{"id":304,"date":"2021-02-04T01:05:00","date_gmt":"2021-02-04T01:05:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=304"},"modified":"2022-03-16T05:22:59","modified_gmt":"2022-03-16T05:22:59","slug":"proving-limit-laws","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/proving-limit-laws\/","title":{"raw":"Proving Limit Laws","rendered":"Proving Limit Laws"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the epsilon-delta definition to prove the limit laws<\/li>\r\n \t<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Describe the epsilon-delta definitions of one-sided limits and infinite limits&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:2304,&quot;11&quot;:4,&quot;14&quot;:{&quot;1&quot;:2,&quot;2&quot;:0}}\">Describe the epsilon-delta definitions of one-sided limits and infinite limits<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170572512411\">We now demonstrate how to use the epsilon-delta definition of a limit to construct a rigorous proof of one of the limit laws. The triangle inequality is used at a key point of the proof, so we first review this key property of absolute value.<\/p>\r\n\r\n<div id=\"fs-id1170572512419\" class=\"textbox shaded\">\r\n<div class=\"title\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\n<p id=\"fs-id1170571712102\">The <strong>triangle inequality<\/strong> states that if [latex]a[\/latex] and [latex]b[\/latex] are any real numbers, then [latex]|a+b|\\le |a|+|b|[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572230013\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1170572230018\">We prove the following limit law: If [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=M[\/latex], then [latex]\\underset{x\\to a}{\\lim}(f(x)+g(x))=L+M[\/latex].<\/p>\r\n<p id=\"fs-id1170572540883\">Let [latex]\\varepsilon &gt;0[\/latex].<\/p>\r\n<p id=\"fs-id1170571562529\">Choose [latex]\\delta_1&gt;0[\/latex] so that if [latex]0&lt;|x-a|&lt;\\delta_1[\/latex], then [latex]|f(x)-L|&lt;\\varepsilon\/2[\/latex].<\/p>\r\n<p id=\"fs-id1170572444498\">Choose [latex]\\delta_2&gt;0[\/latex] so that if [latex]0&lt;|x-a|&lt;\\delta_2[\/latex], then [latex]|g(x)-M|&lt;\\varepsilon\/2[\/latex].<\/p>\r\n<p id=\"fs-id1170572168662\">Choose [latex]\\delta =\\text{min}\\{\\delta_1,\\delta_2\\}[\/latex].<\/p>\r\n<p id=\"fs-id1170572163714\">Assume [latex]0&lt;|x-a|&lt;\\delta[\/latex].<\/p>\r\n<p id=\"fs-id1170572163742\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1170572163746\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]0&lt;|x-a|&lt;\\delta_1[\/latex] and [latex]0&lt;|x-a|&lt;\\delta_2[\/latex]<\/div>\r\n<p id=\"fs-id1170571657220\">Hence,<\/p>\r\n\r\n<div id=\"fs-id1170571657223\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} |(f(x)+g(x))-(L+M)| &amp; =|(f(x)-L)+(g(x)-M)| \\\\ &amp; \\le |f(x)-L|+|g(x)-M| \\\\ &amp; &lt;\\dfrac{\\varepsilon}{2}+\\dfrac{\\varepsilon}{2}=\\varepsilon \\end{array}[\/latex]<\/div>\r\n<div>[latex]\\blacksquare[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572505415\">We now explore what it means for a limit not to exist. The limit [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] does not exist if there is no real number [latex]L[\/latex] for which [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex]. Thus, for all real numbers [latex]L[\/latex], [latex]\\underset{x\\to a}{\\lim}f(x)\\ne L[\/latex]. To understand what this means, we look at each part of the definition of [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex] together with its opposite. A translation of the definition is given in the table below.<\/p>\r\n\r\n<table id=\"fs-id1170571696614\" summary=\"A table with two columns and four rows. The top row contains the headers \u201cdefinition\u201d and \u201copposite.\u201d The second row contains the definition \u201cfor every epsilon &lt; 0\u201d and the opposite \u201cthere exists an epsilon greater than zero so that.\u201d The third row contains the definition \u201cthere exists a delta greater than 0, so that\u201d and the opposite \u201cfor every delta greater than 0.\u201d The last row contains the definition \u201cif 0 is less than the absolute value of x-a, which is less than delta, then the absolute value of f(x) \u2013 L is less than epsilon\u201d and the opposite \u201cthere is an x satisfying 0 is less than the absolute value of x \u2013 a, which is less than delta, so that the absolute value of f() \u2013 L is greater than or equal to epsilon.\"><caption>Translation of the Definition of [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex] and its Opposite<\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th><strong>Definition<\/strong><\/th>\r\n<th><strong>Opposite<\/strong><\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>1. For every [latex]\\varepsilon &gt;0[\/latex],<\/td>\r\n<td>1. There exists [latex]\\varepsilon &gt;0[\/latex] so that<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2. there exists a [latex]\\delta &gt;0[\/latex] so that<\/td>\r\n<td>2. for every [latex]\\delta &gt;0[\/latex],<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>3. if [latex]0&lt;|x-a|&lt;\\delta[\/latex], then [latex]|f(x)-L|&lt;\\varepsilon[\/latex].<\/td>\r\n<td>3. There is an [latex]x[\/latex] satisfying [latex]0&lt;|x-a|&lt;\\delta [\/latex] so that [latex]|f(x)-L|\\ge \\varepsilon[\/latex].<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1170571635904\">Finally, we may state what it means for a limit not to exist. The limit [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] does not exist if for every real number [latex]L[\/latex], there exists a real number [latex]\\varepsilon &gt;0[\/latex] so that for all [latex]\\delta &gt;0[\/latex], there is an [latex]x[\/latex] satisfying [latex]0&lt;|x-a|&lt;\\delta[\/latex], so that [latex]|f(x)-L|\\ge \\varepsilon[\/latex]. Let\u2019s apply this in the example to show that a limit does not exist.<\/p>\r\n\r\n<div id=\"fs-id1170572550055\" class=\"textbook exercises\">\r\n<h3>Example: Showing That a Limit Does Not Exist<\/h3>\r\n<p id=\"fs-id1170572559647\">Show that [latex]\\underset{x\\to 0}{\\lim}\\dfrac{|x|}{x}[\/latex] does not exist. The graph of [latex]f(x)=\\dfrac{|x|}{x}[\/latex] is shown here:<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"342\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203540\/CNX_Calc_Figure_02_05_006.jpg\" alt=\"A graph of a function with two segments. The first exists for x&lt;0, and it is a line with no slope that ends at the y axis in an open circle at (0,-1). The second exists for x&gt;0, and it is a line with no slope that begins at the y axis in an open circle (1,0).\" width=\"342\" height=\"347\" \/> Figure 4.[\/caption]\r\n\r\n[reveal-answer q=\"fs-id1170572601135\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572601135\"]\r\n<p id=\"fs-id1170572601135\">Suppose that [latex]L[\/latex] is a candidate for a limit. Choose [latex]\\varepsilon =\\frac{1}{2}[\/latex].<\/p>\r\n<p id=\"fs-id1170571613379\">Let [latex]\\delta &gt;0[\/latex]. Either [latex]L\\ge 0[\/latex] or [latex]L&lt;0[\/latex]. If [latex]L\\ge 0[\/latex], then let [latex]x=-\\delta\/2[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1170571733792\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|x-0|=|-\\frac{\\delta}{2}-0|=\\frac{\\delta}{2}&lt;\\delta [\/latex]<\/div>\r\n<p id=\"fs-id1170572419028\">and<\/p>\r\n\r\n<div id=\"fs-id1170572419032\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|\\frac{|-\\frac{\\delta}{2}|}{-\\frac{\\delta}{2}}-L|=|-1-L|=L+1\\ge 1&gt;\\frac{1}{2}=\\varepsilon[\/latex]<\/div>\r\n<p id=\"fs-id1170571604788\">On the other hand, if [latex]L&lt;0[\/latex], then let [latex]x=\\delta\/2[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1170572332361\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|x-0|=|\\frac{\\delta}{2}-0|=\\frac{\\delta}{2}&lt;\\delta [\/latex]<\/div>\r\n<p id=\"fs-id1170572243165\">and<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|\\frac{|\\frac{\\delta}{2}|}{\\frac{\\delta}{2}}-L|=|1-L|=|L|+1\\ge 1&gt;\\frac{1}{2}=\\varepsilon[\/latex]<\/div>\r\n<p id=\"fs-id1170571610784\">Thus, for any value of [latex]L[\/latex], [latex]\\underset{x\\to 0}{\\lim}\\frac{|x|}{x}\\ne L[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example: Showing That a Limit Does Not Exist. [\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/5_q7_Zx26RY?controls=0&amp;start=1380&amp;end=1657&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\/2.5PreciseDefinitionOfALimit1380to1657_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.5 Precise Definition of a Limit\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<h2>Finding Deltas Algebraically for Given Epsilons<\/h2>\r\nNow that we have proven limits, we can now apply them with actual numbers for [latex]\\varepsilon[\/latex] and [latex]\\delta[\/latex]. Think of [latex]\\varepsilon[\/latex] as the error in the [latex]x[\/latex]-direction and [latex]\\delta[\/latex] to be the error in the [latex]y[\/latex]-direction. These have applications in engineering when these errors are considered tolerances. We want to know what the error intervals are, and we are trying to minimize these errors.\r\n<div id=\"fs-id1170572550055\" class=\"textbook exercises\">\r\n<h3>Example: Finding deltas algebraically, Part 1<\/h3>\r\n<p id=\"fs-id1170572559647\">Find an open interval about [latex]x_0[\/latex] on which the inequality [latex]|f(x)-L| &lt; 0[\/latex] holds. Then give the largest value [latex]\\delta &amp;gt; 0[\/latex] such that for all [latex]x[\/latex] satisfying [latex]0 &lt; |x-x_0| &lt; \\delta[\/latex] the inequality [latex]|f(x)-L| &lt; \\varepsilon[\/latex] holds.<\/p>\r\n[latex]f(x)=2x-8, \\,\\, L=6, \\,\\, x_0=7, \\,\\, \\varepsilon=0.14[\/latex]\r\n\r\n[reveal-answer q=\"fs-id1170572601140\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572601140\"]First we will need to start with the inequality [latex]|f(x)-L| &lt; \\varepsilon[\/latex] and plug in our numbers. Then we will solve for x.[latex]|2x-8 - 6| &lt; \\varepsilon[\/latex][latex]|2x-14| &lt; \\varepsilon [\/latex][latex]-0.14 &lt; 2x - 14 &amp;lt 0.14[\/latex][latex]13.86 &lt; 2x &lt; 14.14[\/latex][latex]6.93 &lt; x &lt; 7.07[\/latex]Therefore, the interval is [latex](7.93,8.07)[\/latex]For the second answer, we will start with\u00a0[latex]0 &lt; |x-x_0| &lt; \\delta[\/latex].\u00a0 We will plug in our value and solve:[latex]|x-7| &amp;lt; \\delta[\/latex][latex]-\\delta &amp;lt; x-7 &amp;lt; \\delta[\/latex][latex]7-\\delta &amp;lt x &amp;lt; 7+\\delta[\/latex]Now we will set each piece equal to the endpoints we found above.[latex]7-\\delta=7.93[\/latex] and [latex]7+\\delta=8.07[\/latex]After solving we will get the same answer for each equation: [latex]\\delta=0.07[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572550055\" class=\"textbook exercises\">\r\n<h3>Example: Finding deltas algebraically, Part 2<\/h3>\r\n<p id=\"fs-id1170572559647\">Find an open interval about [latex]x_0[\/latex] on which the inequality [latex]|f(x)-L| &lt; 0[\/latex] holds. Then give the largest value [latex]\\delta &amp;gt; 0[\/latex] such that for all [latex]x[\/latex] satisfying [latex]0 &lt; |x-x_0| &lt; \\delta[\/latex] the inequality [latex]|f(x)-L| &lt; \\varepsilon[\/latex] holds.<\/p>\r\n[latex]f(x)=\\sqrt{x+4}, \\,\\, L=3, \\,\\, x_0=5, \\,\\, \\varepsilon=1[\/latex]\r\n\r\n[reveal-answer q=\"fs-id1170572601145\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572601145\"]First we will need to start with the inequality [latex]|f(x)-L| &lt; \\varepsilon[\/latex] and plug in our numbers. Then we will solve for x.[latex]|\\sqrt(x+4) - 3| &lt; \\varepsilon[\/latex][latex]|\\sqrt{x+4} - 3| &lt; 1[\/latex][latex]-1 &lt; \\sqrt{x+4} - 3 &lt; 1[\/latex][latex]2 &lt; \\sqrt{x+4} &lt; 4[\/latex][latex]4 &lt; x + 4 &lt; 16[\/latex][latex]0 &lt; x + 4 &lt; 12[\/latex]Therefore, the interval is [latex](0,12)[\/latex].For the second answer, we will start with\u00a0[latex]0 &lt; |x-x_0| &lt; \\delta[\/latex].\u00a0 We will plug in our value and solve:[latex]|x-5| &amp;lt; \\delta[\/latex][latex]-\\delta &amp;lt; x-5 &amp;lt; \\delta[\/latex][latex]5-\\delta &amp;lt x &amp;lt; 5+\\delta[\/latex]Now we will set each piece equal to the endpoints we found above.[latex]5-\\delta=0[\/latex] and [latex]5+\\delta=12[\/latex]\r\n\r\nAfter solving we will get [latex]\\delta=5 \\text{ and }\\delta=7[\/latex]. Since the question is asking for the smallest interval, we choose the smaller number.\u00a0 Therefore the answer is [latex]\\delta=5[\/latex].\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">One-Sided and Infinite Limits<\/h2>\r\nJust as we first gained an intuitive understanding of limits and then moved on to a more rigorous definition of a limit, we now revisit one-sided limits. To do this, we modify the epsilon-delta definition of a limit to give formal epsilon-delta definitions for limits from the right and left at a point. These definitions only require slight modifications from the definition of the limit. In the definition of the limit from the right, the inequality [latex]0&lt;x-a&lt;\\delta[\/latex] replaces [latex]0&lt;|x-a|&lt;\\delta[\/latex], which ensures that we only consider values of [latex]x[\/latex] that are greater than (to the right of) [latex]a[\/latex]. Similarly, in the definition of the limit from the left, the inequality [latex]-\\delta&lt;x-a&lt;0[\/latex] replaces [latex]0&lt;|x-a|&lt;\\delta[\/latex], which ensures that we only consider values of [latex]x[\/latex] that are less than (to the left of) [latex]a[\/latex].\r\n<div id=\"fs-id1170571597293\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170571597296\"><strong>Limit from the Right:<\/strong> Let [latex]f(x)[\/latex] be defined over an open interval of the form [latex](a,b)[\/latex] where [latex]a&lt;b[\/latex]. Then,<\/p>\r\n\r\n<div id=\"fs-id1170571597343\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a^+}{\\lim}f(x)=L[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1170571652104\" style=\"text-align: center;\">if for every [latex]\\varepsilon &gt;0[\/latex], there exists a [latex]\\delta &gt;0[\/latex] such that if [latex]0&lt;x-a&lt;\\delta[\/latex], then [latex]|f(x)-L|&lt;\\varepsilon[\/latex].<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1170572168879\"><strong>Limit from the Left:<\/strong> Let [latex]f(x)[\/latex] be defined over an open interval of the form [latex](a,b)[\/latex] where [latex]a&lt;b[\/latex]. Then,<\/p>\r\n\r\n<div id=\"fs-id1170572419122\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to b^-}{\\lim}f(x)=L[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1170572540903\" style=\"text-align: center;\">if for every [latex]\\varepsilon &gt;0[\/latex], there exists a [latex]\\delta &gt;0[\/latex] such that if [latex]0&lt;b-x&lt;\\delta[\/latex], then [latex]|f(x)-L|&lt;\\varepsilon[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572330921\" class=\"textbox exercises\">\r\n<h3>Example: Proving a Statement about a Limit From the Right<\/h3>\r\nProve that [latex]\\underset{x\\to 4^+}{\\lim}\\sqrt{x-4}=0[\/latex].\r\n<div id=\"fs-id1170572330924\" class=\"exercise\">[reveal-answer q=\"fs-id1170572601349\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572601349\"]\r\n<p id=\"fs-id1170572601349\">Let [latex]\\varepsilon &gt;0.[\/latex]<\/p>\r\n<p id=\"fs-id1170572601363\">Choose [latex]\\delta =\\varepsilon^2[\/latex]. Since we ultimately want [latex]|\\sqrt{x-4}-0|&lt;\\varepsilon[\/latex], we manipulate this inequality to get [latex]\\sqrt{x-4}&lt;\\varepsilon[\/latex] or, equivalently, [latex]0&lt;x-4&lt;\\varepsilon^2[\/latex], making [latex]\\delta =\\varepsilon^2[\/latex] a clear choice. We may also determine [latex]\\delta[\/latex] geometrically, as shown in Figure 5.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"459\"]<img id=\"20\" src=\"https:\/\/openstax.org\/resources\/53207e09378fd9a465f9a0ae4026f8432751441f\" alt=\"A graph showing how to find delta for the above proof. The function f(x) = sqrt(x-4) is drawn for x &gt; 4. Since the proposed limit is 0, lines y = 0 + epsilon and y = 0 \u2013 epsilon are drawn in blue. Since only the top blue line corresponding to y = 0 + epsilon intersects the function, one red line is drawn from the point of intersection to the x axis. This x value is found by solving sqrt(x-4) = epsilon, or x = epsilon squared + 4. Delta is then the distance between this point and 4, which is epsilon squared.\" width=\"459\" height=\"320\" data-media-type=\"image\/jpeg\" \/> Figure 5. This graph shows how we find [latex]\\delta[\/latex] for the proof of this example.[\/caption]\r\n<p id=\"fs-id1170572376082\">Assume [latex]0&lt;x-4&lt;\\delta[\/latex]. Thus, [latex]0&lt;x-4&lt;\\varepsilon^2[\/latex]. Hence, [latex]0&lt;\\sqrt{x-4}&lt;\\varepsilon[\/latex]. Finally, [latex]|\\sqrt{x-4}-0|&lt;\\varepsilon[\/latex].<\/p>\r\n<p id=\"fs-id1170571586144\">Therefore, [latex]\\underset{x\\to 4^+}{\\lim}\\sqrt{x-4}=0[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example: Proving a Statement about a Limit From the Right. [\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/5_q7_Zx26RY?controls=0&amp;start=1660&amp;end=1870&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\/2.5PreciseDefinitionOfaLimit1660to1870_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.5 Precise Definition of a Limit\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170571711155\" class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170571711163\">Find [latex]\\delta[\/latex] corresponding to [latex]\\varepsilon[\/latex]\u00a0for a proof that [latex]\\underset{x\\to 1^-}{\\lim}\\sqrt{1-x}=0[\/latex].<\/p>\r\n&nbsp;\r\n\r\n[reveal-answer q=\"142584\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"142584\"]\r\n\r\n&nbsp;\r\n\r\nSketch the graph and use Figure 5 as a solving guide.\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"142584\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"142584\"]\r\n\r\n[latex]\\delta =\\varepsilon^2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe conclude the process of converting our intuitive ideas of various types of limits to rigorous formal definitions by pursuing a formal definition of infinite limits. To have [latex]\\underset{x\\to a}{\\lim}f(x)=+\\infty[\/latex], we want the values of the function [latex]f(x)[\/latex] to get larger and larger as [latex]x[\/latex] approaches [latex]a[\/latex]. Instead of the requirement that [latex]|f(x)-L|&lt;\\varepsilon[\/latex] for arbitrarily small [latex]\\varepsilon[\/latex]\u00a0when [latex]0&lt;|x-a|&lt;\\delta[\/latex] for small enough [latex]\\delta[\/latex], we want [latex]f(x)&gt;M[\/latex] for arbitrarily large positive [latex]M[\/latex] when [latex]0&lt;|x-a|&lt;\\delta[\/latex] for small enough [latex]\\delta[\/latex]. Figure 6 illustrates this idea by showing the value of [latex]\\delta[\/latex] for successively larger values of [latex]M[\/latex].\r\n<div id=\"CNX_Calc_Figure_02_05_005\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203546\/CNX_Calc_Figure_02_05_005.jpg\" alt=\"Two graphs side by side. Each graph contains two curves above the x axis separated by an asymptote at x=a. The curves on the left go to infinity as x goes to a and to 0 as x goes to negative infinity. The curves on the right go to infinity as x goes to a and to 0 as x goes to infinity. The first graph has a value M greater than zero marked on the y axis and a horizontal line drawn from there (y=M) to intersect with both curves. Lines are drawn down from the points of intersection to the x axis. Delta is the smaller of the distances between point a and these new spots on the x axis. The same lines are drawn on the second graph, but this M is larger, and the distances from the x axis intersections to point a are smaller.\" width=\"975\" height=\"422\" \/> Figure 6. These graphs plot values of [latex]\\delta[\/latex] for [latex]M[\/latex] to show that [latex]\\underset{x\\to a}{\\lim}f(x)=+\\infty[\/latex].[\/caption]<\/div>\r\n<div id=\"fs-id1170571609486\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170571609489\">Let [latex]f(x)[\/latex] be defined for all [latex]x\\ne a[\/latex] in an open interval containing [latex]a[\/latex]. Then, we have an infinite limit<\/p>\r\n\r\n<div id=\"fs-id1170571635969\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}f(x)=+\\infty[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1170571636004\">if for every [latex]M&gt;0[\/latex], there exists [latex]\\delta &gt;0[\/latex] such that if [latex]0&lt;|x-a|&lt;\\delta[\/latex], then [latex]f(x)&gt;M[\/latex].<\/p>\r\n<p id=\"fs-id1170572216496\">Let [latex]f(x)[\/latex] be defined for all [latex]x\\ne a[\/latex] in an open interval containing [latex]a[\/latex]. Then, we have a negative infinite limit<\/p>\r\n\r\n<div id=\"fs-id1170571600638\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}f(x)=\u2212\\infty[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1170571600673\">if for every [latex]M&gt;0[\/latex], there exists [latex]\\delta &gt;0[\/latex] such that if [latex]0&lt;|x-a|&lt;\\delta[\/latex], then [latex]f(x)&lt;\u2212M[\/latex].<\/p>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the epsilon-delta definition to prove the limit laws<\/li>\n<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Describe the epsilon-delta definitions of one-sided limits and infinite limits&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:2304,&quot;11&quot;:4,&quot;14&quot;:{&quot;1&quot;:2,&quot;2&quot;:0}}\">Describe the epsilon-delta definitions of one-sided limits and infinite limits<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170572512411\">We now demonstrate how to use the epsilon-delta definition of a limit to construct a rigorous proof of one of the limit laws. The triangle inequality is used at a key point of the proof, so we first review this key property of absolute value.<\/p>\n<div id=\"fs-id1170572512419\" class=\"textbox shaded\">\n<div class=\"title\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<\/div>\n<p id=\"fs-id1170571712102\">The <strong>triangle inequality<\/strong> states that if [latex]a[\/latex] and [latex]b[\/latex] are any real numbers, then [latex]|a+b|\\le |a|+|b|[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1170572230013\" class=\"bc-section section\">\n<h3>Proof<\/h3>\n<p id=\"fs-id1170572230018\">We prove the following limit law: If [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=M[\/latex], then [latex]\\underset{x\\to a}{\\lim}(f(x)+g(x))=L+M[\/latex].<\/p>\n<p id=\"fs-id1170572540883\">Let [latex]\\varepsilon >0[\/latex].<\/p>\n<p id=\"fs-id1170571562529\">Choose [latex]\\delta_1>0[\/latex] so that if [latex]0<|x-a|<\\delta_1[\/latex], then [latex]|f(x)-L|<\\varepsilon\/2[\/latex].<\/p>\n<p id=\"fs-id1170572444498\">Choose [latex]\\delta_2>0[\/latex] so that if [latex]0<|x-a|<\\delta_2[\/latex], then [latex]|g(x)-M|<\\varepsilon\/2[\/latex].<\/p>\n<p id=\"fs-id1170572168662\">Choose [latex]\\delta =\\text{min}\\{\\delta_1,\\delta_2\\}[\/latex].<\/p>\n<p id=\"fs-id1170572163714\">Assume [latex]0<|x-a|<\\delta[\/latex].<\/p>\n<p id=\"fs-id1170572163742\">Thus,<\/p>\n<div id=\"fs-id1170572163746\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]0<|x-a|<\\delta_1[\/latex] and [latex]0<|x-a|<\\delta_2[\/latex]<\/div>\n<p id=\"fs-id1170571657220\">Hence,<\/p>\n<div id=\"fs-id1170571657223\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} |(f(x)+g(x))-(L+M)| & =|(f(x)-L)+(g(x)-M)| \\\\ & \\le |f(x)-L|+|g(x)-M| \\\\ & <\\dfrac{\\varepsilon}{2}+\\dfrac{\\varepsilon}{2}=\\varepsilon \\end{array}[\/latex]<\/div>\n<div>[latex]\\blacksquare[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572505415\">We now explore what it means for a limit not to exist. The limit [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] does not exist if there is no real number [latex]L[\/latex] for which [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex]. Thus, for all real numbers [latex]L[\/latex], [latex]\\underset{x\\to a}{\\lim}f(x)\\ne L[\/latex]. To understand what this means, we look at each part of the definition of [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex] together with its opposite. A translation of the definition is given in the table below.<\/p>\n<table id=\"fs-id1170571696614\" summary=\"A table with two columns and four rows. The top row contains the headers \u201cdefinition\u201d and \u201copposite.\u201d The second row contains the definition \u201cfor every epsilon &lt; 0\u201d and the opposite \u201cthere exists an epsilon greater than zero so that.\u201d The third row contains the definition \u201cthere exists a delta greater than 0, so that\u201d and the opposite \u201cfor every delta greater than 0.\u201d The last row contains the definition \u201cif 0 is less than the absolute value of x-a, which is less than delta, then the absolute value of f(x) \u2013 L is less than epsilon\u201d and the opposite \u201cthere is an x satisfying 0 is less than the absolute value of x \u2013 a, which is less than delta, so that the absolute value of f() \u2013 L is greater than or equal to epsilon.\">\n<caption>Translation of the Definition of [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex] and its Opposite<\/caption>\n<thead>\n<tr valign=\"top\">\n<th><strong>Definition<\/strong><\/th>\n<th><strong>Opposite<\/strong><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>1. For every [latex]\\varepsilon >0[\/latex],<\/td>\n<td>1. There exists [latex]\\varepsilon >0[\/latex] so that<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2. there exists a [latex]\\delta >0[\/latex] so that<\/td>\n<td>2. for every [latex]\\delta >0[\/latex],<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>3. if [latex]0<|x-a|<\\delta[\/latex], then [latex]|f(x)-L|<\\varepsilon[\/latex].<\/td>\n<td>3. There is an [latex]x[\/latex] satisfying [latex]0<|x-a|<\\delta[\/latex] so that [latex]|f(x)-L|\\ge \\varepsilon[\/latex].<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1170571635904\">Finally, we may state what it means for a limit not to exist. The limit [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] does not exist if for every real number [latex]L[\/latex], there exists a real number [latex]\\varepsilon >0[\/latex] so that for all [latex]\\delta >0[\/latex], there is an [latex]x[\/latex] satisfying [latex]0<|x-a|<\\delta[\/latex], so that [latex]|f(x)-L|\\ge \\varepsilon[\/latex]. Let\u2019s apply this in the example to show that a limit does not exist.<\/p>\n<div id=\"fs-id1170572550055\" class=\"textbook exercises\">\n<h3>Example: Showing That a Limit Does Not Exist<\/h3>\n<p id=\"fs-id1170572559647\">Show that [latex]\\underset{x\\to 0}{\\lim}\\dfrac{|x|}{x}[\/latex] does not exist. The graph of [latex]f(x)=\\dfrac{|x|}{x}[\/latex] is shown here:<\/p>\n<div style=\"width: 352px\" 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\/11203540\/CNX_Calc_Figure_02_05_006.jpg\" alt=\"A graph of a function with two segments. The first exists for x&lt;0, and it is a line with no slope that ends at the y axis in an open circle at (0,-1). The second exists for x&gt;0, and it is a line with no slope that begins at the y axis in an open circle (1,0).\" width=\"342\" height=\"347\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572601135\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572601135\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572601135\">Suppose that [latex]L[\/latex] is a candidate for a limit. Choose [latex]\\varepsilon =\\frac{1}{2}[\/latex].<\/p>\n<p id=\"fs-id1170571613379\">Let [latex]\\delta >0[\/latex]. Either [latex]L\\ge 0[\/latex] or [latex]L<0[\/latex]. If [latex]L\\ge 0[\/latex], then let [latex]x=-\\delta\/2[\/latex]. Thus,<\/p>\n<div id=\"fs-id1170571733792\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|x-0|=|-\\frac{\\delta}{2}-0|=\\frac{\\delta}{2}<\\delta[\/latex]<\/div>\n<p id=\"fs-id1170572419028\">and<\/p>\n<div id=\"fs-id1170572419032\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|\\frac{|-\\frac{\\delta}{2}|}{-\\frac{\\delta}{2}}-L|=|-1-L|=L+1\\ge 1>\\frac{1}{2}=\\varepsilon[\/latex]<\/div>\n<p id=\"fs-id1170571604788\">On the other hand, if [latex]L<0[\/latex], then let [latex]x=\\delta\/2[\/latex]. Thus,<\/p>\n<div id=\"fs-id1170572332361\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|x-0|=|\\frac{\\delta}{2}-0|=\\frac{\\delta}{2}<\\delta[\/latex]<\/div>\n<p id=\"fs-id1170572243165\">and<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|\\frac{|\\frac{\\delta}{2}|}{\\frac{\\delta}{2}}-L|=|1-L|=|L|+1\\ge 1>\\frac{1}{2}=\\varepsilon[\/latex]<\/div>\n<p id=\"fs-id1170571610784\">Thus, for any value of [latex]L[\/latex], [latex]\\underset{x\\to 0}{\\lim}\\frac{|x|}{x}\\ne L[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Showing That a Limit Does Not Exist. <\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/5_q7_Zx26RY?controls=0&amp;start=1380&amp;end=1657&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\/2.5PreciseDefinitionOfALimit1380to1657_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.5 Precise Definition of a Limit&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<h2>Finding Deltas Algebraically for Given Epsilons<\/h2>\n<p>Now that we have proven limits, we can now apply them with actual numbers for [latex]\\varepsilon[\/latex] and [latex]\\delta[\/latex]. Think of [latex]\\varepsilon[\/latex] as the error in the [latex]x[\/latex]-direction and [latex]\\delta[\/latex] to be the error in the [latex]y[\/latex]-direction. These have applications in engineering when these errors are considered tolerances. We want to know what the error intervals are, and we are trying to minimize these errors.<\/p>\n<div id=\"fs-id1170572550055\" class=\"textbook exercises\">\n<h3>Example: Finding deltas algebraically, Part 1<\/h3>\n<p id=\"fs-id1170572559647\">Find an open interval about [latex]x_0[\/latex] on which the inequality [latex]|f(x)-L| < 0[\/latex] holds. Then give the largest value [latex]\\delta &gt; 0[\/latex] such that for all [latex]x[\/latex] satisfying [latex]0 < |x-x_0| < \\delta[\/latex] the inequality [latex]|f(x)-L| < \\varepsilon[\/latex] holds.<\/p>\n<p>[latex]f(x)=2x-8, \\,\\, L=6, \\,\\, x_0=7, \\,\\, \\varepsilon=0.14[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572601140\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572601140\" class=\"hidden-answer\" style=\"display: none\">First we will need to start with the inequality [latex]|f(x)-L| < \\varepsilon[\/latex] and plug in our numbers. Then we will solve for x.[latex]|2x-8 - 6| < \\varepsilon[\/latex][latex]|2x-14| < \\varepsilon[\/latex][latex]-0.14 < 2x - 14 &lt 0.14[\/latex][latex]13.86 < 2x < 14.14[\/latex][latex]6.93 < x < 7.07[\/latex]Therefore, the interval is [latex](7.93,8.07)[\/latex]For the second answer, we will start with\u00a0[latex]0 < |x-x_0| < \\delta[\/latex].\u00a0 We will plug in our value and solve:[latex]|x-7| &lt; \\delta[\/latex][latex]-\\delta &lt; x-7 &lt; \\delta[\/latex][latex]7-\\delta &lt x &lt; 7+\\delta[\/latex]Now we will set each piece equal to the endpoints we found above.[latex]7-\\delta=7.93[\/latex] and [latex]7+\\delta=8.07[\/latex]After solving we will get the same answer for each equation: [latex]\\delta=0.07[\/latex].<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572550055\" class=\"textbook exercises\">\n<h3>Example: Finding deltas algebraically, Part 2<\/h3>\n<p id=\"fs-id1170572559647\">Find an open interval about [latex]x_0[\/latex] on which the inequality [latex]|f(x)-L| < 0[\/latex] holds. Then give the largest value [latex]\\delta &gt; 0[\/latex] such that for all [latex]x[\/latex] satisfying [latex]0 < |x-x_0| < \\delta[\/latex] the inequality [latex]|f(x)-L| < \\varepsilon[\/latex] holds.<\/p>\n<p>[latex]f(x)=\\sqrt{x+4}, \\,\\, L=3, \\,\\, x_0=5, \\,\\, \\varepsilon=1[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572601145\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572601145\" class=\"hidden-answer\" style=\"display: none\">First we will need to start with the inequality [latex]|f(x)-L| < \\varepsilon[\/latex] and plug in our numbers. Then we will solve for x.[latex]|\\sqrt(x+4) - 3| < \\varepsilon[\/latex][latex]|\\sqrt{x+4} - 3| < 1[\/latex][latex]-1 < \\sqrt{x+4} - 3 < 1[\/latex][latex]2 < \\sqrt{x+4} < 4[\/latex][latex]4 < x + 4 < 16[\/latex][latex]0 < x + 4 < 12[\/latex]Therefore, the interval is [latex](0,12)[\/latex].For the second answer, we will start with\u00a0[latex]0 < |x-x_0| < \\delta[\/latex].\u00a0 We will plug in our value and solve:[latex]|x-5| &lt; \\delta[\/latex][latex]-\\delta &lt; x-5 &lt; \\delta[\/latex][latex]5-\\delta &lt x &lt; 5+\\delta[\/latex]Now we will set each piece equal to the endpoints we found above.[latex]5-\\delta=0[\/latex] and [latex]5+\\delta=12[\/latex]\n\nAfter solving we will get [latex]\\delta=5 \\text{ and }\\delta=7[\/latex]. Since the question is asking for the smallest interval, we choose the smaller number.\u00a0 Therefore the answer is [latex]\\delta=5[\/latex].\n\n&nbsp;\n\n<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">One-Sided and Infinite Limits<\/h2>\n<p>Just as we first gained an intuitive understanding of limits and then moved on to a more rigorous definition of a limit, we now revisit one-sided limits. To do this, we modify the epsilon-delta definition of a limit to give formal epsilon-delta definitions for limits from the right and left at a point. These definitions only require slight modifications from the definition of the limit. In the definition of the limit from the right, the inequality [latex]0<x-a<\\delta[\/latex] replaces [latex]0<|x-a|<\\delta[\/latex], which ensures that we only consider values of [latex]x[\/latex] that are greater than (to the right of) [latex]a[\/latex]. Similarly, in the definition of the limit from the left, the inequality [latex]-\\delta<x-a<0[\/latex] replaces [latex]0<|x-a|<\\delta[\/latex], which ensures that we only consider values of [latex]x[\/latex] that are less than (to the left of) [latex]a[\/latex].\n\n\n<div id=\"fs-id1170571597293\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1170571597296\"><strong>Limit from the Right:<\/strong> Let [latex]f(x)[\/latex] be defined over an open interval of the form [latex](a,b)[\/latex] where [latex]a<b[\/latex]. Then,<\/p>\n<div id=\"fs-id1170571597343\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a^+}{\\lim}f(x)=L[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1170571652104\" style=\"text-align: center;\">if for every [latex]\\varepsilon >0[\/latex], there exists a [latex]\\delta >0[\/latex] such that if [latex]0<x-a<\\delta[\/latex], then [latex]|f(x)-L|<\\varepsilon[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572168879\"><strong>Limit from the Left:<\/strong> Let [latex]f(x)[\/latex] be defined over an open interval of the form [latex](a,b)[\/latex] where [latex]a<b[\/latex]. Then,<\/p>\n<div id=\"fs-id1170572419122\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to b^-}{\\lim}f(x)=L[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1170572540903\" style=\"text-align: center;\">if for every [latex]\\varepsilon >0[\/latex], there exists a [latex]\\delta >0[\/latex] such that if [latex]0<b-x<\\delta[\/latex], then [latex]|f(x)-L|<\\varepsilon[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1170572330921\" class=\"textbox exercises\">\n<h3>Example: Proving a Statement about a Limit From the Right<\/h3>\n<p>Prove that [latex]\\underset{x\\to 4^+}{\\lim}\\sqrt{x-4}=0[\/latex].<\/p>\n<div id=\"fs-id1170572330924\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572601349\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572601349\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572601349\">Let [latex]\\varepsilon >0.[\/latex]<\/p>\n<p id=\"fs-id1170572601363\">Choose [latex]\\delta =\\varepsilon^2[\/latex]. Since we ultimately want [latex]|\\sqrt{x-4}-0|<\\varepsilon[\/latex], we manipulate this inequality to get [latex]\\sqrt{x-4}<\\varepsilon[\/latex] or, equivalently, [latex]0<x-4<\\varepsilon^2[\/latex], making [latex]\\delta =\\varepsilon^2[\/latex] a clear choice. We may also determine [latex]\\delta[\/latex] geometrically, as shown in Figure 5.<\/p>\n<div style=\"width: 469px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" id=\"20\" src=\"https:\/\/openstax.org\/resources\/53207e09378fd9a465f9a0ae4026f8432751441f\" alt=\"A graph showing how to find delta for the above proof. The function f(x) = sqrt(x-4) is drawn for x &gt; 4. Since the proposed limit is 0, lines y = 0 + epsilon and y = 0 \u2013 epsilon are drawn in blue. Since only the top blue line corresponding to y = 0 + epsilon intersects the function, one red line is drawn from the point of intersection to the x axis. This x value is found by solving sqrt(x-4) = epsilon, or x = epsilon squared + 4. Delta is then the distance between this point and 4, which is epsilon squared.\" width=\"459\" height=\"320\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. This graph shows how we find [latex]\\delta[\/latex] for the proof of this example.<\/p>\n<\/div>\n<p id=\"fs-id1170572376082\">Assume [latex]0<x-4<\\delta[\/latex]. Thus, [latex]0<x-4<\\varepsilon^2[\/latex]. Hence, [latex]0<\\sqrt{x-4}<\\varepsilon[\/latex]. Finally, [latex]|\\sqrt{x-4}-0|<\\varepsilon[\/latex].<\/p>\n<p id=\"fs-id1170571586144\">Therefore, [latex]\\underset{x\\to 4^+}{\\lim}\\sqrt{x-4}=0[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Proving a Statement about a Limit From the Right. <\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/5_q7_Zx26RY?controls=0&amp;start=1660&amp;end=1870&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\/2.5PreciseDefinitionOfaLimit1660to1870_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.5 Precise Definition of a Limit&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170571711155\" class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170571711163\">Find [latex]\\delta[\/latex] corresponding to [latex]\\varepsilon[\/latex]\u00a0for a proof that [latex]\\underset{x\\to 1^-}{\\lim}\\sqrt{1-x}=0[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q142584\">Hint<\/span><\/p>\n<div id=\"q142584\" class=\"hidden-answer\" style=\"display: none\">\n<p>&nbsp;<\/p>\n<p>Sketch the graph and use Figure 5 as a solving guide.<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q142584\">Show Solution<\/span><\/p>\n<div id=\"q142584\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\delta =\\varepsilon^2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We conclude the process of converting our intuitive ideas of various types of limits to rigorous formal definitions by pursuing a formal definition of infinite limits. To have [latex]\\underset{x\\to a}{\\lim}f(x)=+\\infty[\/latex], we want the values of the function [latex]f(x)[\/latex] to get larger and larger as [latex]x[\/latex] approaches [latex]a[\/latex]. Instead of the requirement that [latex]|f(x)-L|<\\varepsilon[\/latex] for arbitrarily small [latex]\\varepsilon[\/latex]\u00a0when [latex]0<|x-a|<\\delta[\/latex] for small enough [latex]\\delta[\/latex], we want [latex]f(x)>M[\/latex] for arbitrarily large positive [latex]M[\/latex] when [latex]0<|x-a|<\\delta[\/latex] for small enough [latex]\\delta[\/latex]. Figure 6 illustrates this idea by showing the value of [latex]\\delta[\/latex] for successively larger values of [latex]M[\/latex].\n\n\n<div id=\"CNX_Calc_Figure_02_05_005\" class=\"wp-caption aligncenter\">\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\/11203546\/CNX_Calc_Figure_02_05_005.jpg\" alt=\"Two graphs side by side. Each graph contains two curves above the x axis separated by an asymptote at x=a. The curves on the left go to infinity as x goes to a and to 0 as x goes to negative infinity. The curves on the right go to infinity as x goes to a and to 0 as x goes to infinity. The first graph has a value M greater than zero marked on the y axis and a horizontal line drawn from there (y=M) to intersect with both curves. Lines are drawn down from the points of intersection to the x axis. Delta is the smaller of the distances between point a and these new spots on the x axis. The same lines are drawn on the second graph, but this M is larger, and the distances from the x axis intersections to point a are smaller.\" width=\"975\" height=\"422\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. These graphs plot values of [latex]\\delta[\/latex] for [latex]M[\/latex] to show that [latex]\\underset{x\\to a}{\\lim}f(x)=+\\infty[\/latex].<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571609486\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1170571609489\">Let [latex]f(x)[\/latex] be defined for all [latex]x\\ne a[\/latex] in an open interval containing [latex]a[\/latex]. Then, we have an infinite limit<\/p>\n<div id=\"fs-id1170571635969\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}f(x)=+\\infty[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1170571636004\">if for every [latex]M>0[\/latex], there exists [latex]\\delta >0[\/latex] such that if [latex]0<|x-a|<\\delta[\/latex], then [latex]f(x)>M[\/latex].<\/p>\n<p id=\"fs-id1170572216496\">Let [latex]f(x)[\/latex] be defined for all [latex]x\\ne a[\/latex] in an open interval containing [latex]a[\/latex]. Then, we have a negative infinite limit<\/p>\n<div id=\"fs-id1170571600638\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}f(x)=\u2212\\infty[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1170571600673\">if for every [latex]M>0[\/latex], there exists [latex]\\delta >0[\/latex] such that if [latex]0<|x-a|<\\delta[\/latex], then [latex]f(x)<\u2212M[\/latex].<\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-304\">\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>2.5 Precise Definition of a Limit. <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":22,"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\":\"2.5 Precise Definition of a Limit\",\"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-304","chapter","type-chapter","status-publish","hentry"],"part":28,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/304","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":27,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/304\/revisions"}],"predecessor-version":[{"id":4799,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/304\/revisions\/4799"}],"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\/304\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=304"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=304"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=304"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=304"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}