{"id":287,"date":"2021-02-04T01:00:23","date_gmt":"2021-02-04T01:00:23","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=287"},"modified":"2022-03-11T21:56:45","modified_gmt":"2022-03-11T21:56:45","slug":"continuity-at-a-point","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/continuity-at-a-point\/","title":{"raw":"Continuity at a Point","rendered":"Continuity at a Point"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Explain the three conditions for continuity at a point<\/li>\r\n \t<li>Describe three kinds of discontinuities<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1170573262889\" class=\"bc-section section\">\r\n\r\nWe begin our investigation of continuity by exploring what it means for a function to have\u00a0<em data-effect=\"italics\">continuity at a point<\/em>. Intuitively, a function is continuous at a particular point if there is no break in its graph at that point.\r\n<p id=\"fs-id1170573324888\">Before we look at a formal definition of what it means for a function to be continuous at a point, let\u2019s consider various functions that fail to meet our intuitive notion of what it means to be continuous at a point. We then create a list of conditions that prevent such failures.<\/p>\r\n<p id=\"fs-id1170573390124\">Our first function of interest is shown in Figure 1. We see that the graph of [latex]f(x)[\/latex] has a hole at [latex]a[\/latex]. In fact, [latex]f(a)[\/latex] is undefined. At the very least, for [latex]f(x)[\/latex] to be continuous at [latex]a[\/latex], we need the following conditions:<\/p>\r\n\r\n<div id=\"fs-id1170570997076\" class=\"equation unnumbered\" style=\"text-align: center;\">i. [latex]f(a)[\/latex] is defined.<\/div>\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203502\/CNX_Calc_Figure_02_04_001.jpg\" alt=\"A graph of an increasing linear function f(x) which crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. A point a greater than zero is marked on the x axis. The point on the function f(x) above a is an open circle; the function is not defined at a.\" width=\"325\" height=\"277\" \/> Figure 1. The function [latex]f(x)[\/latex] is not continuous at a because [latex]f(a)[\/latex] is undefined.[\/caption]<\/div>\r\n<p id=\"fs-id1170573750564\">However, as we see in Figure 2, this condition alone is insufficient to guarantee continuity at the point [latex]a[\/latex]. Although [latex]f(a)[\/latex] is defined, the function has a gap at [latex]a[\/latex]. In this example, the gap exists because [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] does not exist. We must add another condition for continuity at [latex]a[\/latex]\u2014namely,<\/p>\r\n\r\n<div id=\"fs-id1170570997548\" class=\"equation unnumbered\" style=\"text-align: center;\">ii. [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] exists.<\/div>\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203504\/CNX_Calc_Figure_02_04_002.jpg\" alt=\"The graph of a piecewise function f(x) with two parts. The first part is an increasing linear function that crosses from quadrant three to quadrant one at the origin. A point a greater than zero is marked on the x axis. At fa. on this segment, there is a solid circle. The other segment is also an increasing linear function. It exists in quadrant one for values of x greater than a. At x=a, this segment has an open circle.\" width=\"325\" height=\"277\" \/> Figure 2. The function [latex]f(x)[\/latex] is not continuous at a because [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] does not exist.[\/caption]<\/div>\r\n<p id=\"fs-id1170573493837\">However, as we see in Figure 3, these two conditions by themselves do not guarantee continuity at a point. The function in this figure satisfies both of our first two conditions, but is still not continuous at [latex]a[\/latex]. We must add a third condition to our list:<\/p>\r\n\r\n<div id=\"fs-id1170573419105\" class=\"equation unnumbered\" style=\"text-align: center;\">iii. [latex]\\underset{x\\to a}{\\lim}f(x)=f(a)[\/latex].<\/div>\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203506\/CNX_Calc_Figure_02_04_003.jpg\" alt=\"The graph of a piecewise function with two parts. The first part is an increasing linear function that crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. A point a greater than zero is marked on the x axis. At this point, there is an open circle on the linear function. The second part is a point at x=a above the line.\" width=\"325\" height=\"277\" \/> Figure 3. The function [latex]f(x)[\/latex] is not continuous at a because [latex]\\underset{x\\to a}{\\lim}f(x)\\ne f(a)[\/latex].[\/caption]<\/div>\r\n<p id=\"fs-id1170573502301\">Now we put our list of conditions together and form a definition of continuity at a point.<\/p>\r\n\r\n<div id=\"fs-id1170571048312\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170573381317\">A function [latex]f(x)[\/latex] is <strong>continuous at a point<\/strong> [latex]a[\/latex] if and only if the following three conditions are satisfied:<\/p>\r\n\r\n<ol id=\"fs-id1170571094915\">\r\n \t<li>[latex]f(a)[\/latex] is defined<\/li>\r\n \t<li>[latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] exists<\/li>\r\n \t<li>[latex]\\underset{x\\to a}{\\lim}f(x)=f(a)[\/latex]<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1170573370662\">A function is <strong>discontinuous at a point<\/strong> [latex]a[\/latex] if it fails to be continuous at [latex]a[\/latex].<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1170573281579\">The following procedure can be used to analyze the continuity of a function at a point using this definition.<\/p>\r\n\r\n<div id=\"fs-id1170573398041\" class=\"textbox examples\">\r\n<h3>Problem-Solving Strategy: Determining Continuity at a Point<\/h3>\r\n<ol id=\"fs-id1170571103215\">\r\n \t<li>Check to see if [latex]f(a)[\/latex] is defined. If [latex]f(a)[\/latex] is undefined, we need go no further. The function is not continuous at [latex]a[\/latex]. If [latex]f(a)[\/latex] is defined, continue to step 2.<\/li>\r\n \t<li>Compute [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex]. In some cases, we may need to do this by first computing [latex]\\underset{x\\to a^-}{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to a^+}{\\lim}f(x)[\/latex]. If [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] does not exist (that is, it is not a real number), then the function is not continuous at [latex]a[\/latex] and the problem is solved. If [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] exists, then continue to step 3.<\/li>\r\n \t<li>Compare [latex]f(a)[\/latex] and [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex]. If [latex]\\underset{x\\to a}{\\lim}f(x)\\ne f(a)[\/latex], then the function is not continuous at [latex]a[\/latex]. If [latex]\\underset{x\\to a}{\\lim}f(x)=f(a)[\/latex], then the function is continuous at [latex]a[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<p id=\"fs-id1170573570770\">The next three examples demonstrate how to apply this definition to determine whether a function is continuous at a given point. These examples illustrate situations in which each of the conditions for continuity in the definition succeed or fail.<\/p>\r\n\r\n<div id=\"fs-id1170573442080\" class=\"textbook exercises\">\r\n<h3>Example: Determining Continuity at a Point, Condition 1<\/h3>\r\n<p id=\"fs-id1170573258704\">Using the definition, determine whether the function [latex]f(x)=\\dfrac{(x^2-4)}{(x-2)}[\/latex] is continuous at [latex]x=2[\/latex]. Justify the conclusion.<\/p>\r\n[reveal-answer q=\"fs-id1170573331408\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573331408\"]\r\n<p id=\"fs-id1170573331408\">Let\u2019s begin by trying to calculate [latex]f(2)[\/latex]. We can see that [latex]f(2)=\\frac{0}{0}[\/latex], which is undefined. Therefore, [latex]f(x)=\\frac{x^2-4}{x-2}[\/latex] is discontinuous at 2 because [latex]f(2)[\/latex] is undefined. The graph of [latex]f(x)[\/latex] is shown in Figure 4.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203508\/CNX_Calc_Figure_02_04_004.jpg\" alt=\"A graph of the given function. There is a line which crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. At a point in quadrant one, there is an open circle where the function is not defined.\" width=\"487\" height=\"425\" \/> Figure 4. The function [latex]f(x)[\/latex] is discontinuous at 2 because [latex]f(2)[\/latex] is undefined.[\/caption][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573389760\" class=\"textbook exercises\">\r\n<h3>Example: Determining Continuity at a Point, Condition 2<\/h3>\r\n<p id=\"fs-id1170573370721\">Using the definition, determine whether the function [latex]f(x)=\\begin{cases} -x^2+4 &amp; \\text{ if } \\, x \\le 3 \\\\ 4x-8 &amp; \\text{ if } \\, x &gt; 3 \\end{cases}[\/latex] is continuous at [latex]x=3[\/latex]. Justify the conclusion.<\/p>\r\n[reveal-answer q=\"fs-id1170573349668\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573349668\"]\r\n<p id=\"fs-id1170573349668\">Let\u2019s begin by trying to calculate [latex]f(3)[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1170573405058\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(3)=-(3^2)+4=-5[\/latex]<\/div>\r\n<p id=\"fs-id1170573307347\">Thus, [latex]f(3)[\/latex] is defined. Next, we calculate [latex]\\underset{x\\to 3}{\\lim}f(x)[\/latex]. To do this, we must compute [latex]\\underset{x\\to 3^-}{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to 3^+}{\\lim}f(x)[\/latex]:<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 3^-}{\\lim}f(x)=-(3^2)+4=-5[\/latex]<\/div>\r\n<p id=\"fs-id1170573400834\">and<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 3^+}{\\lim}f(x)=4(3)-8=4[\/latex]<\/div>\r\n<p id=\"fs-id1170573414763\">Therefore, [latex]\\underset{x\\to 3}{\\lim}f(x)[\/latex] does not exist. Thus, [latex]f(x)[\/latex] is not continuous at 3. The graph of [latex]f(x)[\/latex] is shown in Figure 5.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203511\/CNX_Calc_Figure_02_04_005.jpg\" alt=\"A graph of the given piecewise function, which has two parts. The first is a downward opening parabola which is symmetric about the y axis. Its vertex is on the y axis, greater than zero. There is a closed circle on the parabola for x=3. The second part is an increasing linear function in the first quadrant, which exists for values of x &gt; 3. There is an open circle at the end of the line where x would be 3.\" width=\"487\" height=\"575\" \/> Figure 5 The function \ud835\udc53(\ud835\udc65) is not continuous at 3 because lim\ud835\udc65\u21923\ud835\udc53(\ud835\udc65) does not exist.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170573429945\" class=\"textbook exercises\">\r\n<h3>Example: Determining Continuity at a Point, Condition 3<\/h3>\r\n<p id=\"fs-id1170573502415\">Using the definition, determine whether the function [latex]f(x)=\\begin{cases} \\frac{\\sin x}{x} &amp; \\text{ if } \\, x \\ne 0 \\\\ 1 &amp; \\text{ if } \\, x = 0 \\end{cases}[\/latex] is continuous at [latex]x=0[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170573382461\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573382461\"]\r\n<p id=\"fs-id1170573382461\">First, observe that<\/p>\r\n\r\n<div id=\"fs-id1170573569453\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(0)=1[\/latex]<\/div>\r\n<p id=\"fs-id1170573367184\">Next,<\/p>\r\n\r\n<div id=\"fs-id1170573442773\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}f(x)=\\underset{x\\to 0}{\\lim}\\frac{\\sin x}{x}=1[\/latex]<\/div>\r\n<p id=\"fs-id1170573394612\">Last, compare [latex]f(0)[\/latex] and [latex]\\underset{x\\to 1}{\\lim}f(x)[\/latex]. We see that<\/p>\r\n\r\n<div id=\"fs-id1170573332534\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(0)=1=\\underset{x\\to 0}{\\lim}f(x)[\/latex]<\/div>\r\n<p id=\"fs-id1170573394599\">Since all three of the conditions in the definition of continuity are satisfied, [latex]f(x)[\/latex] is continuous at [latex]x=0[\/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 three Example: Determining Continuity at a Point conditions. [\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/BXUu5bG1CXU?controls=0&amp;start=182&amp;end=412&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\/2.4Continuity182to412_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.4 Continuity\" here (opens in new window).<\/a>[\/hidden-answer]\r\n<div id=\"fs-id1170570976463\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170573406319\">Using the definition, determine whether the function [latex]f(x)=\\begin{cases} 2x+1 &amp; \\text{ if } \\, x &lt; 1 \\\\ 2 &amp; \\text{ if } \\, x = 1 \\\\ -x+4 &amp; \\text{ if } \\, x &gt; 1 \\end{cases}[\/latex] is continuous at [latex]x=1[\/latex]. If the function is not continuous at 1, indicate the condition for continuity at a point that fails to hold.<\/p>\r\n[reveal-answer q=\"98945562\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"98945562\"]\r\n<p id=\"fs-id1170573418814\">Check each condition of the definition.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170573402451\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573402451\"]\r\n<p id=\"fs-id1170573402451\">[latex]f[\/latex] is not continuous at 1 because [latex]f(1)=2\\ne 3=\\underset{x\\to 1}{\\lim}f(x)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]20431[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1170573513079\">By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we can state the following theorem.<\/p>\r\n\r\n<div id=\"fs-id1170573430304\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Continuity of Polynomials and Rational Functions<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170573255225\">Polynomials and rational functions are continuous at every point in their domains.<\/p>\r\n\r\n<\/div>\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1170573326736\">Previously, we showed that if [latex]p(x)[\/latex] and [latex]q(x)[\/latex] are polynomials, [latex]\\underset{x\\to a}{\\lim}p(x)=p(a)[\/latex] for every polynomial [latex]p(x)[\/latex] and [latex]\\underset{x\\to a}{\\lim}\\dfrac{p(x)}{q(x)}=\\dfrac{p(a)}{q(a)}[\/latex] as long as [latex]q(a)\\ne 0[\/latex]. Therefore, polynomials and rational functions are continuous on their domains.<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Domain of polynomial and rational functions<\/h3>\r\nThe domain of every polynomial function is <strong>all real numbers.\u00a0<\/strong>\r\n\r\nThe domain of a rational functional can be found by:\r\n<ol>\r\n \t<li>Set the denominator equal to zero.<\/li>\r\n \t<li>Solve to find the values of the variable that cause the denominator to equal zero.<\/li>\r\n \t<li>The domain contains all real numbers except those found in Step 2.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<p id=\"fs-id1170573335424\">We now apply continuity of polynomials and rational functions to determine the points at which a given rational function is continuous.<\/p>\r\n\r\n<div id=\"fs-id1170573381194\" class=\"textbook exercises\">\r\n<h3>Example: Continuity of a Rational Function<\/h3>\r\n<p id=\"fs-id1170573365757\">For what values of [latex]x[\/latex] is [latex]f(x)=\\dfrac{x+1}{x-5}[\/latex] continuous?<\/p>\r\n[reveal-answer q=\"fs-id1170573397031\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573397031\"]\r\n<p id=\"fs-id1170573397031\">The rational function [latex]f(x)=\\frac{x+1}{x-5}[\/latex] is continuous for every value of [latex]x[\/latex] except [latex]x=5[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170573434500\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170570999873\">For what values of [latex]x[\/latex] is [latex]f(x)=3x^4-4x^2[\/latex] continuous?<\/p>\r\n[reveal-answer q=\"3768441\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3768441\"]\r\n<p id=\"fs-id1170573431750\">Use continuity of polynomials and rational functions<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170573246209\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573246209\"]\r\n<p id=\"fs-id1170573246209\">[latex]f(x)[\/latex] is continuous at every real number.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Types of Discontinuities<\/h2>\r\n<p id=\"fs-id1170573435662\">As we have seen in the earlier examples, discontinuities take on several different appearances. We classify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or jump discontinuities. Intuitively, a<strong> removable discontinuity<\/strong> is a discontinuity for which there is a hole in the graph, a<strong> jump discontinuity<\/strong> is a noninfinite discontinuity for which the sections of the function do not meet up, and an <strong>infinite discontinuity<\/strong> is a discontinuity located at a vertical asymptote. Figure 6 illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep in mind that not all discontinuities fit neatly into these categories.<\/p>\r\n\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\/11203514\/CNX_Calc_Figure_02_04_006.jpg\" alt=\"&quot;Three 0, there is an open circle on the line and a closed circle a few units above the line. The second is a jump discontinuity. Here, there are two lines with positive slope. The first line exists for x&lt;=a, and the second exists for x&gt;a, where a&gt;0. The first line ends at a solid circle where x=a, and the second begins a few units up with an open circle at x=a. The third discontinuity type is infinite discontinuity. Here, the function has two parts separated by an asymptote x=a. The first segment is a curve stretching along the x axis to 0 as x goes to negative infinity and along the y axis to infinity as x goes to zero. The second segment is a curve stretching along the y axis to negative infinity as x goes to zero and along the x axis to 0 as x goes to infinity.&quot; width=&quot;975&quot; height=&quot;315&quot;\" width=\"975\" height=\"315\" \/> Figure 6. Discontinuities are classified as (a) removable, (b) jump, or (c) infinite.[\/caption]\r\n<p id=\"fs-id1170573400557\">These three discontinuities are formally defined as follows:<\/p>\r\n\r\n<div id=\"fs-id1170573419172\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nIf [latex]f(x)[\/latex] is discontinuous at [latex]a[\/latex], then\r\n<ol id=\"fs-id1170573216342\">\r\n \t<li>[latex]f[\/latex] has a <strong>removable discontinuity<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] exists. (Note: When we state that [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] exists, we mean that [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex], where [latex]L[\/latex] is a real number.)<\/li>\r\n \t<li>[latex]f[\/latex] has a <strong>jump discontinuity<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to a^-}{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to a^+}{\\lim}f(x)[\/latex] both exist, but [latex]\\underset{x\\to a^-}{\\lim}f(x)\\ne \\underset{x\\to a^+}{\\lim}f(x)[\/latex]. (Note: When we state that [latex]\\underset{x\\to a^-}{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to a^+}{\\lim}f(x)[\/latex] both exist, we mean that both are real-valued and that neither take on the values [latex]\\pm \\infty[\/latex].)<\/li>\r\n \t<li>[latex]f[\/latex] has an <strong>infinite discontinuity<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to a^-}{\\lim}f(x)=\\pm \\infty [\/latex] or [latex]\\underset{x\\to a^+}{\\lim}f(x)=\\pm \\infty[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1170570976348\" class=\"textbook exercises\">\r\n<h3>Example: Classifying a Discontinuity 1<\/h3>\r\n<p id=\"fs-id1170573424344\">In an earlier example, we showed that [latex]f(x)=\\dfrac{x^2-4}{x-2}[\/latex] is discontinuous at [latex]x=2[\/latex]. Classify this discontinuity as removable, jump, or infinite.<\/p>\r\n[reveal-answer q=\"fs-id1170573402024\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573402024\"]\r\n<p id=\"fs-id1170573402024\">To classify the discontinuity at 2 we must evaluate [latex]\\underset{x\\to 2}{\\lim}f(x)[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1170571103554\" class=\"equation unnumbered\" style=\"text-align: left;\">[latex]\\begin{array}{cc}\\underset{x\\to 2}{\\lim}f(x) &amp; =\\underset{x\\to 2}{\\lim}\\frac{x^2-4}{x-2} \\\\ &amp; =\\underset{x\\to 2}{\\lim}\\frac{(x-2)(x+2)}{x-2} \\\\ &amp; =\\underset{x\\to 2}{\\lim}(x+2) \\\\ &amp; = 4 \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571136107\">Since [latex]f[\/latex] is discontinuous at 2 and [latex]\\underset{x\\to 2}{\\lim}f(x)[\/latex] exists, [latex]f[\/latex] has a removable discontinuity at [latex]x=2[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170573426544\" class=\"textbook exercises\">\r\n<h3>Example: Classifying a Discontinuity 2<\/h3>\r\n<p id=\"fs-id1170573586256\">In an earlier example, we showed that [latex]f(x)=\\begin{cases} -x^2+4 &amp; \\text{ if } \\, x \\le 3 \\\\ 4x-8 &amp; \\text{ if } \\, x &gt; 3 \\end{cases}[\/latex] is discontinuous at [latex]x=3[\/latex]. Classify this discontinuity as removable, jump, or infinite.<\/p>\r\n[reveal-answer q=\"fs-id1170571095481\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571095481\"]\r\n<p id=\"fs-id1170571095481\">Earlier, we showed that [latex]f[\/latex] is discontinuous at 3 because [latex]\\underset{x\\to 3}{\\lim}f(x)[\/latex] does not exist. However, since [latex]\\underset{x\\to 3^-}{\\lim}f(x)=-5[\/latex] and [latex]\\underset{x\\to 3^+}{\\lim}f(x)=4[\/latex] both exist, we conclude that the function has a jump discontinuity at 3.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170573355402\" class=\"textbook exercises\">\r\n<h3>Example: Classifying a Discontinuity 3<\/h3>\r\n<p id=\"fs-id1170573732461\">Determine whether [latex]f(x)=\\dfrac{x+2}{x+1}[\/latex] is continuous at \u22121. If the function is discontinuous at \u22121, classify the discontinuity as removable, jump, or infinite.<\/p>\r\n[reveal-answer q=\"fs-id1170571000111\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571000111\"]\r\n<p id=\"fs-id1170571000111\">The function value [latex]f(-1)[\/latex] is undefined. Therefore, the function is not continuous at \u22121. To determine the type of discontinuity, we must determine the limit at \u22121. We see that [latex]\\underset{x\\to -1^-}{\\lim}\\frac{x+2}{x+1}=\u2212\\infty [\/latex] and [latex]\\underset{x\\to -1^+}{\\lim}\\frac{x+2}{x+1}=+\\infty [\/latex]. Therefore, the function has an infinite discontinuity at \u22121.<\/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 three Example: Classifying a Discontinuity conditions. [\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/BXUu5bG1CXU?controls=0&amp;start=529&amp;end=591&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\/2.4Continuity529to591_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.4 Continuity\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170571047778\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170573361594\">For [latex]f(x)=\\begin{cases} x^2 &amp; \\text{ if } \\, x \\ne 1 \\\\ 3 &amp; \\text{ if } \\, x = 1 \\end{cases}[\/latex], decide whether [latex]f[\/latex] is continuous at 1. If [latex]f[\/latex] is not continuous at 1, classify the discontinuity as removable, jump, or infinite.<\/p>\r\n[reveal-answer q=\"3388654\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3388654\"]\r\n<p id=\"fs-id1170573436674\">Follow the steps in the Problem-Solving Strategy: Determining Continuity at a Point. If the function is discontinuous at 1, look at [latex]\\underset{x\\to 1}{\\lim}f(x)[\/latex] and use the definition to determine the type of discontinuity.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170573355512\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573355512\"]\r\n<p id=\"fs-id1170573355512\">Discontinuous at 1; removable<\/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>Explain the three conditions for continuity at a point<\/li>\n<li>Describe three kinds of discontinuities<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1170573262889\" class=\"bc-section section\">\n<p>We begin our investigation of continuity by exploring what it means for a function to have\u00a0<em data-effect=\"italics\">continuity at a point<\/em>. Intuitively, a function is continuous at a particular point if there is no break in its graph at that point.<\/p>\n<p id=\"fs-id1170573324888\">Before we look at a formal definition of what it means for a function to be continuous at a point, let\u2019s consider various functions that fail to meet our intuitive notion of what it means to be continuous at a point. We then create a list of conditions that prevent such failures.<\/p>\n<p id=\"fs-id1170573390124\">Our first function of interest is shown in Figure 1. We see that the graph of [latex]f(x)[\/latex] has a hole at [latex]a[\/latex]. In fact, [latex]f(a)[\/latex] is undefined. At the very least, for [latex]f(x)[\/latex] to be continuous at [latex]a[\/latex], we need the following conditions:<\/p>\n<div id=\"fs-id1170570997076\" class=\"equation unnumbered\" style=\"text-align: center;\">i. [latex]f(a)[\/latex] is defined.<\/div>\n<div>\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\/11203502\/CNX_Calc_Figure_02_04_001.jpg\" alt=\"A graph of an increasing linear function f(x) which crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. A point a greater than zero is marked on the x axis. The point on the function f(x) above a is an open circle; the function is not defined at a.\" width=\"325\" height=\"277\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The function [latex]f(x)[\/latex] is not continuous at a because [latex]f(a)[\/latex] is undefined.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573750564\">However, as we see in Figure 2, this condition alone is insufficient to guarantee continuity at the point [latex]a[\/latex]. Although [latex]f(a)[\/latex] is defined, the function has a gap at [latex]a[\/latex]. In this example, the gap exists because [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] does not exist. We must add another condition for continuity at [latex]a[\/latex]\u2014namely,<\/p>\n<div id=\"fs-id1170570997548\" class=\"equation unnumbered\" style=\"text-align: center;\">ii. [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] exists.<\/div>\n<div>\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\/11203504\/CNX_Calc_Figure_02_04_002.jpg\" alt=\"The graph of a piecewise function f(x) with two parts. The first part is an increasing linear function that crosses from quadrant three to quadrant one at the origin. A point a greater than zero is marked on the x axis. At fa. on this segment, there is a solid circle. The other segment is also an increasing linear function. It exists in quadrant one for values of x greater than a. At x=a, this segment has an open circle.\" width=\"325\" height=\"277\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The function [latex]f(x)[\/latex] is not continuous at a because [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] does not exist.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573493837\">However, as we see in Figure 3, these two conditions by themselves do not guarantee continuity at a point. The function in this figure satisfies both of our first two conditions, but is still not continuous at [latex]a[\/latex]. We must add a third condition to our list:<\/p>\n<div id=\"fs-id1170573419105\" class=\"equation unnumbered\" style=\"text-align: center;\">iii. [latex]\\underset{x\\to a}{\\lim}f(x)=f(a)[\/latex].<\/div>\n<div>\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\/11203506\/CNX_Calc_Figure_02_04_003.jpg\" alt=\"The graph of a piecewise function with two parts. The first part is an increasing linear function that crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. A point a greater than zero is marked on the x axis. At this point, there is an open circle on the linear function. The second part is a point at x=a above the line.\" width=\"325\" height=\"277\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. The function [latex]f(x)[\/latex] is not continuous at a because [latex]\\underset{x\\to a}{\\lim}f(x)\\ne f(a)[\/latex].<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573502301\">Now we put our list of conditions together and form a definition of continuity at a point.<\/p>\n<div id=\"fs-id1170571048312\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1170573381317\">A function [latex]f(x)[\/latex] is <strong>continuous at a point<\/strong> [latex]a[\/latex] if and only if the following three conditions are satisfied:<\/p>\n<ol id=\"fs-id1170571094915\">\n<li>[latex]f(a)[\/latex] is defined<\/li>\n<li>[latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] exists<\/li>\n<li>[latex]\\underset{x\\to a}{\\lim}f(x)=f(a)[\/latex]<\/li>\n<\/ol>\n<p id=\"fs-id1170573370662\">A function is <strong>discontinuous at a point<\/strong> [latex]a[\/latex] if it fails to be continuous at [latex]a[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1170573281579\">The following procedure can be used to analyze the continuity of a function at a point using this definition.<\/p>\n<div id=\"fs-id1170573398041\" class=\"textbox examples\">\n<h3>Problem-Solving Strategy: Determining Continuity at a Point<\/h3>\n<ol id=\"fs-id1170571103215\">\n<li>Check to see if [latex]f(a)[\/latex] is defined. If [latex]f(a)[\/latex] is undefined, we need go no further. The function is not continuous at [latex]a[\/latex]. If [latex]f(a)[\/latex] is defined, continue to step 2.<\/li>\n<li>Compute [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex]. In some cases, we may need to do this by first computing [latex]\\underset{x\\to a^-}{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to a^+}{\\lim}f(x)[\/latex]. If [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] does not exist (that is, it is not a real number), then the function is not continuous at [latex]a[\/latex] and the problem is solved. If [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] exists, then continue to step 3.<\/li>\n<li>Compare [latex]f(a)[\/latex] and [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex]. If [latex]\\underset{x\\to a}{\\lim}f(x)\\ne f(a)[\/latex], then the function is not continuous at [latex]a[\/latex]. If [latex]\\underset{x\\to a}{\\lim}f(x)=f(a)[\/latex], then the function is continuous at [latex]a[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p id=\"fs-id1170573570770\">The next three examples demonstrate how to apply this definition to determine whether a function is continuous at a given point. These examples illustrate situations in which each of the conditions for continuity in the definition succeed or fail.<\/p>\n<div id=\"fs-id1170573442080\" class=\"textbook exercises\">\n<h3>Example: Determining Continuity at a Point, Condition 1<\/h3>\n<p id=\"fs-id1170573258704\">Using the definition, determine whether the function [latex]f(x)=\\dfrac{(x^2-4)}{(x-2)}[\/latex] is continuous at [latex]x=2[\/latex]. Justify the conclusion.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573331408\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573331408\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573331408\">Let\u2019s begin by trying to calculate [latex]f(2)[\/latex]. We can see that [latex]f(2)=\\frac{0}{0}[\/latex], which is undefined. Therefore, [latex]f(x)=\\frac{x^2-4}{x-2}[\/latex] is discontinuous at 2 because [latex]f(2)[\/latex] is undefined. The graph of [latex]f(x)[\/latex] is shown in Figure 4.<\/p>\n<div style=\"width: 497px\" 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\/11203508\/CNX_Calc_Figure_02_04_004.jpg\" alt=\"A graph of the given function. There is a line which crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. At a point in quadrant one, there is an open circle where the function is not defined.\" width=\"487\" height=\"425\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. The function [latex]f(x)[\/latex] is discontinuous at 2 because [latex]f(2)[\/latex] is undefined.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573389760\" class=\"textbook exercises\">\n<h3>Example: Determining Continuity at a Point, Condition 2<\/h3>\n<p id=\"fs-id1170573370721\">Using the definition, determine whether the function [latex]f(x)=\\begin{cases} -x^2+4 & \\text{ if } \\, x \\le 3 \\\\ 4x-8 & \\text{ if } \\, x > 3 \\end{cases}[\/latex] is continuous at [latex]x=3[\/latex]. Justify the conclusion.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573349668\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573349668\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573349668\">Let\u2019s begin by trying to calculate [latex]f(3)[\/latex].<\/p>\n<div id=\"fs-id1170573405058\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(3)=-(3^2)+4=-5[\/latex]<\/div>\n<p id=\"fs-id1170573307347\">Thus, [latex]f(3)[\/latex] is defined. Next, we calculate [latex]\\underset{x\\to 3}{\\lim}f(x)[\/latex]. To do this, we must compute [latex]\\underset{x\\to 3^-}{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to 3^+}{\\lim}f(x)[\/latex]:<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 3^-}{\\lim}f(x)=-(3^2)+4=-5[\/latex]<\/div>\n<p id=\"fs-id1170573400834\">and<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 3^+}{\\lim}f(x)=4(3)-8=4[\/latex]<\/div>\n<p id=\"fs-id1170573414763\">Therefore, [latex]\\underset{x\\to 3}{\\lim}f(x)[\/latex] does not exist. Thus, [latex]f(x)[\/latex] is not continuous at 3. The graph of [latex]f(x)[\/latex] is shown in Figure 5.<\/p>\n<div style=\"width: 497px\" 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\/11203511\/CNX_Calc_Figure_02_04_005.jpg\" alt=\"A graph of the given piecewise function, which has two parts. The first is a downward opening parabola which is symmetric about the y axis. Its vertex is on the y axis, greater than zero. There is a closed circle on the parabola for x=3. The second part is an increasing linear function in the first quadrant, which exists for values of x &gt; 3. There is an open circle at the end of the line where x would be 3.\" width=\"487\" height=\"575\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5 The function \ud835\udc53(\ud835\udc65) is not continuous at 3 because lim\ud835\udc65\u21923\ud835\udc53(\ud835\udc65) does not exist.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573429945\" class=\"textbook exercises\">\n<h3>Example: Determining Continuity at a Point, Condition 3<\/h3>\n<p id=\"fs-id1170573502415\">Using the definition, determine whether the function [latex]f(x)=\\begin{cases} \\frac{\\sin x}{x} & \\text{ if } \\, x \\ne 0 \\\\ 1 & \\text{ if } \\, x = 0 \\end{cases}[\/latex] is continuous at [latex]x=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573382461\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573382461\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573382461\">First, observe that<\/p>\n<div id=\"fs-id1170573569453\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(0)=1[\/latex]<\/div>\n<p id=\"fs-id1170573367184\">Next,<\/p>\n<div id=\"fs-id1170573442773\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0}{\\lim}f(x)=\\underset{x\\to 0}{\\lim}\\frac{\\sin x}{x}=1[\/latex]<\/div>\n<p id=\"fs-id1170573394612\">Last, compare [latex]f(0)[\/latex] and [latex]\\underset{x\\to 1}{\\lim}f(x)[\/latex]. We see that<\/p>\n<div id=\"fs-id1170573332534\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(0)=1=\\underset{x\\to 0}{\\lim}f(x)[\/latex]<\/div>\n<p id=\"fs-id1170573394599\">Since all three of the conditions in the definition of continuity are satisfied, [latex]f(x)[\/latex] is continuous at [latex]x=0[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the three Example: Determining Continuity at a Point conditions. <\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/BXUu5bG1CXU?controls=0&amp;start=182&amp;end=412&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\/2.4Continuity182to412_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.4 Continuity&#8221; here (opens in new window).<\/a><\/div>\n<\/div>\n<div id=\"fs-id1170570976463\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170573406319\">Using the definition, determine whether the function [latex]f(x)=\\begin{cases} 2x+1 & \\text{ if } \\, x < 1 \\\\ 2 & \\text{ if } \\, x = 1 \\\\ -x+4 & \\text{ if } \\, x > 1 \\end{cases}[\/latex] is continuous at [latex]x=1[\/latex]. If the function is not continuous at 1, indicate the condition for continuity at a point that fails to hold.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q98945562\">Hint<\/span><\/p>\n<div id=\"q98945562\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573418814\">Check each condition of the definition.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573402451\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573402451\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573402451\">[latex]f[\/latex] is not continuous at 1 because [latex]f(1)=2\\ne 3=\\underset{x\\to 1}{\\lim}f(x)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm20431\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20431&theme=oea&iframe_resize_id=ohm20431&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1170573513079\">By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we can state the following theorem.<\/p>\n<div id=\"fs-id1170573430304\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Continuity of Polynomials and Rational Functions<\/h3>\n<hr \/>\n<p id=\"fs-id1170573255225\">Polynomials and rational functions are continuous at every point in their domains.<\/p>\n<\/div>\n<h3>Proof<\/h3>\n<p id=\"fs-id1170573326736\">Previously, we showed that if [latex]p(x)[\/latex] and [latex]q(x)[\/latex] are polynomials, [latex]\\underset{x\\to a}{\\lim}p(x)=p(a)[\/latex] for every polynomial [latex]p(x)[\/latex] and [latex]\\underset{x\\to a}{\\lim}\\dfrac{p(x)}{q(x)}=\\dfrac{p(a)}{q(a)}[\/latex] as long as [latex]q(a)\\ne 0[\/latex]. Therefore, polynomials and rational functions are continuous on their domains.<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Domain of polynomial and rational functions<\/h3>\n<p>The domain of every polynomial function is <strong>all real numbers.\u00a0<\/strong><\/p>\n<p>The domain of a rational functional can be found by:<\/p>\n<ol>\n<li>Set the denominator equal to zero.<\/li>\n<li>Solve to find the values of the variable that cause the denominator to equal zero.<\/li>\n<li>The domain contains all real numbers except those found in Step 2.<\/li>\n<\/ol>\n<\/div>\n<p id=\"fs-id1170573335424\">We now apply continuity of polynomials and rational functions to determine the points at which a given rational function is continuous.<\/p>\n<div id=\"fs-id1170573381194\" class=\"textbook exercises\">\n<h3>Example: Continuity of a Rational Function<\/h3>\n<p id=\"fs-id1170573365757\">For what values of [latex]x[\/latex] is [latex]f(x)=\\dfrac{x+1}{x-5}[\/latex] continuous?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573397031\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573397031\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573397031\">The rational function [latex]f(x)=\\frac{x+1}{x-5}[\/latex] is continuous for every value of [latex]x[\/latex] except [latex]x=5[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573434500\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170570999873\">For what values of [latex]x[\/latex] is [latex]f(x)=3x^4-4x^2[\/latex] continuous?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3768441\">Hint<\/span><\/p>\n<div id=\"q3768441\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573431750\">Use continuity of polynomials and rational functions<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573246209\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573246209\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573246209\">[latex]f(x)[\/latex] is continuous at every real number.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Types of Discontinuities<\/h2>\n<p id=\"fs-id1170573435662\">As we have seen in the earlier examples, discontinuities take on several different appearances. We classify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or jump discontinuities. Intuitively, a<strong> removable discontinuity<\/strong> is a discontinuity for which there is a hole in the graph, a<strong> jump discontinuity<\/strong> is a noninfinite discontinuity for which the sections of the function do not meet up, and an <strong>infinite discontinuity<\/strong> is a discontinuity located at a vertical asymptote. Figure 6 illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep in mind that not all discontinuities fit neatly into these categories.<\/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\/11203514\/CNX_Calc_Figure_02_04_006.jpg\" alt=\"&quot;Three 0, there is an open circle on the line and a closed circle a few units above the line. The second is a jump discontinuity. Here, there are two lines with positive slope. The first line exists for x&lt;=a, and the second exists for x&gt;a, where a&gt;0. The first line ends at a solid circle where x=a, and the second begins a few units up with an open circle at x=a. The third discontinuity type is infinite discontinuity. Here, the function has two parts separated by an asymptote x=a. The first segment is a curve stretching along the x axis to 0 as x goes to negative infinity and along the y axis to infinity as x goes to zero. The second segment is a curve stretching along the y axis to negative infinity as x goes to zero and along the x axis to 0 as x goes to infinity.&quot; width=&quot;975&quot; height=&quot;315&quot;\" width=\"975\" height=\"315\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. Discontinuities are classified as (a) removable, (b) jump, or (c) infinite.<\/p>\n<\/div>\n<p id=\"fs-id1170573400557\">These three discontinuities are formally defined as follows:<\/p>\n<div id=\"fs-id1170573419172\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p>If [latex]f(x)[\/latex] is discontinuous at [latex]a[\/latex], then<\/p>\n<ol id=\"fs-id1170573216342\">\n<li>[latex]f[\/latex] has a <strong>removable discontinuity<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] exists. (Note: When we state that [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] exists, we mean that [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex], where [latex]L[\/latex] is a real number.)<\/li>\n<li>[latex]f[\/latex] has a <strong>jump discontinuity<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to a^-}{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to a^+}{\\lim}f(x)[\/latex] both exist, but [latex]\\underset{x\\to a^-}{\\lim}f(x)\\ne \\underset{x\\to a^+}{\\lim}f(x)[\/latex]. (Note: When we state that [latex]\\underset{x\\to a^-}{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to a^+}{\\lim}f(x)[\/latex] both exist, we mean that both are real-valued and that neither take on the values [latex]\\pm \\infty[\/latex].)<\/li>\n<li>[latex]f[\/latex] has an <strong>infinite discontinuity<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to a^-}{\\lim}f(x)=\\pm \\infty[\/latex] or [latex]\\underset{x\\to a^+}{\\lim}f(x)=\\pm \\infty[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1170570976348\" class=\"textbook exercises\">\n<h3>Example: Classifying a Discontinuity 1<\/h3>\n<p id=\"fs-id1170573424344\">In an earlier example, we showed that [latex]f(x)=\\dfrac{x^2-4}{x-2}[\/latex] is discontinuous at [latex]x=2[\/latex]. Classify this discontinuity as removable, jump, or infinite.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573402024\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573402024\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573402024\">To classify the discontinuity at 2 we must evaluate [latex]\\underset{x\\to 2}{\\lim}f(x)[\/latex]:<\/p>\n<div id=\"fs-id1170571103554\" class=\"equation unnumbered\" style=\"text-align: left;\">[latex]\\begin{array}{cc}\\underset{x\\to 2}{\\lim}f(x) & =\\underset{x\\to 2}{\\lim}\\frac{x^2-4}{x-2} \\\\ & =\\underset{x\\to 2}{\\lim}\\frac{(x-2)(x+2)}{x-2} \\\\ & =\\underset{x\\to 2}{\\lim}(x+2) \\\\ & = 4 \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571136107\">Since [latex]f[\/latex] is discontinuous at 2 and [latex]\\underset{x\\to 2}{\\lim}f(x)[\/latex] exists, [latex]f[\/latex] has a removable discontinuity at [latex]x=2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573426544\" class=\"textbook exercises\">\n<h3>Example: Classifying a Discontinuity 2<\/h3>\n<p id=\"fs-id1170573586256\">In an earlier example, we showed that [latex]f(x)=\\begin{cases} -x^2+4 & \\text{ if } \\, x \\le 3 \\\\ 4x-8 & \\text{ if } \\, x > 3 \\end{cases}[\/latex] is discontinuous at [latex]x=3[\/latex]. Classify this discontinuity as removable, jump, or infinite.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571095481\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571095481\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571095481\">Earlier, we showed that [latex]f[\/latex] is discontinuous at 3 because [latex]\\underset{x\\to 3}{\\lim}f(x)[\/latex] does not exist. However, since [latex]\\underset{x\\to 3^-}{\\lim}f(x)=-5[\/latex] and [latex]\\underset{x\\to 3^+}{\\lim}f(x)=4[\/latex] both exist, we conclude that the function has a jump discontinuity at 3.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573355402\" class=\"textbook exercises\">\n<h3>Example: Classifying a Discontinuity 3<\/h3>\n<p id=\"fs-id1170573732461\">Determine whether [latex]f(x)=\\dfrac{x+2}{x+1}[\/latex] is continuous at \u22121. If the function is discontinuous at \u22121, classify the discontinuity as removable, jump, or infinite.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571000111\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571000111\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571000111\">The function value [latex]f(-1)[\/latex] is undefined. Therefore, the function is not continuous at \u22121. To determine the type of discontinuity, we must determine the limit at \u22121. We see that [latex]\\underset{x\\to -1^-}{\\lim}\\frac{x+2}{x+1}=\u2212\\infty[\/latex] and [latex]\\underset{x\\to -1^+}{\\lim}\\frac{x+2}{x+1}=+\\infty[\/latex]. Therefore, the function has an infinite discontinuity at \u22121.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the three Example: Classifying a Discontinuity conditions. <\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/BXUu5bG1CXU?controls=0&amp;start=529&amp;end=591&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\/2.4Continuity529to591_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.4 Continuity&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170571047778\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170573361594\">For [latex]f(x)=\\begin{cases} x^2 & \\text{ if } \\, x \\ne 1 \\\\ 3 & \\text{ if } \\, x = 1 \\end{cases}[\/latex], decide whether [latex]f[\/latex] is continuous at 1. If [latex]f[\/latex] is not continuous at 1, classify the discontinuity as removable, jump, or infinite.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3388654\">Hint<\/span><\/p>\n<div id=\"q3388654\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573436674\">Follow the steps in the Problem-Solving Strategy: Determining Continuity at a Point. If the function is discontinuous at 1, look at [latex]\\underset{x\\to 1}{\\lim}f(x)[\/latex] and use the definition to determine the type of discontinuity.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573355512\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573355512\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573355512\">Discontinuous at 1; removable<\/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-287\">\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>2.4 Continuity. <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":17,"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\":\"2.4 Continuity\",\"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-287","chapter","type-chapter","status-publish","hentry"],"part":28,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/287","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":29,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/287\/revisions"}],"predecessor-version":[{"id":4793,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/287\/revisions\/4793"}],"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\/287\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=287"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=287"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=287"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=287"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}