{"id":3000,"date":"2025-08-15T19:32:56","date_gmt":"2025-08-15T19:32:56","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=3000"},"modified":"2025-08-15T19:32:56","modified_gmt":"2025-08-15T19:32:56","slug":"continuity-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/continuity-learn-it-2\/","title":{"raw":"Continuity: Learn It 2","rendered":"Continuity: Learn It 2"},"content":{"raw":"

Identifying Discontinuities<\/h2>\r\nDiscontinuity can occur in different ways. We saw in the previous section that a function could have a left-hand limit<\/strong> and a right-hand limit<\/strong> even if they are not equal. If the left- and right-hand limits exist but are different, the graph \"jumps\" at [latex]x=a[\/latex] . The function is said to have a jump discontinuity.\r\n\r\nAs an example, look at the graph of the function [latex]y=f\\left(x\\right)[\/latex]. Notice as [latex]x[\/latex] approaches [latex]a[\/latex] how the output approaches different values from the left and from the right.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Graph of a function with a jump discontinuity.[\/caption]\r\n\r\n
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jump discontinuity<\/h3>\r\nA function [latex]f\\left(x\\right)[\/latex] has a jump discontinuity<\/strong> at [latex]x=a[\/latex] if the left- and right-hand limits both exist but are not equal: [latex]\\underset{x\\to {a}^{-}}{\\mathrm{lim}}f\\left(x\\right)\\ne \\underset{x\\to {a}^{+}}{\\mathrm{lim}}f\\left(x\\right)[\/latex] .\r\n\r\n<\/section>\r\n

Identifying Removable Discontinuity<\/h2>\r\nSome functions have a discontinuity, but it is possible to redefine the function at that point to make it continuous. This type of function is said to have a removable discontinuity. Let\u2019s look at the function [latex]y=f\\left(x\\right)[\/latex] represented by the graph below. The function has a limit. However, there is a hole at [latex]x=a[\/latex] . The hole can be filled by extending the domain to include the input [latex]x=a[\/latex] and defining the corresponding output of the function at that value as the limit of the function at [latex]x=a[\/latex] .\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Graph of function [latex]f[\/latex] with a removable discontinuity at [latex]x=a[\/latex] .[\/caption]
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removable discontinuity<\/h3>\r\nA function [latex]f\\left(x\\right)[\/latex] has a removable discontinuity<\/strong> at [latex]x=a[\/latex] if the limit, [latex]\\underset{x\\to a}{\\mathrm{lim}}f\\left(x\\right)[\/latex], exists, but either\r\n
\r\n
    \r\n \t
  1. [latex]f\\left(a\\right)[\/latex] does not exist or<\/em><\/li>\r\n \t
  2. [latex]f\\left(a\\right)[\/latex], the value of the function at [latex]x=a[\/latex] does not equal the limit, [latex]f\\left(a\\right)\\ne \\underset{x\\to a}{\\mathrm{lim}}f\\left(x\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/section>
    Identify all discontinuities for the following functions as either a jump or a removable discontinuity.\r\n
      \r\n \t
    1. [latex]f\\left(x\\right)=\\frac{{x}^{2}-2x - 15}{x - 5}[\/latex]<\/li>\r\n \t
    2. [latex]g\\left(x\\right)=\\begin{cases}x+1, \\hfill& x<2 \\\\ -x, \\hfill& x\\geq2\\end{cases}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"277675\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"277675\"]\r\n
        \r\n \t
      1. Notice that the function is defined everywhere except at [latex]x=5[\/latex].Thus, [latex]f\\left(5\\right)[\/latex] does not exist, Condition 2 is not satisfied. Since Condition 1 is satisfied, the limit as [latex]x[\/latex] approaches 5 is 8, and Condition 2 is not satisfied.This means there is a removable discontinuity at [latex]x=5[\/latex].<\/li>\r\n \t
      2. Condition 2 is satisfied because [latex]g\\left(2\\right)=-2[\/latex].Notice that the function is a piecewise function<\/strong>, and for each piece, the function is defined everywhere on its domain. Let\u2019s examine Condition 1 by determining the left- and right-hand limits as [latex]x[\/latex] approaches 2.Left-hand limit: [latex]\\underset{x\\to {2}^{-}}{\\mathrm{lim}}\\left(x+1\\right)=2+1=3[\/latex]. The left-hand limit exists.Right-hand limit: [latex]\\underset{x\\to {2}^{+}}{\\mathrm{lim}}\\left(-x\\right)=-2[\/latex]. The right-hand limit exists. But\r\n
        [latex]\\underset{x\\to {2}^{-}}{\\mathrm{lim}}f\\left(x\\right)\\ne \\underset{x\\to {2}^{+}}{\\mathrm{lim}}f\\left(x\\right)[\/latex].<\/div>\r\nSo, [latex]\\underset{x\\to 2}{\\mathrm{lim}}f\\left(x\\right)[\/latex] does not exist, and Condition 2 fails: There is no removable discontinuity. However, since both left- and right-hand limits exist but are not equal, the conditions are satisfied for a jump discontinuity at [latex]x=2[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>
        Identify all discontinuities for the following functions as either a jump or a removable discontinuity.\r\n

        a. [latex]f\\left(x\\right)=\\frac{{x}^{2}-6x}{x - 6}[\/latex]<\/p>\r\n

        b. [latex]g\\left(x\\right)=\\begin{cases}\\sqrt{x}, \\hfill& 0\\leq x<4 \\\\ 2x, \\hfill& x\\geq4\\end{cases}[\/latex]<\/p>\r\n[reveal-answer q=\"284168\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"284168\"]\r\n\r\na.\u00a0removable discontinuity at [latex]x=6[\/latex];\r\nb. jump discontinuity at [latex]x=4[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

        Identifying Discontinuities<\/h2>\n

        Discontinuity can occur in different ways. We saw in the previous section that a function could have a left-hand limit<\/strong> and a right-hand limit<\/strong> even if they are not equal. If the left- and right-hand limits exist but are different, the graph “jumps” at [latex]x=a[\/latex] . The function is said to have a jump discontinuity.<\/p>\n

        As an example, look at the graph of the function [latex]y=f\\left(x\\right)[\/latex]. Notice as [latex]x[\/latex] approaches [latex]a[\/latex] how the output approaches different values from the left and from the right.<\/p>\n

        \"Graph
        Graph of a function with a jump discontinuity.<\/figcaption><\/figure>\n
        \n

        jump discontinuity<\/h3>\n

        A function [latex]f\\left(x\\right)[\/latex] has a jump discontinuity<\/strong> at [latex]x=a[\/latex] if the left- and right-hand limits both exist but are not equal: [latex]\\underset{x\\to {a}^{-}}{\\mathrm{lim}}f\\left(x\\right)\\ne \\underset{x\\to {a}^{+}}{\\mathrm{lim}}f\\left(x\\right)[\/latex] .<\/p>\n<\/section>\n

        Identifying Removable Discontinuity<\/h2>\n

        Some functions have a discontinuity, but it is possible to redefine the function at that point to make it continuous. This type of function is said to have a removable discontinuity. Let\u2019s look at the function [latex]y=f\\left(x\\right)[\/latex] represented by the graph below. The function has a limit. However, there is a hole at [latex]x=a[\/latex] . The hole can be filled by extending the domain to include the input [latex]x=a[\/latex] and defining the corresponding output of the function at that value as the limit of the function at [latex]x=a[\/latex] .<\/p>\n

        \"Graph
        Graph of function [latex]f[\/latex] with a removable discontinuity at [latex]x=a[\/latex] .<\/figcaption><\/figure>\n
        \n

        removable discontinuity<\/h3>\n

        A function [latex]f\\left(x\\right)[\/latex] has a removable discontinuity<\/strong> at [latex]x=a[\/latex] if the limit, [latex]\\underset{x\\to a}{\\mathrm{lim}}f\\left(x\\right)[\/latex], exists, but either<\/p>\n

        \n
          \n
        1. [latex]f\\left(a\\right)[\/latex] does not exist or<\/em><\/li>\n
        2. [latex]f\\left(a\\right)[\/latex], the value of the function at [latex]x=a[\/latex] does not equal the limit, [latex]f\\left(a\\right)\\ne \\underset{x\\to a}{\\mathrm{lim}}f\\left(x\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/section>\n
          Identify all discontinuities for the following functions as either a jump or a removable discontinuity.<\/p>\n
            \n
          1. [latex]f\\left(x\\right)=\\frac{{x}^{2}-2x - 15}{x - 5}[\/latex]<\/li>\n
          2. [latex]g\\left(x\\right)=\\begin{cases}x+1, \\hfill& x<2 \\\\ -x, \\hfill& x\\geq2\\end{cases}[\/latex]<\/li>\n<\/ol>\n