{"id":300,"date":"2021-02-04T01:00:54","date_gmt":"2021-02-04T01:00:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=300"},"modified":"2022-03-11T21:57:36","modified_gmt":"2022-03-11T21:57:36","slug":"continuity-over-an-interval","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/continuity-over-an-interval\/","title":{"raw":"Continuity Over an Interval","rendered":"Continuity Over an Interval"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li><span class=\"os-abstract-content\">Define continuity on an interval<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">State the theorem for limits of composite functions<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Provide an example of the intermediate value theorem<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170573612125\">Now that we have explored the concept of continuity at a point, we extend that idea to <strong>continuity over an interval<\/strong>. As we develop this idea for different types of intervals, it may be useful to keep in mind the intuitive idea that a function is continuous over an interval if we can use a pencil to trace the function between any two points in the interval without lifting the pencil from the paper. In preparation for defining continuity on an interval, we begin by looking at the definition of what it means for a function to be continuous from the right at a point and continuous from the left at a point.<\/p>\r\n\r\n<div id=\"fs-id1170573581958\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Continuity from the Right and from the Left<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170573352232\">A function [latex]f(x)[\/latex] is said to be <strong>continuous from the right<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to a^+}{\\lim}f(x)=f(a)[\/latex].<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1170571120467\">A function [latex]f(x)[\/latex] is said to be<strong> continuous from the left<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to a^-}{\\lim}f(x)=f(a)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1170570967081\">A function is continuous over an open interval if it is continuous at every point in the interval. A function [latex]f(x)[\/latex] is continuous over a closed interval of the form [latex][a,b][\/latex] if it is continuous at every point in [latex](a,b)[\/latex] and is continuous from the right at [latex]a[\/latex] and is continuous from the left at [latex]b[\/latex]. Analogously, a function [latex]f(x)[\/latex] is continuous over an interval of the form [latex](a,b][\/latex] if it is continuous over [latex](a,b)[\/latex] and is continuous from the left at [latex]b[\/latex]. Continuity over other types of intervals are defined in a similar fashion.<\/p>\r\n<p id=\"fs-id1170573533852\">Requiring that [latex]\\underset{x\\to a^+}{\\lim}f(x)=f(a)[\/latex] and [latex]\\underset{x\\to b^-}{\\lim}f(x)=f(b)[\/latex] ensures that we can trace the graph of the function from the point [latex](a,f(a))[\/latex] to the point [latex](b,f(b))[\/latex] without lifting the pencil. If, for example, [latex]\\underset{x\\to a^+}{\\lim}f(x)\\ne f(a)[\/latex], we would need to lift our pencil to jump from [latex]f(a)[\/latex] to the graph of the rest of the function over [latex](a,b][\/latex].<\/p>\r\n\r\n<div id=\"fs-id1170573395559\" class=\"textbook exercises\">\r\n<h3>Example: Continuity on an Interval<\/h3>\r\n<p id=\"fs-id1170570990867\">State the interval(s) over which the function [latex]f(x)=\\dfrac{x-1}{x^2+2x}[\/latex] is continuous.<\/p>\r\n[reveal-answer q=\"fs-id1170573426588\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573426588\"]\r\n<p id=\"fs-id1170573426588\">Since [latex]f(x)=\\frac{x-1}{x^2+2x}[\/latex] is a rational function, it is continuous at every point in its domain. The domain of [latex]f(x)[\/latex] is the set [latex](\u2212\\infty ,-2) \\cup (-2,0) \\cup (0,+\\infty)[\/latex]. Thus, [latex]f(x)[\/latex] is continuous over each of the intervals [latex](\u2212\\infty ,-2), \\, (-2,0)[\/latex], and [latex](0,+\\infty)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170573387892\" class=\"textbook exercises\">\r\n<h3>Example: Continuity over an Interval<\/h3>\r\n<p id=\"fs-id1170573406820\">State the interval(s) over which the function [latex]f(x)=\\sqrt{4-x^2}[\/latex] is continuous.<\/p>\r\n[reveal-answer q=\"fs-id1170573439420\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573439420\"]\r\n<p id=\"fs-id1170573439420\">From the limit laws, we know that [latex]\\underset{x\\to a}{\\lim}\\sqrt{4-x^2}=\\sqrt{4-a^2}[\/latex] for all values of [latex]a[\/latex] in [latex](-2,2)[\/latex]. We also know that [latex]\\underset{x\\to -2^+}{\\lim}\\sqrt{4-x^2}=0[\/latex] exists and [latex]\\underset{x\\to 2^-}{\\lim}\\sqrt{4-x^2}=0[\/latex] exists. Therefore, [latex]f(x)[\/latex] is continuous over the interval [latex][-2,2][\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170573403253\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170571120474\">State the interval(s) over which the function [latex]f(x)=\\sqrt{x+3}[\/latex] is continuous.<\/p>\r\n[reveal-answer q=\"2282726\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"2282726\"]\r\n<p id=\"fs-id1170570999934\">Use the example above as a guide for solving.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170571287079\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571287079\"]\r\n<p id=\"fs-id1170571287079\">[latex][-3,+\\infty)[\/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]204644[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1170573361765\">The Composite Function Theorem allows us to expand our ability to compute limits. In particular, this theorem ultimately allows us to demonstrate that trigonometric functions are continuous over their domains.<\/p>\r\n\r\n<div id=\"fs-id1170573352212\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Composite Function Theorem<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170573762934\">If [latex]f(x)[\/latex] is continuous at [latex]L[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=L[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1170573400410\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}f(g(x))=f(\\underset{x\\to a}{\\lim}g(x))=f(L)[\/latex].<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see examples of solving limits of composite functions. [\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/RgfKNIkpFWc?controls=0\" width=\"560\" height=\"315\" 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\/LimitsOfCompositeFunctions_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Limits of composite functions | Limits and continuity | AP Calculus AB | Khan Academy\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<p id=\"fs-id1170573573603\">Before we move on to the next example, recall that earlier, in the section on limit laws, we showed [latex]\\underset{x\\to 0}{\\lim} \\cos x=1= \\cos (0)[\/latex]. Consequently, we know that [latex]f(x)= \\cos x[\/latex] is continuous at 0. In the next example we see how to combine this result with the composite function theorem.<\/p>\r\n\r\n<div id=\"fs-id1170573718134\" class=\"textbook exercises\">\r\n<h3>Example: Limit of a Composite Cosine Function<\/h3>\r\nEvaluate [latex]\\underset{x\\to \\pi\/2}{\\lim}\\cos(x-\\dfrac{\\pi }{2})[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1170573408578\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573408578\"]\r\n<p id=\"fs-id1170573408578\">The given function is a composite of [latex]\\cos x[\/latex] and [latex]x-\\frac{\\pi}{2}[\/latex]. Since [latex]\\underset{x\\to \\pi\/2}{\\lim}(x-\\frac{\\pi}{2})=0[\/latex] and [latex]\\cos x[\/latex] is continuous at 0, we may apply the composite function theorem. Thus,<\/p>\r\n\r\n<div id=\"fs-id1170573570975\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\pi\/2}{\\lim}\\cos(x-\\frac{\\pi}{2})= \\cos (\\underset{x\\to \\pi\/2}{\\lim}(x-\\frac{\\pi}{2}))= \\cos (0)=1[\/latex].\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573732417\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170571131881\">Evaluate [latex]\\underset{x\\to \\pi}{\\lim}\\sin(x-\\pi)[\/latex].<\/p>\r\n[reveal-answer q=\"790361\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"790361\"]\r\n<p id=\"fs-id1170573359398\">[latex]f(x)= \\sin x[\/latex] is continuous at 0. Use the example above as a guide.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170573362583\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573362583\"]\r\n<p id=\"fs-id1170573362583\">0<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1170571101566\">The proof of the next theorem uses the composite function theorem as well as the continuity of [latex]f(x)= \\sin x[\/latex] and [latex]g(x)= \\cos x[\/latex] at the point 0 to show that trigonometric functions are continuous over their entire domains.<\/p>\r\n\r\n<div id=\"fs-id1170573418920\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Continuity of Trigonometric Functions<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170573512311\">Trigonometric functions are continuous over their entire domains.<\/p>\r\n\r\n<\/div>\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1170573429466\">We begin by demonstrating that [latex]\\cos x[\/latex] is continuous at every real number. To do this, we must show that [latex]\\underset{x\\to a}{\\lim}\\cos x = \\cos a[\/latex] for all values of [latex]a[\/latex].<\/p>\r\n<p id=\"fs-id1170571099777\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}\\underset{x\\to a}{\\lim}\\cos x &amp; =\\underset{x\\to a}{\\lim}\\cos((x-a)+a) &amp; &amp; &amp; \\text{rewrite} \\, x \\, \\text{as} \\, x-a+a \\, \\text{and group} \\, (x-a) \\\\ &amp; =\\underset{x\\to a}{\\lim}(\\cos(x-a)\\cos a - \\sin(x-a)\\sin a) &amp; &amp; &amp; \\text{apply the identity for the cosine of the sum of two angles} \\\\ &amp; = \\cos(\\underset{x\\to a}{\\lim}(x-a)) \\cos a - \\sin(\\underset{x\\to a}{\\lim}(x-a))\\sin a &amp; &amp; &amp; \\underset{x\\to a}{\\lim}(x-a)=0, \\, \\text{and} \\, \\sin x \\, \\text{and} \\, \\cos x \\, \\text{are continuous at 0} \\\\ &amp; = \\cos(0)\\cos a - \\sin(0)\\sin a &amp; &amp; &amp; \\text{evaluate cos(0) and sin(0) and simplify} \\\\ &amp; =1 \\cdot \\cos a - 0 \\cdot \\sin a = \\cos a \\end{array}[\/latex]<\/p>\r\n<p id=\"fs-id1170571216459\">The proof that [latex] \\sin x[\/latex] is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of [latex] \\sin x[\/latex] and [latex] \\cos x,[\/latex] their continuity follows from the quotient limit law.<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n<p id=\"fs-id1170573370285\">As you can see, the composite function theorem is invaluable in demonstrating the continuity of trigonometric functions. As we continue our study of calculus, we revisit this theorem many times.<\/p>\r\n\r\n<div id=\"fs-id1170571100154\" class=\"bc-section section\">\r\n<h2>The Intermediate Value Theorem<\/h2>\r\n<p id=\"fs-id1170573717714\">Functions that are continuous over intervals of the form [latex][a,b][\/latex], where [latex]a[\/latex] and [latex]b[\/latex] are real numbers, exhibit many useful properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. The first of these theorems is the <strong>Intermediate Value Theorem<\/strong>.<\/p>\r\n\r\n<div id=\"fs-id1170571120551\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">The Intermediate Value Theorem<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170573534418\">Let [latex]f[\/latex] be continuous over a closed, bounded interval [latex][a,b][\/latex]. If [latex]z[\/latex] is any real number between [latex]f(a)[\/latex] and [latex]f(b)[\/latex], then there is a number [latex]c[\/latex] in [latex][a,b][\/latex] satisfying [latex]f(c)=z[\/latex]. (See Figure 7).<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"400\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203518\/CNX_Calc_Figure_02_04_007.jpg\" alt=\"A diagram illustrating the intermediate value theorem. There is a generic continuous curved function shown over the interval [a,b]. The points fa. and fb. are marked, and dotted lines are drawn from a, b, fa., and fb. to the points (a, fa.) and (b, fb.). A third point, c, is plotted between a and b. Since the function is continuous, there is a value for fc. along the curve, and a line is drawn from c to (c, fc.) and from (c, fc.) to fc., which is labeled as z on the y axis.\" width=\"400\" height=\"370\" \/> Figure 7. There is a number [latex]c \\in [a,b][\/latex] that satisfies [latex]f(c)=z[\/latex].[\/caption]<\/div>\r\n<div id=\"fs-id1170571120881\" class=\"textbook exercises\">\r\n<h3>Example: Application of the Intermediate Value Theorem<\/h3>\r\n<p id=\"fs-id1170570997458\">Show that [latex]f(x)=x- \\cos x[\/latex] has at least one zero.<\/p>\r\n[reveal-answer q=\"fs-id1170571136342\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571136342\"]\r\n<p id=\"fs-id1170571136342\">Since [latex]f(x)=x-\\cos x[\/latex] is continuous over [latex](\u2212\\infty,+\\infty)[\/latex], it is continuous over any closed interval of the form [latex][a,b][\/latex]. If you can find an interval [latex][a,b][\/latex] such that [latex]f(a)[\/latex] and [latex]f(b)[\/latex] have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number [latex]c[\/latex] in [latex](a,b)[\/latex] that satisfies [latex]f(c)=0[\/latex]. Note that<\/p>\r\n\r\n<div id=\"fs-id1170571239008\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(0)=0 - \\cos (0)=-1&lt;0[\/latex]<\/div>\r\n<p id=\"fs-id1170573424328\">and<\/p>\r\n\r\n<div id=\"fs-id1170573424331\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(\\frac{\\pi}{2})=\\frac{\\pi}{2} - \\cos \\frac{\\pi}{2}=\\frac{\\pi}{2}&gt;0[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573753266\">Using the Intermediate Value Theorem, we can see that there must be a real number [latex]c[\/latex] in [latex][0,\\pi\/2][\/latex] that satisfies [latex]f(c)=0[\/latex]. Therefore, [latex]f(x)=x- \\cos x[\/latex] has at least one zero.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573439386\" class=\"textbook exercises\">\r\n<h3>Example: When Can You Apply the Intermediate Value Theorem?<\/h3>\r\n<p id=\"fs-id1170573413598\">If [latex]f(x)[\/latex] is continuous over [latex][0,2], \\, f(0)&gt;0[\/latex], and [latex]f(2)&gt;0,[\/latex] can we use the Intermediate Value Theorem to conclude that [latex]f(x)[\/latex] has no zeros in the interval [latex][0,2][\/latex]? Explain.<\/p>\r\n[reveal-answer q=\"fs-id1170571262087\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571262087\"]\r\n<p id=\"fs-id1170571262087\">No. The Intermediate Value Theorem only allows us to conclude that we can find a value between [latex]f(0)[\/latex] and [latex]f(2)[\/latex]; it doesn\u2019t allow us to conclude that we can\u2019t find other values. To see this more clearly, consider the function [latex]f(x)=(x-1)^2[\/latex]. It satisfies [latex]f(0)=1&gt;0, \\, f(2)=1&gt;0[\/latex], and [latex]f(1)=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 Example: When Can You Apply the Intermediate Value Theorem?. [\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/BXUu5bG1CXU?controls=0&amp;start=804&amp;end=860&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\/2.4Continuity804to860_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-id1170570996536\" class=\"textbook exercises\">\r\n<h3>Example: When Can You Apply the Intermediate Value Theorem?<\/h3>\r\n<p id=\"fs-id1170573586819\">For [latex]f(x)=1\/x, \\, f(-1)=-1&lt;0[\/latex] and [latex]f(1)=1&gt;0[\/latex]. Can we conclude that [latex]f(x)[\/latex] has a zero in the interval [latex][-1,1][\/latex]?<\/p>\r\n[reveal-answer q=\"fs-id1170571138880\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571138880\"]\r\n<p id=\"fs-id1170571138880\">No. The function is not continuous over [latex][-1,1][\/latex]. The Intermediate Value Theorem does not apply here.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170571100300\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170571048766\">Show that [latex]f(x)=x^3-x^2-3x+1[\/latex] has a zero over the interval [latex][0,1][\/latex].<\/p>\r\n[reveal-answer q=\"371299\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"371299\"]\r\n<p id=\"fs-id1170573569887\">Find [latex]f(0)[\/latex] and [latex]f(1)[\/latex]. Apply the Intermediate Value Theorem.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170573382710\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573382710\"]\r\n<p id=\"fs-id1170573382710\">[latex]f(0)=1&gt;0, \\, f(1)=-2&lt;0[\/latex]; [latex]f(x)[\/latex] is continuous over [latex][0,1][\/latex]. It must have a zero on this interval.<\/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]4922[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li><span class=\"os-abstract-content\">Define continuity on an interval<\/span><\/li>\n<li><span class=\"os-abstract-content\">State the theorem for limits of composite functions<\/span><\/li>\n<li><span class=\"os-abstract-content\">Provide an example of the intermediate value theorem<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170573612125\">Now that we have explored the concept of continuity at a point, we extend that idea to <strong>continuity over an interval<\/strong>. As we develop this idea for different types of intervals, it may be useful to keep in mind the intuitive idea that a function is continuous over an interval if we can use a pencil to trace the function between any two points in the interval without lifting the pencil from the paper. In preparation for defining continuity on an interval, we begin by looking at the definition of what it means for a function to be continuous from the right at a point and continuous from the left at a point.<\/p>\n<div id=\"fs-id1170573581958\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Continuity from the Right and from the Left<\/h3>\n<hr \/>\n<p id=\"fs-id1170573352232\">A function [latex]f(x)[\/latex] is said to be <strong>continuous from the right<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to a^+}{\\lim}f(x)=f(a)[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571120467\">A function [latex]f(x)[\/latex] is said to be<strong> continuous from the left<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to a^-}{\\lim}f(x)=f(a)[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1170570967081\">A function is continuous over an open interval if it is continuous at every point in the interval. A function [latex]f(x)[\/latex] is continuous over a closed interval of the form [latex][a,b][\/latex] if it is continuous at every point in [latex](a,b)[\/latex] and is continuous from the right at [latex]a[\/latex] and is continuous from the left at [latex]b[\/latex]. Analogously, a function [latex]f(x)[\/latex] is continuous over an interval of the form [latex](a,b][\/latex] if it is continuous over [latex](a,b)[\/latex] and is continuous from the left at [latex]b[\/latex]. Continuity over other types of intervals are defined in a similar fashion.<\/p>\n<p id=\"fs-id1170573533852\">Requiring that [latex]\\underset{x\\to a^+}{\\lim}f(x)=f(a)[\/latex] and [latex]\\underset{x\\to b^-}{\\lim}f(x)=f(b)[\/latex] ensures that we can trace the graph of the function from the point [latex](a,f(a))[\/latex] to the point [latex](b,f(b))[\/latex] without lifting the pencil. If, for example, [latex]\\underset{x\\to a^+}{\\lim}f(x)\\ne f(a)[\/latex], we would need to lift our pencil to jump from [latex]f(a)[\/latex] to the graph of the rest of the function over [latex](a,b][\/latex].<\/p>\n<div id=\"fs-id1170573395559\" class=\"textbook exercises\">\n<h3>Example: Continuity on an Interval<\/h3>\n<p id=\"fs-id1170570990867\">State the interval(s) over which the function [latex]f(x)=\\dfrac{x-1}{x^2+2x}[\/latex] is continuous.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573426588\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573426588\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573426588\">Since [latex]f(x)=\\frac{x-1}{x^2+2x}[\/latex] is a rational function, it is continuous at every point in its domain. The domain of [latex]f(x)[\/latex] is the set [latex](\u2212\\infty ,-2) \\cup (-2,0) \\cup (0,+\\infty)[\/latex]. Thus, [latex]f(x)[\/latex] is continuous over each of the intervals [latex](\u2212\\infty ,-2), \\, (-2,0)[\/latex], and [latex](0,+\\infty)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573387892\" class=\"textbook exercises\">\n<h3>Example: Continuity over an Interval<\/h3>\n<p id=\"fs-id1170573406820\">State the interval(s) over which the function [latex]f(x)=\\sqrt{4-x^2}[\/latex] is continuous.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573439420\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573439420\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573439420\">From the limit laws, we know that [latex]\\underset{x\\to a}{\\lim}\\sqrt{4-x^2}=\\sqrt{4-a^2}[\/latex] for all values of [latex]a[\/latex] in [latex](-2,2)[\/latex]. We also know that [latex]\\underset{x\\to -2^+}{\\lim}\\sqrt{4-x^2}=0[\/latex] exists and [latex]\\underset{x\\to 2^-}{\\lim}\\sqrt{4-x^2}=0[\/latex] exists. Therefore, [latex]f(x)[\/latex] is continuous over the interval [latex][-2,2][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573403253\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170571120474\">State the interval(s) over which the function [latex]f(x)=\\sqrt{x+3}[\/latex] is continuous.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q2282726\">Hint<\/span><\/p>\n<div id=\"q2282726\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170570999934\">Use the example above as a guide for solving.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571287079\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571287079\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571287079\">[latex][-3,+\\infty)[\/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=\"ohm204644\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=204644&theme=oea&iframe_resize_id=ohm204644&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1170573361765\">The Composite Function Theorem allows us to expand our ability to compute limits. In particular, this theorem ultimately allows us to demonstrate that trigonometric functions are continuous over their domains.<\/p>\n<div id=\"fs-id1170573352212\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Composite Function Theorem<\/h3>\n<hr \/>\n<p id=\"fs-id1170573762934\">If [latex]f(x)[\/latex] is continuous at [latex]L[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=L[\/latex], then<\/p>\n<div id=\"fs-id1170573400410\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}f(g(x))=f(\\underset{x\\to a}{\\lim}g(x))=f(L)[\/latex].<\/div>\n<\/div>\n<p>Watch the following video to see examples of solving limits of composite functions. <\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/RgfKNIkpFWc?controls=0\" width=\"560\" height=\"315\" 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\/LimitsOfCompositeFunctions_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Limits of composite functions | Limits and continuity | AP Calculus AB | Khan Academy&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<p id=\"fs-id1170573573603\">Before we move on to the next example, recall that earlier, in the section on limit laws, we showed [latex]\\underset{x\\to 0}{\\lim} \\cos x=1= \\cos (0)[\/latex]. Consequently, we know that [latex]f(x)= \\cos x[\/latex] is continuous at 0. In the next example we see how to combine this result with the composite function theorem.<\/p>\n<div id=\"fs-id1170573718134\" class=\"textbook exercises\">\n<h3>Example: Limit of a Composite Cosine Function<\/h3>\n<p>Evaluate [latex]\\underset{x\\to \\pi\/2}{\\lim}\\cos(x-\\dfrac{\\pi }{2})[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573408578\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573408578\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573408578\">The given function is a composite of [latex]\\cos x[\/latex] and [latex]x-\\frac{\\pi}{2}[\/latex]. Since [latex]\\underset{x\\to \\pi\/2}{\\lim}(x-\\frac{\\pi}{2})=0[\/latex] and [latex]\\cos x[\/latex] is continuous at 0, we may apply the composite function theorem. Thus,<\/p>\n<div id=\"fs-id1170573570975\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\pi\/2}{\\lim}\\cos(x-\\frac{\\pi}{2})= \\cos (\\underset{x\\to \\pi\/2}{\\lim}(x-\\frac{\\pi}{2}))= \\cos (0)=1[\/latex].\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573732417\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170571131881\">Evaluate [latex]\\underset{x\\to \\pi}{\\lim}\\sin(x-\\pi)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q790361\">Hint<\/span><\/p>\n<div id=\"q790361\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573359398\">[latex]f(x)= \\sin x[\/latex] is continuous at 0. Use the example above as a guide.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573362583\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573362583\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573362583\">0<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571101566\">The proof of the next theorem uses the composite function theorem as well as the continuity of [latex]f(x)= \\sin x[\/latex] and [latex]g(x)= \\cos x[\/latex] at the point 0 to show that trigonometric functions are continuous over their entire domains.<\/p>\n<div id=\"fs-id1170573418920\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Continuity of Trigonometric Functions<\/h3>\n<hr \/>\n<p id=\"fs-id1170573512311\">Trigonometric functions are continuous over their entire domains.<\/p>\n<\/div>\n<h3>Proof<\/h3>\n<p id=\"fs-id1170573429466\">We begin by demonstrating that [latex]\\cos x[\/latex] is continuous at every real number. To do this, we must show that [latex]\\underset{x\\to a}{\\lim}\\cos x = \\cos a[\/latex] for all values of [latex]a[\/latex].<\/p>\n<p id=\"fs-id1170571099777\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}\\underset{x\\to a}{\\lim}\\cos x & =\\underset{x\\to a}{\\lim}\\cos((x-a)+a) & & & \\text{rewrite} \\, x \\, \\text{as} \\, x-a+a \\, \\text{and group} \\, (x-a) \\\\ & =\\underset{x\\to a}{\\lim}(\\cos(x-a)\\cos a - \\sin(x-a)\\sin a) & & & \\text{apply the identity for the cosine of the sum of two angles} \\\\ & = \\cos(\\underset{x\\to a}{\\lim}(x-a)) \\cos a - \\sin(\\underset{x\\to a}{\\lim}(x-a))\\sin a & & & \\underset{x\\to a}{\\lim}(x-a)=0, \\, \\text{and} \\, \\sin x \\, \\text{and} \\, \\cos x \\, \\text{are continuous at 0} \\\\ & = \\cos(0)\\cos a - \\sin(0)\\sin a & & & \\text{evaluate cos(0) and sin(0) and simplify} \\\\ & =1 \\cdot \\cos a - 0 \\cdot \\sin a = \\cos a \\end{array}[\/latex]<\/p>\n<p id=\"fs-id1170571216459\">The proof that [latex]\\sin x[\/latex] is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of [latex]\\sin x[\/latex] and [latex]\\cos x,[\/latex] their continuity follows from the quotient limit law.<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1170573370285\">As you can see, the composite function theorem is invaluable in demonstrating the continuity of trigonometric functions. As we continue our study of calculus, we revisit this theorem many times.<\/p>\n<div id=\"fs-id1170571100154\" class=\"bc-section section\">\n<h2>The Intermediate Value Theorem<\/h2>\n<p id=\"fs-id1170573717714\">Functions that are continuous over intervals of the form [latex][a,b][\/latex], where [latex]a[\/latex] and [latex]b[\/latex] are real numbers, exhibit many useful properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. The first of these theorems is the <strong>Intermediate Value Theorem<\/strong>.<\/p>\n<div id=\"fs-id1170571120551\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">The Intermediate Value Theorem<\/h3>\n<hr \/>\n<p id=\"fs-id1170573534418\">Let [latex]f[\/latex] be continuous over a closed, bounded interval [latex][a,b][\/latex]. If [latex]z[\/latex] is any real number between [latex]f(a)[\/latex] and [latex]f(b)[\/latex], then there is a number [latex]c[\/latex] in [latex][a,b][\/latex] satisfying [latex]f(c)=z[\/latex]. (See Figure 7).<\/p>\n<div style=\"width: 410px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203518\/CNX_Calc_Figure_02_04_007.jpg\" alt=\"A diagram illustrating the intermediate value theorem. There is a generic continuous curved function shown over the interval [a,b]. The points fa. and fb. are marked, and dotted lines are drawn from a, b, fa., and fb. to the points (a, fa.) and (b, fb.). A third point, c, is plotted between a and b. Since the function is continuous, there is a value for fc. along the curve, and a line is drawn from c to (c, fc.) and from (c, fc.) to fc., which is labeled as z on the y axis.\" width=\"400\" height=\"370\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. There is a number [latex]c \\in [a,b][\/latex] that satisfies [latex]f(c)=z[\/latex].<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571120881\" class=\"textbook exercises\">\n<h3>Example: Application of the Intermediate Value Theorem<\/h3>\n<p id=\"fs-id1170570997458\">Show that [latex]f(x)=x- \\cos x[\/latex] has at least one zero.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571136342\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571136342\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571136342\">Since [latex]f(x)=x-\\cos x[\/latex] is continuous over [latex](\u2212\\infty,+\\infty)[\/latex], it is continuous over any closed interval of the form [latex][a,b][\/latex]. If you can find an interval [latex][a,b][\/latex] such that [latex]f(a)[\/latex] and [latex]f(b)[\/latex] have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number [latex]c[\/latex] in [latex](a,b)[\/latex] that satisfies [latex]f(c)=0[\/latex]. Note that<\/p>\n<div id=\"fs-id1170571239008\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(0)=0 - \\cos (0)=-1<0[\/latex]<\/div>\n<p id=\"fs-id1170573424328\">and<\/p>\n<div id=\"fs-id1170573424331\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(\\frac{\\pi}{2})=\\frac{\\pi}{2} - \\cos \\frac{\\pi}{2}=\\frac{\\pi}{2}>0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573753266\">Using the Intermediate Value Theorem, we can see that there must be a real number [latex]c[\/latex] in [latex][0,\\pi\/2][\/latex] that satisfies [latex]f(c)=0[\/latex]. Therefore, [latex]f(x)=x- \\cos x[\/latex] has at least one zero.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573439386\" class=\"textbook exercises\">\n<h3>Example: When Can You Apply the Intermediate Value Theorem?<\/h3>\n<p id=\"fs-id1170573413598\">If [latex]f(x)[\/latex] is continuous over [latex][0,2], \\, f(0)>0[\/latex], and [latex]f(2)>0,[\/latex] can we use the Intermediate Value Theorem to conclude that [latex]f(x)[\/latex] has no zeros in the interval [latex][0,2][\/latex]? Explain.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571262087\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571262087\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571262087\">No. The Intermediate Value Theorem only allows us to conclude that we can find a value between [latex]f(0)[\/latex] and [latex]f(2)[\/latex]; it doesn\u2019t allow us to conclude that we can\u2019t find other values. To see this more clearly, consider the function [latex]f(x)=(x-1)^2[\/latex]. It satisfies [latex]f(0)=1>0, \\, f(2)=1>0[\/latex], and [latex]f(1)=0[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: When Can You Apply the Intermediate Value Theorem?. <\/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=804&amp;end=860&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\/2.4Continuity804to860_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-id1170570996536\" class=\"textbook exercises\">\n<h3>Example: When Can You Apply the Intermediate Value Theorem?<\/h3>\n<p id=\"fs-id1170573586819\">For [latex]f(x)=1\/x, \\, f(-1)=-1<0[\/latex] and [latex]f(1)=1>0[\/latex]. Can we conclude that [latex]f(x)[\/latex] has a zero in the interval [latex][-1,1][\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571138880\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571138880\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571138880\">No. The function is not continuous over [latex][-1,1][\/latex]. The Intermediate Value Theorem does not apply here.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571100300\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170571048766\">Show that [latex]f(x)=x^3-x^2-3x+1[\/latex] has a zero over the interval [latex][0,1][\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q371299\">Hint<\/span><\/p>\n<div id=\"q371299\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573569887\">Find [latex]f(0)[\/latex] and [latex]f(1)[\/latex]. Apply the Intermediate Value Theorem.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573382710\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573382710\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573382710\">[latex]f(0)=1>0, \\, f(1)=-2<0[\/latex]; [latex]f(x)[\/latex] is continuous over [latex][0,1][\/latex]. It must have a zero on this interval.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm4922\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=4922&theme=oea&iframe_resize_id=ohm4922&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/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-300\">\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":18,"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-300","chapter","type-chapter","status-publish","hentry"],"part":28,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/300","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":21,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/300\/revisions"}],"predecessor-version":[{"id":4795,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/300\/revisions\/4795"}],"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\/300\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=300"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=300"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=300"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=300"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}