we say that [latex]\\frac{f(x)}{g(x)}[\/latex] has the <strong>indeterminate form<\/strong>\u00a0[latex]\\frac{0}{0}[\/latex].) The following Problem-Solving Strategy provides a general outline for evaluating limits of this type.<\/p>\r\n\r\n<div id=\"fs-id1170571611384\" class=\"textbox examples\">\r\n<h3>Problem-Solving Strategy: Calculating a Limit When [latex]f(x)\/g(x)[\/latex] has the Indeterminate Form 0\/0<\/h3>\r\n<ol id=\"fs-id1170572627081\">\r\n \t<li>First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately using the limit laws.<\/li>\r\n \t<li>We then need to find a function that is equal to [latex]h(x)=f(x)\/g(x)[\/latex] for all [latex]x\\ne a[\/latex] over some interval containing [latex]a[\/latex]. To do this, we may need to try one or more of the following steps:\r\n<ol id=\"fs-id1170572627147\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>If [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are polynomials, we should factor each function and cancel out any common factors.<\/li>\r\n \t<li>If the numerator or denominator contains a difference involving a square root, we should try multiplying the numerator and denominator by the conjugate of the expression involving the square root.<\/li>\r\n \t<li>If [latex]f(x)\/g(x)[\/latex] is a complex fraction, we begin by simplifying it.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Last, we apply the limit laws.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<p id=\"fs-id1170571669700\">The next examples demonstrate the use of this Problem-Solving Strategy. It may be helpful to revisit the following algebraic techniques.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Factoring polynomials<\/h3>\r\n<h2>Key Equations<\/h2>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><strong>difference of squares<\/strong><\/td>\r\n<td>[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>perfect square trinomial<\/strong><\/td>\r\n<td>[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>sum of cubes<\/strong><\/td>\r\n<td>[latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>difference of cubes<\/strong><\/td>\r\n<td>[latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>The greatest common factor, or GCF, can be factored out of a polynomial. Checking for a GCF should be the first step in any factoring problem.<\/li>\r\n \t<li>Trinomials with leading coefficient 1 can be factored by finding numbers that have a product of the third term and a sum of the second term.<\/li>\r\n \t<li>Trinomials can be factored using a process called factoring by grouping.<\/li>\r\n \t<li>Perfect square trinomials and the difference of squares are special products and can be factored using equations.<\/li>\r\n \t<li>The sum of cubes and the difference of cubes can be factored using equations.<\/li>\r\n \t<li>Polynomials containing fractional and negative exponents can be factored by pulling out a GCF.<\/li>\r\n<\/ul>\r\n<dl id=\"fs-id1165133085661\" class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165133085661\" class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165133085661\" class=\"definition\">\r\n \t<dt><strong>factor by grouping<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165133085665\">a method for factoring a trinomial of the form [latex]a{x}^{2}+bx+c[\/latex] by dividing the <em>x<\/em> term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression<\/dd>\r\n<\/dl>\r\n<\/dt>\r\n<\/dl>\r\n<\/dt>\r\n \t<dt>\r\n<dl id=\"fs-id1165133085661\" class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165137644987\" class=\"definition\">\r\n \t<dt><strong>greatest common factor<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137644990\">the largest polynomial that divides evenly into each polynomial<\/dd>\r\n<\/dl>\r\n<\/dt>\r\n<\/dl>\r\n<\/dt>\r\n<\/dl>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: The conjugate of a sum or difference of a rational and irrational term<\/h3>\r\nThe conjugate of the sum or difference of a rational and an irrational term can be found by changing the sign of the radical portion of the denominator. If the denominator is [latex]a+b\\sqrt{c}[\/latex], then the conjugate is [latex]a-b\\sqrt{c}[\/latex].\r\n\r\n<\/div>\r\nThe example below illustrates the factor-and-cancel technique; the next shows multiplying by a conjugate. In the one after that, we look at simplifying a complex fraction.\r\n<div id=\"fs-id1170571669713\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating a Limit by Factoring and Canceling<\/h3>\r\n<p id=\"fs-id1170571669723\">Evaluate [latex]\\underset{x\\to 3}{\\lim}\\dfrac{x^2-3x}{2x^2-5x-3}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572335183\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572335183\"]\r\n<p id=\"fs-id1170572335183\"><strong>Step 1.<\/strong> The function [latex]f(x)=\\frac{x^2-3x}{2x^2-5x-3}[\/latex] is undefined for [latex]x=3[\/latex]. In fact, if we substitute 3 into the function we get 0\/0, which is undefined. Factoring and canceling is a good strategy:<\/p>\r\n\r\n<div id=\"fs-id1170572560619\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 3}{\\lim}\\dfrac{x^2-3x}{2x^2-5x-3}=\\underset{x\\to 3}{\\lim}\\dfrac{x(x-3)}{(x-3)(2x+1)}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572548178\"><strong>Step 2.<\/strong> For all [latex]x\\ne 3, \\, \\frac{x^2-3x}{2x^2-5x-3}=\\frac{x}{2x+1}[\/latex]. Therefore,<\/p>\r\n\r\n<div id=\"fs-id1170572548248\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 3}{\\lim}\\dfrac{x(x-3)}{(x-3)(2x+1)}=\\underset{x\\to 3}{\\lim}\\dfrac{x}{2x+1}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572347050\"><strong>Step 3.<\/strong> Evaluate using the limit laws:<\/p>\r\n\r\n<div id=\"fs-id1170572347056\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 3}{\\lim}\\dfrac{x}{2x+1}=\\dfrac{3}{7}[\/latex]\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571597999\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170571598007\">Evaluate [latex]\\underset{x\\to -3}{\\lim}\\dfrac{x^2+4x+3}{x^2-9}[\/latex]<\/p>\r\n[reveal-answer q=\"57755345\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"57755345\"]\r\n<p id=\"fs-id1170571598058\">Follow the steps in the Problem-Solving Strategy and the example above.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170571598067\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571598067\"]\r\n<p id=\"fs-id1170571598067\">[latex]\\dfrac{1}{3}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It as well as another worked example of evaluating a limit by factoring and canceling. [\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Jv0Wi-JERjo?controls=0&amp;start=341&amp;end=454&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.3LimitLaws341to454_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.3 Limit Laws\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]16095[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]18749[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572307613\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating a Limit by Multiplying by a Conjugate<\/h3>\r\n<p id=\"fs-id1170572307623\">Evaluate [latex]\\underset{x\\to -1}{\\lim}\\dfrac{\\sqrt{x+2}-1}{x+1}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572307671\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572307671\"]\r\n<p id=\"fs-id1170572307671\"><strong>Step 1.<\/strong>[latex]\\frac{\\sqrt{x+2}-1}{x+1}[\/latex] has the form 0\/0 at \u22121. Let\u2019s begin by multiplying by [latex]\\sqrt{x+2}+1[\/latex], the conjugate of [latex]\\sqrt{x+2}-1[\/latex], on the numerator and denominator:<\/p>\r\n\r\n<div id=\"fs-id1170571648338\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to -1}{\\lim}\\frac{\\sqrt{x+2}-1}{x+1}=\\underset{x\\to -1}{\\lim}\\frac{\\sqrt{x+2}-1}{x+1}\\cdot \\frac{\\sqrt{x+2}+1}{\\sqrt{x+2}+1}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572306393\"><strong>Step 2.<\/strong> We then multiply out the numerator. We don\u2019t multiply out the denominator because we are hoping that the [latex](x+1)[\/latex] in the denominator cancels out in the end:<\/p>\r\n\r\n<div id=\"fs-id1170572306418\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]=\\underset{x\\to -1}{\\lim}\\frac{x+1}{(x+1)(\\sqrt{x+2}+1)}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571562568\"><strong>Step 3.<\/strong> Then we cancel:<\/p>\r\n\r\n<div id=\"fs-id1170571562574\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]=\\underset{x\\to -1}{\\lim}\\frac{1}{\\sqrt{x+2}+1}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571562617\"><strong>Step 4.<\/strong> Last, we apply the limit laws:<\/p>\r\n\r\n<div id=\"fs-id1170571562624\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to -1}{\\lim}\\frac{1}{\\sqrt{x+2}+1}=\\frac{1}{2}[\/latex]\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571611949\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170571611956\">Evaluate [latex]\\underset{x\\to 5}{\\lim}\\dfrac{\\sqrt{x-1}-2}{x-5}[\/latex]<\/p>\r\n[reveal-answer q=\"26658901\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"26658901\"]\r\n<p id=\"fs-id1170571611999\">Follow the steps in the Problem-Solving Strategy and the example above.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170571612008\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571612008\"]\r\n<p id=\"fs-id1170571612008\">[latex]\\dfrac{1}{4}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe next few examples involve complex fractions that will require combining rational expressions to simplify complex fractions.\r\n<div class=\"textbox examples\">\r\n<h3>Recall:\u00a0Given two rational expressions, add or subtract them<\/h3>\r\n<ol>\r\n \t<li>Factor the numerator and denominator.<\/li>\r\n \t<li>Find the LCD of the expressions.<\/li>\r\n \t<li>Multiply the expressions by a form of 1 that changes the denominators to the LCD.<\/li>\r\n \t<li>Add or subtract the numerators.<\/li>\r\n \t<li>Simplify.\r\n<div style=\"text-align: center;\"><\/div><\/li>\r\n<\/ol>\r\n<em>Example:<\/em>\r\n<div style=\"text-align: center;\">[latex]\\dfrac{5}{x}+\\dfrac{6}{y}[\/latex]<\/div>\r\nFirst, we have to find the LCD. In this case, the LCD will be [latex]xy[\/latex]. We then multiply each expression by the appropriate form of 1 to obtain [latex]xy[\/latex] as the denominator for each fraction.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\dfrac{5}{x}\\cdot \\dfrac{y}{y}+\\dfrac{6}{y}\\cdot \\dfrac{x}{x}\\\\ \\dfrac{5y}{xy}+\\dfrac{6x}{xy}\\end{array}[\/latex]<\/div>\r\nNow that the expressions have the same denominator, we simply add the numerators to find the sum.\r\n<div style=\"text-align: center;\">[latex]\\dfrac{6x+5y}{xy}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Given a complex rational expression, simplify it<\/h3>\r\n<ol>\r\n \t<li>Combine the expressions in the numerator into a single rational expression by adding or subtracting.<\/li>\r\n \t<li>Combine the expressions in the denominator into a single rational expression by adding or subtracting.<\/li>\r\n \t<li>Rewrite as the numerator divided by the denominator.<\/li>\r\n \t<li>Rewrite as multiplication.<\/li>\r\n \t<li>Multiply.<\/li>\r\n \t<li>Simplify.\r\n<em>Example:<\/em> [latex]\\dfrac{y+\\dfrac{1}{x}}{\\dfrac{x}{y}}[\/latex] .Begin by combining the expressions in the numerator into one expression.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}y\\cdot \\dfrac{x}{x}+\\dfrac{1}{x}\\hfill &amp; \\text{Multiply by }\\dfrac{x}{x}\\text{to get LCD as denominator}.\\hfill \\\\ \\dfrac{xy}{x}+\\dfrac{1}{x}\\hfill &amp; \\\\ \\dfrac{xy+1}{x}\\hfill &amp; \\text{Add numerators}.\\hfill \\end{array}[\/latex]<\/div>\r\nNow the numerator is a single rational expression and the denominator is a single rational expression.\r\n<div style=\"text-align: center;\">[latex]\\dfrac{\\dfrac{xy+1}{x}}{\\dfrac{x}{y}}[\/latex]<\/div>\r\nWe can rewrite this as division and then multiplication.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}\\dfrac{xy+1}{x}\\div \\dfrac{x}{y}\\hfill &amp; \\\\ \\dfrac{xy+1}{x}\\cdot \\dfrac{y}{x}\\hfill &amp; \\text{Rewrite as multiplication}\\text{.}\\hfill \\\\ \\dfrac{y\\left(xy+1\\right)}{{x}^{2}}\\hfill &amp; \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\r\n<div><\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1170571612021\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating a Limit by Simplifying a Complex Fraction<\/h3>\r\n<p id=\"fs-id1170571612031\">Evaluate [latex]\\underset{x\\to 1}{\\lim}\\dfrac{\\frac{1}{x+1}-\\frac{1}{2}}{x-1}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571681059\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571681059\"]\r\n<p id=\"fs-id1170571681059\"><strong>Step 1.\u00a0<\/strong>[latex]\\frac{\\frac{1}{x+1}-\\frac{1}{2}}{x-1}[\/latex] has the form 0\/0 at 1. We simplify the algebraic fraction by multiplying by [latex]\\frac{2(x+1)}{2(x+1)}[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1170571681146\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1}{\\lim}\\dfrac{\\frac{1}{x+1}-\\frac{1}{2}}{x-1}=\\underset{x\\to 1}{\\lim}\\dfrac{\\frac{1}{x+1}-\\frac{1}{2}}{x-1} \\cdot \\dfrac{2(x+1)}{2(x+1)}[\/latex]<\/div>\r\n<p id=\"fs-id1170571596311\"><strong>Step 2.<\/strong> Next, we multiply through the numerators. Do not multiply the denominators because we want to be able to cancel the factor [latex](x-1)[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1170571622080\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]=\\underset{x\\to 1}{\\lim}\\frac{2-(x+1)}{2(x-1)(x+1)}[\/latex]<\/div>\r\n<p id=\"fs-id1170571622151\"><strong>Step 3.<\/strong> Then, we simplify the numerator:<\/p>\r\n\r\n<div id=\"fs-id1170571622157\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]=\\underset{x\\to 1}{\\lim}\\frac{-x+1}{2(x-1)(x+1)}[\/latex]<\/div>\r\n<p id=\"fs-id1170571650203\"><strong>Step 4.<\/strong> Now we factor out \u22121 from the numerator:<\/p>\r\n\r\n<div id=\"fs-id1170571650209\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]=\\underset{x\\to 1}{\\lim}\\frac{-(x-1)}{2(x-1)(x+1)}[\/latex]<\/div>\r\n<p id=\"fs-id1170571650278\"><strong>Step 5.<\/strong> Then, we cancel the common factors of [latex](x-1)[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1170571650301\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]=\\underset{x\\to 1}{\\lim}\\frac{-1}{2(x+1)}[\/latex]<\/div>\r\n<p id=\"fs-id1170572394292\"><strong>Step 6.<\/strong> Last, we evaluate using the limit laws:<\/p>\r\n\r\n<div id=\"fs-id1170572394298\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1}{\\lim}\\frac{-1}{2(x+1)}=-\\frac{1}{4}[\/latex]\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572394353\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572394360\">Evaluate [latex]\\underset{x\\to -3}{\\lim}\\dfrac{\\frac{1}{x+2}+1}{x+3}[\/latex]<\/p>\r\n[reveal-answer q=\"2119010\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"2119010\"]\r\n<p id=\"fs-id1170572394407\">Follow the steps in the Problem-Solving Strategy and the example above.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170571648126\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571648126\"]\r\n<p id=\"fs-id1170571648126\">[latex]\u22121[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It. [\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Jv0Wi-JERjo?controls=0&amp;start=667&amp;end=744&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.3LimitLaws667to744_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.3 Limit Laws\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<p id=\"fs-id1170571648132\">The example below does not fall neatly into any of the patterns established in the previous examples. However, with a little creativity, we can still use these same techniques.<\/p>\r\n\r\n<div id=\"fs-id1170571648139\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating a Limit When the Limit Laws Do Not Apply<\/h3>\r\n<p id=\"fs-id1170571648149\">Evaluate [latex]\\underset{x\\to 0}{\\lim}\\left(\\dfrac{1}{x}+\\dfrac{5}{x(x-5)}\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571648205\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571648205\"]\r\n<p id=\"fs-id1170571648205\">Both [latex]\\frac{1}{x}[\/latex] and [latex]\\frac{5}{x(x-5)}[\/latex] fail to have a limit at zero. Since neither of the two functions has a limit at zero, we cannot apply the sum law for limits; we must use a different strategy. In this case, we find the limit by performing addition and then applying one of our previous strategies. Observe that<\/p>\r\n\r\n<div id=\"fs-id1170571648246\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc} \\dfrac{1}{x}+\\dfrac{5}{x(x-5)}&amp; =\\dfrac{x-5+5}{x(x-5)} \\\\ &amp; =\\dfrac{x}{x(x-5)}\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571649584\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1170571649587\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\underset{x\\to 0}{\\lim}\\big(\\frac{1}{x}+\\frac{5}{x(x-5)}\\big)&amp; =\\underset{x\\to 0}{\\lim}\\frac{x}{x(x-5)} \\\\ &amp; =\\underset{x\\to 0}{\\lim}\\frac{1}{x-5} \\\\ &amp; =-\\frac{1}{5} \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571681422\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170571681429\">Evaluate [latex]\\underset{x\\to 3}{\\lim}\\left(\\dfrac{1}{x-3}-\\dfrac{4}{x^2-2x-3}\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"80944622\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"80944622\"]\r\n<p id=\"fs-id1170572233797\">Use the same technique as the example above. Don\u2019t forget to factor [latex]x^2-2x-3[\/latex] before getting a common denominator.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572233826\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572233826\"]\r\n<p id=\"fs-id1170572233826\">[latex]\\frac{1}{4}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572233839\">Simple modifications in the limit laws allow us to apply them to one-sided limits. For example, to apply the limit laws to a limit of the form [latex]\\underset{x\\to a^-}{\\lim}h(x)[\/latex], we require the function [latex]h(x)[\/latex] to be defined over an open interval of the form [latex](b,a)[\/latex]; for a limit of the form [latex]\\underset{x\\to a^+}{\\lim}h(x)[\/latex], we require the function [latex]h(x)[\/latex] to be defined over an open interval of the form [latex](a,c)[\/latex]. The example illustrates this point.<\/p>\r\n\r\n<div id=\"fs-id1170571679268\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating a One-Sided Limit Using the Limit Laws<\/h3>\r\n<p id=\"fs-id1170571679278\">Evaluate each of the following limits, if possible.<\/p>\r\n\r\n<ol id=\"fs-id1170571679281\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\underset{x\\to 3^-}{\\lim}\\sqrt{x-3}[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to 3^+}{\\lim}\\sqrt{x-3}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170571679347\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571679347\"]\r\n<p id=\"fs-id1170571679347\">Figure 2 illustrates the function [latex]f(x)=\\sqrt{x-3}[\/latex] and aids in our understanding of these limits.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203420\/CNX_Calc_Figure_02_03_002.jpg\" alt=\"A graph of the function f(x) = sqrt(x-3). Visually, the function looks like the top half of a parabola opening to the right with vertex at (3,0).\" width=\"325\" height=\"162\" \/> Figure 2. The graph shows the function [latex]f(x)=\\sqrt{x-3}[\/latex].[\/caption]\r\n<ol id=\"fs-id1170571680900\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>The function [latex]f(x)=\\sqrt{x-3}[\/latex] is defined over the interval [latex][3,+\\infty)[\/latex]. Since this function is not defined to the left of 3, we cannot apply the limit laws to compute [latex]\\underset{x\\to 3^-}{\\lim}\\sqrt{x-3}[\/latex]. In fact, since [latex]f(x)=\\sqrt{x-3}[\/latex] is undefined to the left of 3, [latex]\\underset{x\\to 3^-}{\\lim}\\sqrt{x-3}[\/latex] does not exist.<\/li>\r\n \t<li>Since [latex]f(x)=\\sqrt{x-3}[\/latex] is defined to the right of 3, the limit laws do apply to [latex]\\underset{x\\to 3^+}{\\lim}\\sqrt{x-3}[\/latex]. By applying these limit laws we obtain [latex]\\underset{x\\to 3^+}{\\lim}\\sqrt{x-3}=0[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1170571558874\">In the example below, we look at one-sided limits of a piecewise-defined function and use these limits to draw a conclusion about a two-sided limit of the same function.<\/p>\r\n\r\n<div id=\"fs-id1170571558882\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating a Two-Sided Limit Using the Limit Laws<\/h3>\r\n<p id=\"fs-id1170571558891\">For [latex]f(x)=\\begin{cases} 4x-3 &amp; \\text{ if } \\, x&lt;2 \\\\ (x-3)^2 &amp; \\text{ if } \\, x \\ge 2 \\end{cases}[\/latex] evaluate each of the following limits:<\/p>\r\n\r\n<ol id=\"fs-id1170571649889\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\underset{x\\to 2^-}{\\lim}f(x)[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to 2^+}{\\lim}f(x)[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to 2}{\\lim}f(x)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170571573824\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571573824\"]\r\n<p id=\"fs-id1170571573824\">Figure 3 illustrates the function [latex]f(x)[\/latex] and aids in our understanding of these limits.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img id=\"24\" src=\"https:\/\/openstax.org\/resources\/42329e49b363445e47e79b9689883b2017ba83a6\" alt=\"The graph of a piecewise function with two segments. For x&lt;2, the function is linear with the equation 4x-3. There is an open circle at (2,5). The second segment is a parabola and exists for x&gt;=2, with the equation (x-3)^2. There is a closed circle at (2,1). The vertex of the parabola is at (3,0).\" width=\"325\" height=\"350\" data-media-type=\"image\/jpeg\" \/> Figure 3. This graph shows a function [latex](x)[\/latex][\/caption]\r\n<ol id=\"fs-id1170571573876\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Since [latex]f(x)=4x-3[\/latex] for all [latex]x[\/latex] in [latex](\u2212\\infty,2)[\/latex], replace [latex]f(x)[\/latex] in the limit with [latex]4x-3[\/latex] and apply the limit laws:\r\n<div id=\"fs-id1170571573954\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 2^-}{\\lim}f(x)=\\underset{x\\to 2^-}{\\lim}(4x-3)=5[\/latex].<\/div><\/li>\r\n \t<li>Since [latex]f(x)=(x-3)^2[\/latex] for all [latex]x[\/latex] in [latex](2,+\\infty)[\/latex], replace [latex]f(x)[\/latex] in the limit with [latex](x-3)^2[\/latex] and apply the limit laws:\r\n<div id=\"fs-id1170571644376\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 2^+}{\\lim}f(x)=\\underset{x\\to 2^+}{\\lim}(x-3)^2=1[\/latex].<\/div><\/li>\r\n \t<li>Since [latex]\\underset{x\\to 2^-}{\\lim}f(x)=5[\/latex] and [latex]\\underset{x\\to 2^+}{\\lim}f(x)=1[\/latex], we conclude that [latex]\\underset{x\\to 2}{\\lim}f(x)[\/latex] does not exist.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572235169\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572235178\">Graph [latex]f(x)=\\begin{cases} -x-2 &amp; \\text{ if } \\, x&lt;-1 \\\\ 2 &amp; \\text{ if } \\, x = -1 \\\\ x^3 &amp; \\text{ if } \\, x &gt; -1 \\end{cases}[\/latex] and evaluate [latex]\\underset{x\\to -1^-}{\\lim}f(x)[\/latex]<\/p>\r\n[reveal-answer q=\"30056723\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"30056723\"]\r\n<p id=\"fs-id1170572235312\">Use the method in the example above to evaluate the limit.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572559753\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572559753\"]\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203427\/CNX_Calc_Figure_02_03_004.jpg\" alt=\"The graph of a piecewise function with three segments. The first is a linear function, -x-2, for x&lt;-1. The x intercept is at (-2,0), and there is an open circle at (-1,-1). The next segment is simply the point (-1, 2). The third segment is the function x^3 for x &gt; -1, which crossed the x axis and y axis at the origin.\" width=\"325\" height=\"350\" \/> Figure 4.[\/caption]\r\n\r\n<span id=\"fs-id1170572559760\"><\/span>\r\n[latex]\\underset{x\\to -1^-}{\\lim}f(x)=-1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe now turn our attention to evaluating a limit of the form [latex]\\underset{x\\to a}{\\lim}\\frac{f(x)}{g(x)}[\/latex], where [latex]\\underset{x\\to a}{\\lim}f(x) = K[\/latex], where [latex]K\\ne{0}[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x) = 0[\/latex]. That is [latex]f(x)\\text{\/}g(x)[\/latex] has the form [latex]K\\text{\/}0[\/latex],\u00a0[latex]K\\ne{0}[\/latex] at [latex]a[\/latex].\r\n<div id=\"fs-id1170571611196\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating a Limit of the Form [latex]K\/0, \\, K\\ne 0[\/latex] Using the Limit Laws<\/h3>\r\n<p id=\"fs-id1170571611224\">Evaluate [latex]\\underset{x\\to 2^-}{\\lim}\\dfrac{x-3}{x^2-2x}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571611272\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571611272\"]\r\n<p id=\"fs-id1170571611272\"><strong>Step 1.<\/strong> After substituting in [latex]x=2[\/latex], we see that this limit has the form [latex]-\\frac{1}{0}[\/latex]. That is, as [latex]x[\/latex] approaches 2 from the left, the numerator approaches \u22121 and the denominator approaches 0. Consequently, the magnitude of [latex]\\frac{x-3}{x(x-2)}[\/latex] becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:<\/p>\r\n\r\n<div id=\"fs-id1170572420265\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 2^-}{\\lim}\\frac{x-3}{x^2-2x}=\\underset{x\\to 2^-}{\\lim}\\frac{x-3}{x(x-2)}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572420356\"><strong>Step 2.<\/strong> Since [latex]x-2[\/latex] is the only part of the denominator that is zero when 2 is substituted, we then separate [latex]\\frac{1}{(x-2)}[\/latex] from the rest of the function:<\/p>\r\n\r\n<div id=\"fs-id1170571612873\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]=\\underset{x\\to 2^-}{\\lim}\\frac{x-3}{x} \\cdot \\frac{1}{x-2}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571612925\"><strong>Step 3.<\/strong>[latex]\\underset{x\\to 2^-}{\\lim}\\frac{x-3}{x}=-\\frac{1}{2}[\/latex] and [latex]\\underset{x\\to 2^-}{\\lim}\\frac{1}{x-2}=\u2212\\infty[\/latex]. Therefore, the product of [latex]\\frac{(x-3)}{x}[\/latex] and [latex]\\frac{1}{(x-2)}[\/latex] has a limit of [latex]+\\infty[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1170571650460\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 2^-}{\\lim}\\frac{x-3}{x^2-2x}=+\\infty[\/latex].\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571650517\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170571650526\">Evaluate [latex]\\underset{x\\to 1}{\\lim}\\dfrac{x+2}{(x-1)^2}[\/latex]<\/p>\r\n[reveal-answer q=\"6003722\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"6003722\"]\r\n<p id=\"fs-id1170572611875\">Use the methods from the example above.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572611885\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572611885\"]\r\n<p id=\"fs-id1170572611885\">[latex]+\\infty[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Evaluate the limit of a function by factoring<\/li>\n<li>Use the limit laws to evaluate the limit of a polynomial or rational function<\/li>\n<li><span class=\"os-abstract-content\">Evaluate the limit of a function by factoring or by using conjugates<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170571688085\">As we have seen, we may evaluate easily the limits of polynomials and limits of some (but not all) rational functions by direct substitution. However, as we saw in the introductory section on limits, it is certainly possible for [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] to exist when [latex]f(a)[\/latex] is undefined. The following observation allows us to evaluate many limits of this type:<\/p>\n<p id=\"fs-id1170571688129\">If for all [latex]x\\ne a, \\, f(x)=g(x)[\/latex] over some open interval containing [latex]a[\/latex], then [latex]\\underset{x\\to a}{\\lim}f(x)=\\underset{x\\to a}{\\lim}g(x)[\/latex].<\/p>\n<p id=\"fs-id1170572404968\">To understand this idea better, consider the limit [latex]\\underset{x\\to 1}{\\lim}\\frac{x^2-1}{x-1}[\/latex].<\/p>\n<p id=\"fs-id1170572405008\">The function<\/p>\n<div id=\"fs-id1170572405011\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill f(x)& =\\dfrac{x^2-1}{x-1}\\hfill \\\\ & =\\dfrac{(x-1)(x+1)}{x-1}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571689728\">and the function [latex]g(x)=x+1[\/latex] are identical for all values of [latex]x\\ne 1.[\/latex] The graphs of these two functions are shown in Figure 1.<\/p>\n<div style=\"width: 985px\" 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\/11203417\/CNX_Calc_Figure_02_03_001.jpg\" alt=\"Two graphs side by side. The first is a graph of g(x) = x + 1, a linear function with y intercept at (0,1) and x intercept at (-1,0). The second is a graph of f(x) = (x^2 \u2013 1) \/ (x \u2013 1). This graph is identical to the first for all x not equal to 1, as there is an open circle at (1,2) in the second graph.\" width=\"975\" height=\"468\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The graphs of [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are identical for all [latex]x\\ne 1[\/latex]. Their limits at 1 are equal.<\/p>\n<\/div>\n<p id=\"fs-id1170571650070\">We see that<\/p>\n<div id=\"fs-id1170571650073\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\underset{x\\to 1}{\\lim}\\dfrac{x^2-1}{x-1}& =\\underset{x\\to 1}{\\lim}\\dfrac{(x-1)(x+1)}{x-1}\\hfill \\\\ & =\\underset{x\\to 1}{\\lim}(x+1)\\hfill \\\\ & =2\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571649704\">The limit has the form [latex]\\underset{x\\to a}{\\lim}\\frac{f(x)}{g(x)}[\/latex], where [latex]\\underset{x\\to a}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=0[\/latex]. (In this case, we say that [latex]\\frac{f(x)}{g(x)}[\/latex] has the <strong>indeterminate form<\/strong>\u00a0[latex]\\frac{0}{0}[\/latex].) The following Problem-Solving Strategy provides a general outline for evaluating limits of this type.<\/p>\n<div id=\"fs-id1170571611384\" class=\"textbox examples\">\n<h3>Problem-Solving Strategy: Calculating a Limit When [latex]f(x)\/g(x)[\/latex] has the Indeterminate Form 0\/0<\/h3>\n<ol id=\"fs-id1170572627081\">\n<li>First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately using the limit laws.<\/li>\n<li>We then need to find a function that is equal to [latex]h(x)=f(x)\/g(x)[\/latex] for all [latex]x\\ne a[\/latex] over some interval containing [latex]a[\/latex]. To do this, we may need to try one or more of the following steps:\n<ol id=\"fs-id1170572627147\" style=\"list-style-type: lower-alpha;\">\n<li>If [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are polynomials, we should factor each function and cancel out any common factors.<\/li>\n<li>If the numerator or denominator contains a difference involving a square root, we should try multiplying the numerator and denominator by the conjugate of the expression involving the square root.<\/li>\n<li>If [latex]f(x)\/g(x)[\/latex] is a complex fraction, we begin by simplifying it.<\/li>\n<\/ol>\n<\/li>\n<li>Last, we apply the limit laws.<\/li>\n<\/ol>\n<\/div>\n<p id=\"fs-id1170571669700\">The next examples demonstrate the use of this Problem-Solving Strategy. It may be helpful to revisit the following algebraic techniques.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Factoring polynomials<\/h3>\n<h2>Key Equations<\/h2>\n<table>\n<tbody>\n<tr>\n<td><strong>difference of squares<\/strong><\/td>\n<td>[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>perfect square trinomial<\/strong><\/td>\n<td>[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>sum of cubes<\/strong><\/td>\n<td>[latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>difference of cubes<\/strong><\/td>\n<td>[latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>The greatest common factor, or GCF, can be factored out of a polynomial. Checking for a GCF should be the first step in any factoring problem.<\/li>\n<li>Trinomials with leading coefficient 1 can be factored by finding numbers that have a product of the third term and a sum of the second term.<\/li>\n<li>Trinomials can be factored using a process called factoring by grouping.<\/li>\n<li>Perfect square trinomials and the difference of squares are special products and can be factored using equations.<\/li>\n<li>The sum of cubes and the difference of cubes can be factored using equations.<\/li>\n<li>Polynomials containing fractional and negative exponents can be factored by pulling out a GCF.<\/li>\n<\/ul>\n<dl id=\"fs-id1165133085661\" class=\"definition\">\n<dt>\n<\/dt>\n<dt>\n<\/dt>\n<dt><strong>factor by grouping<\/strong><\/dt>\n<dd id=\"fs-id1165133085665\">a method for factoring a trinomial of the form [latex]a{x}^{2}+bx+c[\/latex] by dividing the <em>x<\/em> term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression<\/dd>\n<\/dl>\n<dl id=\"fs-id1165133085661\" class=\"definition\">\n<dt>\n<\/dt>\n<dt><strong>greatest common factor<\/strong><\/dt>\n<dd id=\"fs-id1165137644990\">the largest polynomial that divides evenly into each polynomial<\/dd>\n<\/dl>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: The conjugate of a sum or difference of a rational and irrational term<\/h3>\n<p>The conjugate of the sum or difference of a rational and an irrational term can be found by changing the sign of the radical portion of the denominator. If the denominator is [latex]a+b\\sqrt{c}[\/latex], then the conjugate is [latex]a-b\\sqrt{c}[\/latex].<\/p>\n<\/div>\n<p>The example below illustrates the factor-and-cancel technique; the next shows multiplying by a conjugate. In the one after that, we look at simplifying a complex fraction.<\/p>\n<div id=\"fs-id1170571669713\" class=\"textbook exercises\">\n<h3>Example: Evaluating a Limit by Factoring and Canceling<\/h3>\n<p id=\"fs-id1170571669723\">Evaluate [latex]\\underset{x\\to 3}{\\lim}\\dfrac{x^2-3x}{2x^2-5x-3}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572335183\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572335183\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572335183\"><strong>Step 1.<\/strong> The function [latex]f(x)=\\frac{x^2-3x}{2x^2-5x-3}[\/latex] is undefined for [latex]x=3[\/latex]. In fact, if we substitute 3 into the function we get 0\/0, which is undefined. Factoring and canceling is a good strategy:<\/p>\n<div id=\"fs-id1170572560619\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 3}{\\lim}\\dfrac{x^2-3x}{2x^2-5x-3}=\\underset{x\\to 3}{\\lim}\\dfrac{x(x-3)}{(x-3)(2x+1)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572548178\"><strong>Step 2.<\/strong> For all [latex]x\\ne 3, \\, \\frac{x^2-3x}{2x^2-5x-3}=\\frac{x}{2x+1}[\/latex]. Therefore,<\/p>\n<div id=\"fs-id1170572548248\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 3}{\\lim}\\dfrac{x(x-3)}{(x-3)(2x+1)}=\\underset{x\\to 3}{\\lim}\\dfrac{x}{2x+1}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572347050\"><strong>Step 3.<\/strong> Evaluate using the limit laws:<\/p>\n<div id=\"fs-id1170572347056\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 3}{\\lim}\\dfrac{x}{2x+1}=\\dfrac{3}{7}[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571597999\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170571598007\">Evaluate [latex]\\underset{x\\to -3}{\\lim}\\dfrac{x^2+4x+3}{x^2-9}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q57755345\">Hint<\/span><\/p>\n<div id=\"q57755345\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571598058\">Follow the steps in the Problem-Solving Strategy and the example above.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571598067\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571598067\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571598067\">[latex]\\dfrac{1}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It as well as another worked example of evaluating a limit by factoring and canceling. <\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Jv0Wi-JERjo?controls=0&amp;start=341&amp;end=454&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.3LimitLaws341to454_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.3 Limit Laws&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm16095\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16095&theme=oea&iframe_resize_id=ohm16095&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm18749\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=18749&theme=oea&iframe_resize_id=ohm18749&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1170572307613\" class=\"textbook exercises\">\n<h3>Example: Evaluating a Limit by Multiplying by a Conjugate<\/h3>\n<p id=\"fs-id1170572307623\">Evaluate [latex]\\underset{x\\to -1}{\\lim}\\dfrac{\\sqrt{x+2}-1}{x+1}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572307671\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572307671\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572307671\"><strong>Step 1.<\/strong>[latex]\\frac{\\sqrt{x+2}-1}{x+1}[\/latex] has the form 0\/0 at \u22121. Let\u2019s begin by multiplying by [latex]\\sqrt{x+2}+1[\/latex], the conjugate of [latex]\\sqrt{x+2}-1[\/latex], on the numerator and denominator:<\/p>\n<div id=\"fs-id1170571648338\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to -1}{\\lim}\\frac{\\sqrt{x+2}-1}{x+1}=\\underset{x\\to -1}{\\lim}\\frac{\\sqrt{x+2}-1}{x+1}\\cdot \\frac{\\sqrt{x+2}+1}{\\sqrt{x+2}+1}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572306393\"><strong>Step 2.<\/strong> We then multiply out the numerator. We don\u2019t multiply out the denominator because we are hoping that the [latex](x+1)[\/latex] in the denominator cancels out in the end:<\/p>\n<div id=\"fs-id1170572306418\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]=\\underset{x\\to -1}{\\lim}\\frac{x+1}{(x+1)(\\sqrt{x+2}+1)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571562568\"><strong>Step 3.<\/strong> Then we cancel:<\/p>\n<div id=\"fs-id1170571562574\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]=\\underset{x\\to -1}{\\lim}\\frac{1}{\\sqrt{x+2}+1}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571562617\"><strong>Step 4.<\/strong> Last, we apply the limit laws:<\/p>\n<div id=\"fs-id1170571562624\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to -1}{\\lim}\\frac{1}{\\sqrt{x+2}+1}=\\frac{1}{2}[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571611949\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170571611956\">Evaluate [latex]\\underset{x\\to 5}{\\lim}\\dfrac{\\sqrt{x-1}-2}{x-5}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q26658901\">Hint<\/span><\/p>\n<div id=\"q26658901\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571611999\">Follow the steps in the Problem-Solving Strategy and the example above.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571612008\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571612008\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571612008\">[latex]\\dfrac{1}{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The next few examples involve complex fractions that will require combining rational expressions to simplify complex fractions.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall:\u00a0Given two rational expressions, add or subtract them<\/h3>\n<ol>\n<li>Factor the numerator and denominator.<\/li>\n<li>Find the LCD of the expressions.<\/li>\n<li>Multiply the expressions by a form of 1 that changes the denominators to the LCD.<\/li>\n<li>Add or subtract the numerators.<\/li>\n<li>Simplify.\n<div style=\"text-align: center;\"><\/div>\n<\/li>\n<\/ol>\n<p><em>Example:<\/em><\/p>\n<div style=\"text-align: center;\">[latex]\\dfrac{5}{x}+\\dfrac{6}{y}[\/latex]<\/div>\n<p>First, we have to find the LCD. In this case, the LCD will be [latex]xy[\/latex]. We then multiply each expression by the appropriate form of 1 to obtain [latex]xy[\/latex] as the denominator for each fraction.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\dfrac{5}{x}\\cdot \\dfrac{y}{y}+\\dfrac{6}{y}\\cdot \\dfrac{x}{x}\\\\ \\dfrac{5y}{xy}+\\dfrac{6x}{xy}\\end{array}[\/latex]<\/div>\n<p>Now that the expressions have the same denominator, we simply add the numerators to find the sum.<\/p>\n<div style=\"text-align: center;\">[latex]\\dfrac{6x+5y}{xy}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Given a complex rational expression, simplify it<\/h3>\n<ol>\n<li>Combine the expressions in the numerator into a single rational expression by adding or subtracting.<\/li>\n<li>Combine the expressions in the denominator into a single rational expression by adding or subtracting.<\/li>\n<li>Rewrite as the numerator divided by the denominator.<\/li>\n<li>Rewrite as multiplication.<\/li>\n<li>Multiply.<\/li>\n<li>Simplify.<br \/>\n<em>Example:<\/em> [latex]\\dfrac{y+\\dfrac{1}{x}}{\\dfrac{x}{y}}[\/latex] .Begin by combining the expressions in the numerator into one expression.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}y\\cdot \\dfrac{x}{x}+\\dfrac{1}{x}\\hfill & \\text{Multiply by }\\dfrac{x}{x}\\text{to get LCD as denominator}.\\hfill \\\\ \\dfrac{xy}{x}+\\dfrac{1}{x}\\hfill & \\\\ \\dfrac{xy+1}{x}\\hfill & \\text{Add numerators}.\\hfill \\end{array}[\/latex]<\/div>\n<p>Now the numerator is a single rational expression and the denominator is a single rational expression.<\/p>\n<div style=\"text-align: center;\">[latex]\\dfrac{\\dfrac{xy+1}{x}}{\\dfrac{x}{y}}[\/latex]<\/div>\n<p>We can rewrite this as division and then multiplication.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}\\dfrac{xy+1}{x}\\div \\dfrac{x}{y}\\hfill & \\\\ \\dfrac{xy+1}{x}\\cdot \\dfrac{y}{x}\\hfill & \\text{Rewrite as multiplication}\\text{.}\\hfill \\\\ \\dfrac{y\\left(xy+1\\right)}{{x}^{2}}\\hfill & \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1170571612021\" class=\"textbook exercises\">\n<h3>Example: Evaluating a Limit by Simplifying a Complex Fraction<\/h3>\n<p id=\"fs-id1170571612031\">Evaluate [latex]\\underset{x\\to 1}{\\lim}\\dfrac{\\frac{1}{x+1}-\\frac{1}{2}}{x-1}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571681059\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571681059\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571681059\"><strong>Step 1.\u00a0<\/strong>[latex]\\frac{\\frac{1}{x+1}-\\frac{1}{2}}{x-1}[\/latex] has the form 0\/0 at 1. We simplify the algebraic fraction by multiplying by [latex]\\frac{2(x+1)}{2(x+1)}[\/latex]:<\/p>\n<div id=\"fs-id1170571681146\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1}{\\lim}\\dfrac{\\frac{1}{x+1}-\\frac{1}{2}}{x-1}=\\underset{x\\to 1}{\\lim}\\dfrac{\\frac{1}{x+1}-\\frac{1}{2}}{x-1} \\cdot \\dfrac{2(x+1)}{2(x+1)}[\/latex]<\/div>\n<p id=\"fs-id1170571596311\"><strong>Step 2.<\/strong> Next, we multiply through the numerators. Do not multiply the denominators because we want to be able to cancel the factor [latex](x-1)[\/latex]:<\/p>\n<div id=\"fs-id1170571622080\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]=\\underset{x\\to 1}{\\lim}\\frac{2-(x+1)}{2(x-1)(x+1)}[\/latex]<\/div>\n<p id=\"fs-id1170571622151\"><strong>Step 3.<\/strong> Then, we simplify the numerator:<\/p>\n<div id=\"fs-id1170571622157\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]=\\underset{x\\to 1}{\\lim}\\frac{-x+1}{2(x-1)(x+1)}[\/latex]<\/div>\n<p id=\"fs-id1170571650203\"><strong>Step 4.<\/strong> Now we factor out \u22121 from the numerator:<\/p>\n<div id=\"fs-id1170571650209\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]=\\underset{x\\to 1}{\\lim}\\frac{-(x-1)}{2(x-1)(x+1)}[\/latex]<\/div>\n<p id=\"fs-id1170571650278\"><strong>Step 5.<\/strong> Then, we cancel the common factors of [latex](x-1)[\/latex]:<\/p>\n<div id=\"fs-id1170571650301\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]=\\underset{x\\to 1}{\\lim}\\frac{-1}{2(x+1)}[\/latex]<\/div>\n<p id=\"fs-id1170572394292\"><strong>Step 6.<\/strong> Last, we evaluate using the limit laws:<\/p>\n<div id=\"fs-id1170572394298\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1}{\\lim}\\frac{-1}{2(x+1)}=-\\frac{1}{4}[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572394353\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572394360\">Evaluate [latex]\\underset{x\\to -3}{\\lim}\\dfrac{\\frac{1}{x+2}+1}{x+3}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q2119010\">Hint<\/span><\/p>\n<div id=\"q2119010\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572394407\">Follow the steps in the Problem-Solving Strategy and the example above.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571648126\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571648126\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571648126\">[latex]\u22121[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It. <\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Jv0Wi-JERjo?controls=0&amp;start=667&amp;end=744&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.3LimitLaws667to744_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.3 Limit Laws&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<p id=\"fs-id1170571648132\">The example below does not fall neatly into any of the patterns established in the previous examples. However, with a little creativity, we can still use these same techniques.<\/p>\n<div id=\"fs-id1170571648139\" class=\"textbook exercises\">\n<h3>Example: Evaluating a Limit When the Limit Laws Do Not Apply<\/h3>\n<p id=\"fs-id1170571648149\">Evaluate [latex]\\underset{x\\to 0}{\\lim}\\left(\\dfrac{1}{x}+\\dfrac{5}{x(x-5)}\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571648205\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571648205\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571648205\">Both [latex]\\frac{1}{x}[\/latex] and [latex]\\frac{5}{x(x-5)}[\/latex] fail to have a limit at zero. Since neither of the two functions has a limit at zero, we cannot apply the sum law for limits; we must use a different strategy. In this case, we find the limit by performing addition and then applying one of our previous strategies. Observe that<\/p>\n<div id=\"fs-id1170571648246\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc} \\dfrac{1}{x}+\\dfrac{5}{x(x-5)}& =\\dfrac{x-5+5}{x(x-5)} \\\\ & =\\dfrac{x}{x(x-5)}\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571649584\">Thus,<\/p>\n<div id=\"fs-id1170571649587\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\underset{x\\to 0}{\\lim}\\big(\\frac{1}{x}+\\frac{5}{x(x-5)}\\big)& =\\underset{x\\to 0}{\\lim}\\frac{x}{x(x-5)} \\\\ & =\\underset{x\\to 0}{\\lim}\\frac{1}{x-5} \\\\ & =-\\frac{1}{5} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571681422\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170571681429\">Evaluate [latex]\\underset{x\\to 3}{\\lim}\\left(\\dfrac{1}{x-3}-\\dfrac{4}{x^2-2x-3}\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q80944622\">Hint<\/span><\/p>\n<div id=\"q80944622\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572233797\">Use the same technique as the example above. Don\u2019t forget to factor [latex]x^2-2x-3[\/latex] before getting a common denominator.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572233826\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572233826\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572233826\">[latex]\\frac{1}{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170572233839\">Simple modifications in the limit laws allow us to apply them to one-sided limits. For example, to apply the limit laws to a limit of the form [latex]\\underset{x\\to a^-}{\\lim}h(x)[\/latex], we require the function [latex]h(x)[\/latex] to be defined over an open interval of the form [latex](b,a)[\/latex]; for a limit of the form [latex]\\underset{x\\to a^+}{\\lim}h(x)[\/latex], we require the function [latex]h(x)[\/latex] to be defined over an open interval of the form [latex](a,c)[\/latex]. The example illustrates this point.<\/p>\n<div id=\"fs-id1170571679268\" class=\"textbook exercises\">\n<h3>Example: Evaluating a One-Sided Limit Using the Limit Laws<\/h3>\n<p id=\"fs-id1170571679278\">Evaluate each of the following limits, if possible.<\/p>\n<ol id=\"fs-id1170571679281\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\underset{x\\to 3^-}{\\lim}\\sqrt{x-3}[\/latex]<\/li>\n<li>[latex]\\underset{x\\to 3^+}{\\lim}\\sqrt{x-3}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571679347\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571679347\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571679347\">Figure 2 illustrates the function [latex]f(x)=\\sqrt{x-3}[\/latex] and aids in our understanding of these limits.<\/p>\n<div style=\"width: 335px\" 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\/11203420\/CNX_Calc_Figure_02_03_002.jpg\" alt=\"A graph of the function f(x) = sqrt(x-3). Visually, the function looks like the top half of a parabola opening to the right with vertex at (3,0).\" width=\"325\" height=\"162\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The graph shows the function [latex]f(x)=\\sqrt{x-3}[\/latex].<\/p>\n<\/div>\n<ol id=\"fs-id1170571680900\" style=\"list-style-type: lower-alpha;\">\n<li>The function [latex]f(x)=\\sqrt{x-3}[\/latex] is defined over the interval [latex][3,+\\infty)[\/latex]. Since this function is not defined to the left of 3, we cannot apply the limit laws to compute [latex]\\underset{x\\to 3^-}{\\lim}\\sqrt{x-3}[\/latex]. In fact, since [latex]f(x)=\\sqrt{x-3}[\/latex] is undefined to the left of 3, [latex]\\underset{x\\to 3^-}{\\lim}\\sqrt{x-3}[\/latex] does not exist.<\/li>\n<li>Since [latex]f(x)=\\sqrt{x-3}[\/latex] is defined to the right of 3, the limit laws do apply to [latex]\\underset{x\\to 3^+}{\\lim}\\sqrt{x-3}[\/latex]. By applying these limit laws we obtain [latex]\\underset{x\\to 3^+}{\\lim}\\sqrt{x-3}=0[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571558874\">In the example below, we look at one-sided limits of a piecewise-defined function and use these limits to draw a conclusion about a two-sided limit of the same function.<\/p>\n<div id=\"fs-id1170571558882\" class=\"textbook exercises\">\n<h3>Example: Evaluating a Two-Sided Limit Using the Limit Laws<\/h3>\n<p id=\"fs-id1170571558891\">For [latex]f(x)=\\begin{cases} 4x-3 & \\text{ if } \\, x<2 \\\\ (x-3)^2 & \\text{ if } \\, x \\ge 2 \\end{cases}[\/latex] evaluate each of the following limits:<\/p>\n<ol id=\"fs-id1170571649889\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\underset{x\\to 2^-}{\\lim}f(x)[\/latex]<\/li>\n<li>[latex]\\underset{x\\to 2^+}{\\lim}f(x)[\/latex]<\/li>\n<li>[latex]\\underset{x\\to 2}{\\lim}f(x)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571573824\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571573824\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571573824\">Figure 3 illustrates the function [latex]f(x)[\/latex] and aids in our understanding of these limits.<\/p>\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" id=\"24\" src=\"https:\/\/openstax.org\/resources\/42329e49b363445e47e79b9689883b2017ba83a6\" alt=\"The graph of a piecewise function with two segments. For x&lt;2, the function is linear with the equation 4x-3. There is an open circle at (2,5). The second segment is a parabola and exists for x&gt;=2, with the equation (x-3)^2. There is a closed circle at (2,1). The vertex of the parabola is at (3,0).\" width=\"325\" height=\"350\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. This graph shows a function [latex](x)[\/latex]<\/p>\n<\/div>\n<ol id=\"fs-id1170571573876\" style=\"list-style-type: lower-alpha;\">\n<li>Since [latex]f(x)=4x-3[\/latex] for all [latex]x[\/latex] in [latex](\u2212\\infty,2)[\/latex], replace [latex]f(x)[\/latex] in the limit with [latex]4x-3[\/latex] and apply the limit laws:\n<div id=\"fs-id1170571573954\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 2^-}{\\lim}f(x)=\\underset{x\\to 2^-}{\\lim}(4x-3)=5[\/latex].<\/div>\n<\/li>\n<li>Since [latex]f(x)=(x-3)^2[\/latex] for all [latex]x[\/latex] in [latex](2,+\\infty)[\/latex], replace [latex]f(x)[\/latex] in the limit with [latex](x-3)^2[\/latex] and apply the limit laws:\n<div id=\"fs-id1170571644376\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 2^+}{\\lim}f(x)=\\underset{x\\to 2^+}{\\lim}(x-3)^2=1[\/latex].<\/div>\n<\/li>\n<li>Since [latex]\\underset{x\\to 2^-}{\\lim}f(x)=5[\/latex] and [latex]\\underset{x\\to 2^+}{\\lim}f(x)=1[\/latex], we conclude that [latex]\\underset{x\\to 2}{\\lim}f(x)[\/latex] does not exist.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572235169\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572235178\">Graph [latex]f(x)=\\begin{cases} -x-2 & \\text{ if } \\, x<-1 \\\\ 2 & \\text{ if } \\, x = -1 \\\\ x^3 & \\text{ if } \\, x > -1 \\end{cases}[\/latex] and evaluate [latex]\\underset{x\\to -1^-}{\\lim}f(x)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q30056723\">Show Solution<\/span><\/p>\n<div id=\"q30056723\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572235312\">Use the method in the example above to evaluate the limit.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572559753\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572559753\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"width: 335px\" 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\/11203427\/CNX_Calc_Figure_02_03_004.jpg\" alt=\"The graph of a piecewise function with three segments. The first is a linear function, -x-2, for x&lt;-1. The x intercept is at (-2,0), and there is an open circle at (-1,-1). The next segment is simply the point (-1, 2). The third segment is the function x^3 for x &gt; -1, which crossed the x axis and y axis at the origin.\" width=\"325\" height=\"350\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4.<\/p>\n<\/div>\n<p><span id=\"fs-id1170572559760\"><\/span><br \/>\n[latex]\\underset{x\\to -1^-}{\\lim}f(x)=-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We now turn our attention to evaluating a limit of the form [latex]\\underset{x\\to a}{\\lim}\\frac{f(x)}{g(x)}[\/latex], where [latex]\\underset{x\\to a}{\\lim}f(x) = K[\/latex], where [latex]K\\ne{0}[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x) = 0[\/latex]. That is [latex]f(x)\\text{\/}g(x)[\/latex] has the form [latex]K\\text{\/}0[\/latex],\u00a0[latex]K\\ne{0}[\/latex] at [latex]a[\/latex].<\/p>\n<div id=\"fs-id1170571611196\" class=\"textbook exercises\">\n<h3>Example: Evaluating a Limit of the Form [latex]K\/0, \\, K\\ne 0[\/latex] Using the Limit Laws<\/h3>\n<p id=\"fs-id1170571611224\">Evaluate [latex]\\underset{x\\to 2^-}{\\lim}\\dfrac{x-3}{x^2-2x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571611272\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571611272\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571611272\"><strong>Step 1.<\/strong> After substituting in [latex]x=2[\/latex], we see that this limit has the form [latex]-\\frac{1}{0}[\/latex]. That is, as [latex]x[\/latex] approaches 2 from the left, the numerator approaches \u22121 and the denominator approaches 0. Consequently, the magnitude of [latex]\\frac{x-3}{x(x-2)}[\/latex] becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:<\/p>\n<div id=\"fs-id1170572420265\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 2^-}{\\lim}\\frac{x-3}{x^2-2x}=\\underset{x\\to 2^-}{\\lim}\\frac{x-3}{x(x-2)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572420356\"><strong>Step 2.<\/strong> Since [latex]x-2[\/latex] is the only part of the denominator that is zero when 2 is substituted, we then separate [latex]\\frac{1}{(x-2)}[\/latex] from the rest of the function:<\/p>\n<div id=\"fs-id1170571612873\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]=\\underset{x\\to 2^-}{\\lim}\\frac{x-3}{x} \\cdot \\frac{1}{x-2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571612925\"><strong>Step 3.<\/strong>[latex]\\underset{x\\to 2^-}{\\lim}\\frac{x-3}{x}=-\\frac{1}{2}[\/latex] and [latex]\\underset{x\\to 2^-}{\\lim}\\frac{1}{x-2}=\u2212\\infty[\/latex]. Therefore, the product of [latex]\\frac{(x-3)}{x}[\/latex] and [latex]\\frac{1}{(x-2)}[\/latex] has a limit of [latex]+\\infty[\/latex]:<\/p>\n<div id=\"fs-id1170571650460\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 2^-}{\\lim}\\frac{x-3}{x^2-2x}=+\\infty[\/latex].\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571650517\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170571650526\">Evaluate [latex]\\underset{x\\to 1}{\\lim}\\dfrac{x+2}{(x-1)^2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q6003722\">Hint<\/span><\/p>\n<div id=\"q6003722\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572611875\">Use the methods from the example above.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572611885\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572611885\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572611885\">[latex]+\\infty[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-285\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>2.3 Limit Laws. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"2.3 Limit Laws\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-285","chapter","type-chapter","status-publish","hentry"],"part":28,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/285","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":47,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/285\/revisions"}],"predecessor-version":[{"id":4789,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/285\/revisions\/4789"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/28"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/285\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=285"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=285"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=285"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=285"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}