{"id":2464,"date":"2024-05-22T18:34:31","date_gmt":"2024-05-22T18:34:31","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=2464"},"modified":"2025-05-08T15:55:31","modified_gmt":"2025-05-08T15:55:31","slug":"the-precise-definition-of-a-limit-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-precise-definition-of-a-limit-fresh-take\/","title":{"raw":"The Precise Definition of a Limit: Fresh Take","rendered":"The Precise Definition of a Limit: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Use the epsilon-delta method to determine the limit of a function<\/li>\r\n\t<li>Explain the epsilon-delta definitions of one-sided limits and infinite limits<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Epsilon-Delta Definition of the Limit<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\r\n<ul>\r\n\t<li class=\"whitespace-normal break-words\">Formal Definition:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">For [latex]f(x)[\/latex] defined near [latex]a[\/latex], [latex]\\lim_{x \\to a} f(x) = L[\/latex] if:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">For every [latex]\\varepsilon &gt; 0[\/latex], there exists [latex]\\delta &gt; 0[\/latex] such that:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">If [latex]0 &lt; |x - a| &lt; \\delta[\/latex], then [latex]|f(x) - L| &lt; \\varepsilon[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Geometric Interpretation:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">[latex]\\varepsilon[\/latex]: Vertical distance from [latex]L[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]\\delta[\/latex]: Horizontal distance from [latex]a[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">As [latex]\\varepsilon[\/latex] gets smaller, [latex]\\delta[\/latex] typically needs to be smaller<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Key Equivalences:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">[latex]|f(x) - L| &lt; \\varepsilon[\/latex] is equivalent to [latex]L - \\varepsilon &lt; f(x) &lt; L + \\varepsilon[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]0 &lt; |x - a| &lt; \\delta[\/latex] is equivalent to [latex]a - \\delta &lt; x &lt; a + \\delta[\/latex] and [latex]x \\neq a[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170571712682\">Complete the proof that [latex]\\underset{x\\to 2}{\\lim}(3x-2)=4[\/latex] by filling in the blanks.<\/p>\r\n<p id=\"fs-id1170572444308\">Let _____.<\/p>\r\n<p id=\"fs-id1170572444311\">Choose [latex]\\delta =[\/latex] ________.<\/p>\r\n<p id=\"fs-id1170572444323\">Assume [latex]0&lt;|x-\\text{___}|&lt;\\delta[\/latex].<\/p>\r\n<p id=\"fs-id1170572560044\">Thus, [latex]|\\text{________}-\\text{___}| =|\\text{_________}| = |\\text{___}||\\text{_________}| = \\text{___} \\, |\\text{_______}| &lt; \\text{______} = \\text{_______} = \\varepsilon[\/latex].<\/p>\r\n<p id=\"fs-id1170572292953\">Therefore, [latex]\\underset{x\\to 2}{\\lim}(3x-2)=4[\/latex].<\/p>\r\n<p>[reveal-answer q=\"4567890\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"4567890\"]<\/p>\r\n<p id=\"fs-id1170572448141\">Follow the outline in the Problem-Solving Strategy that we worked out in full in the example above.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1170571572023\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170571572023\"]<\/p>\r\n<p id=\"fs-id1170571572023\">Let [latex]\\varepsilon &gt;0[\/latex]; choose [latex]\\delta =\\frac{\\varepsilon}{3}[\/latex]; assume [latex]0&lt;|x-2|&lt;\\delta[\/latex].<\/p>\r\n<p id=\"fs-id1170572311214\">Thus, [latex]|(3x-2)-4|=|3x-6|=|3| \\cdot |x-2|&lt;3 \\cdot \\delta =3 \\cdot (\\varepsilon\/3)=\\varepsilon[\/latex].<\/p>\r\n<p>Therefore, [latex]\\underset{x\\to 2}{\\lim}3x-2=4[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170571571830\">Find [latex]\\delta[\/latex] corresponding to [latex]\\varepsilon &gt;0[\/latex] for a proof that [latex]\\underset{x\\to 9}{\\lim}\\sqrt{x}=3[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1170572512097\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170572512097\"]<\/p>\r\n<p id=\"fs-id1170572512097\">Choose [latex]\\delta =\\text{min}\\{9-(3-\\varepsilon)^2,(3+\\varepsilon)^2-9\\}[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170572243120\">Complete the proof that [latex]\\underset{x\\to 1}{\\lim}x^2=1[\/latex].<\/p>\r\n<p id=\"fs-id1170572334818\">Let [latex]\\varepsilon &gt;0[\/latex]; choose [latex]\\delta =\\text{min}\\{1,\\frac{\\varepsilon}{3}\\}[\/latex]; assume [latex]0&lt;|x-1|&lt;\\delta[\/latex].<\/p>\r\n<p id=\"fs-id1170571543218\">Since [latex]|x-1|&lt;1[\/latex], we may conclude that [latex]-1&lt;x-1&lt;1[\/latex]. Thus, [latex]1&lt;x+1&lt;3[\/latex]. Hence, [latex]|x+1|&lt;3[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1170571712301\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170571712301\"]<\/p>\r\n<p id=\"fs-id1170571712301\">[latex]|x^2-1|=|x-1| \\cdot |x+1|&lt;\\varepsilon\/3 \\cdot 3=\\varepsilon [\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<h2>Advanced Applications of the Epsilon-Delta Definition: Proofs, Non-Existence, and Algebraic Calculations<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\r\n<ul>\r\n\t<li class=\"whitespace-normal break-words\">Triangle Inequality:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">For any real numbers [latex]a[\/latex] and [latex]b[\/latex], [latex]|a+b| \\leq |a| + |b|[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Key tool in epsilon-delta proofs<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Proving Limit Laws:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Use epsilon-delta definition to prove properties of limits<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Example: Sum of limits is limit of sum<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Non-Existence of Limits:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">A limit doesn't exist if no real number satisfies the epsilon-delta definition<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Requires finding an [latex]\\varepsilon &gt; 0[\/latex] that works for all [latex]\\delta &gt; 0[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Algebraic Approach to Finding Deltas:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Solve [latex]|f(x) - L| &lt; \\varepsilon[\/latex] for [latex]x[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Find [latex]\\delta[\/latex] that ensures [latex]|x - x_0| &lt; \\delta[\/latex] implies [latex]|f(x) - L| &lt; \\varepsilon[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170572559647\">Find an open interval about [latex]x_0[\/latex] on which the inequality [latex]|f(x)-L| &lt; 0[\/latex] holds. Then give the largest value [latex]\\delta &amp;gt; 0[\/latex] such that for all [latex]x[\/latex] satisfying [latex]0 &lt; |x-x_0| &lt; \\delta[\/latex] the inequality [latex]|f(x)-L| &lt; \\varepsilon[\/latex] holds.<\/p>\r\n<center>[latex]f(x)=\\sqrt{x+4}, \\,\\, L=3, \\,\\, x_0=5, \\,\\, \\varepsilon=1[\/latex]<\/center>\r\n<p>[reveal-answer q=\"fs-id1170572601145\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170572601145\"]First we will need to start with the inequality [latex]|f(x)-L| &lt; \\varepsilon[\/latex] and plug in our numbers. Then we will solve for [latex]x[\/latex].<\/p>\r\n<center>[latex]|\\sqrt(x+4) - 3| &lt; \\varepsilon[\/latex]<\/center><center>[latex]|\\sqrt{x+4} - 3| &lt; 1[\/latex]<\/center><center>[latex]-1 &lt; \\sqrt{x+4} - 3 &lt; 1[\/latex]<\/center><center>[latex]2 &lt; \\sqrt{x+4} &lt; 4[\/latex]<\/center><center>[latex]4 &lt; x + 4 &lt; 16[\/latex]<\/center><center>[latex]0 &lt; x + 4 &lt; 12[\/latex]<\/center>\r\n<p>Therefore, the interval is [latex](0,12)[\/latex]. For the second answer, we will start with\u00a0[latex]0 &lt; |x-x_0| &lt; \\delta[\/latex].\u00a0 We will plug in our value and solve:<\/p>\r\n<center>[latex]|x-5| &amp;lt; \\delta[\/latex]<\/center><center>[latex]-\\delta &amp;lt; x-5 &amp;lt; \\delta[\/latex]<\/center><center>[latex]5-\\delta &amp;lt x &amp;lt; 5+\\delta[\/latex]<\/center>\r\n<p>Now we will set each piece equal to the endpoints we found above.<\/p>\r\n<center>[latex]5-\\delta=0[\/latex] and [latex]5+\\delta=12[\/latex]<\/center>\r\n<p>After solving we will get [latex]\\delta=5 \\text{ and }\\delta=7[\/latex]. Since the question is asking for the smallest interval, we choose the smaller number.\u00a0 Therefore the answer is [latex]\\delta=5[\/latex].<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<h2 data-type=\"title\">One-Sided and Infinite Limits<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\r\n<ul>\r\n\t<li class=\"whitespace-normal break-words\">One-Sided Limits:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Limit from the right: [latex]\\lim_{x \\to a^+} f(x) = L[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Limit from the left: [latex]\\lim_{x \\to a^-} f(x) = L[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Modifications to standard epsilon-delta definition<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Infinite Limits:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Positive infinite limit: [latex]\\lim_{x \\to a} f(x) = +\\infty[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Negative infinite limit: [latex]\\lim_{x \\to a} f(x) = -\\infty[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Replace [latex]\\varepsilon[\/latex] with [latex]M[\/latex] for arbitrarily large values<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<p class=\"font-bold\"><strong>Formal Definitions<\/strong><\/p>\r\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Limit from the Right: For every [latex]\\varepsilon &gt; 0[\/latex], there exists [latex]\\delta &gt; 0[\/latex] such that: If [latex]0 &lt; x - a &lt; \\delta[\/latex], then [latex]|f(x) - L| &lt; \\varepsilon[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Limit from the Left: For every [latex]\\varepsilon &gt; 0[\/latex], there exists [latex]\\delta &gt; 0[\/latex] such that: If [latex]0 &lt; a - x &lt; \\delta[\/latex], then [latex]|f(x) - L| &lt; \\varepsilon[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Positive Infinite Limit: For every [latex]M &gt; 0[\/latex], there exists [latex]\\delta &gt; 0[\/latex] such that: If [latex]0 &lt; |x - a| &lt; \\delta[\/latex], then [latex]f(x) &gt; M[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Negative Infinite Limit: For every [latex]M &gt; 0[\/latex], there exists [latex]\\delta &gt; 0[\/latex] such that: If [latex]0 &lt; |x - a| &lt; \\delta[\/latex], then [latex]f(x) &lt; -M[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170571711163\">Find [latex]\\delta[\/latex] corresponding to [latex]\\varepsilon[\/latex]\u00a0for a proof that [latex]\\underset{x\\to 1^-}{\\lim}\\sqrt{1-x}=0[\/latex].<\/p>\r\n<p>[reveal-answer q=\"142584\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"142584\"]<\/p>\r\n<p>[latex]\\delta =\\varepsilon^2[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p>Prove that [latex]\\lim_{x \\to 4^+} \\sqrt{x-4} = 0[\/latex].<\/p>\r\n<p><br \/>\r\n[reveal-answer q=\"789575\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"789575\"]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Let [latex]\\varepsilon &gt; 0[\/latex]. Choose [latex]\\delta = \\varepsilon^2[\/latex]. Assume [latex]0 &lt; x - 4 &lt; \\delta[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Then:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex] \\begin{array}{rcl} 0 &lt; x - 4 &lt; \\delta &amp;=&amp; \\varepsilon^2 \\ 0 &lt; \\sqrt{x-4} &lt; \\varepsilon \\ |\\sqrt{x-4} - 0| &lt; \\varepsilon \\end{array} [\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Therefore, [latex]\\lim_{x \\to 4^+} \\sqrt{x-4} = 0[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p>&nbsp;<\/p>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Use the epsilon-delta method to determine the limit of a function<\/li>\n<li>Explain the epsilon-delta definitions of one-sided limits and infinite limits<\/li>\n<\/ul>\n<\/section>\n<h2>Epsilon-Delta Definition of the Limit<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Formal Definition:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">For [latex]f(x)[\/latex] defined near [latex]a[\/latex], [latex]\\lim_{x \\to a} f(x) = L[\/latex] if:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">For every [latex]\\varepsilon > 0[\/latex], there exists [latex]\\delta > 0[\/latex] such that:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">If [latex]0 < |x - a| < \\delta[\/latex], then [latex]|f(x) - L| < \\varepsilon[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Geometric Interpretation:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]\\varepsilon[\/latex]: Vertical distance from [latex]L[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\delta[\/latex]: Horizontal distance from [latex]a[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">As [latex]\\varepsilon[\/latex] gets smaller, [latex]\\delta[\/latex] typically needs to be smaller<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Key Equivalences:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]|f(x) - L| < \\varepsilon[\/latex] is equivalent to [latex]L - \\varepsilon < f(x) < L + \\varepsilon[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]0 < |x - a| < \\delta[\/latex] is equivalent to [latex]a - \\delta < x < a + \\delta[\/latex] and [latex]x \\neq a[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1170571712682\">Complete the proof that [latex]\\underset{x\\to 2}{\\lim}(3x-2)=4[\/latex] by filling in the blanks.<\/p>\n<p id=\"fs-id1170572444308\">Let _____.<\/p>\n<p id=\"fs-id1170572444311\">Choose [latex]\\delta =[\/latex] ________.<\/p>\n<p id=\"fs-id1170572444323\">Assume [latex]0<|x-\\text{___}|<\\delta[\/latex].<\/p>\n<p id=\"fs-id1170572560044\">Thus, [latex]|\\text{________}-\\text{___}| =|\\text{_________}| = |\\text{___}||\\text{_________}| = \\text{___} \\, |\\text{_______}| < \\text{______} = \\text{_______} = \\varepsilon[\/latex].<\/p>\n<p id=\"fs-id1170572292953\">Therefore, [latex]\\underset{x\\to 2}{\\lim}(3x-2)=4[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q4567890\">Hint<\/button><\/p>\n<div id=\"q4567890\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572448141\">Follow the outline in the Problem-Solving Strategy that we worked out in full in the example above.<\/p>\n<\/div>\n<\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571572023\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170571572023\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571572023\">Let [latex]\\varepsilon >0[\/latex]; choose [latex]\\delta =\\frac{\\varepsilon}{3}[\/latex]; assume [latex]0<|x-2|<\\delta[\/latex].<\/p>\n<p id=\"fs-id1170572311214\">Thus, [latex]|(3x-2)-4|=|3x-6|=|3| \\cdot |x-2|<3 \\cdot \\delta =3 \\cdot (\\varepsilon\/3)=\\varepsilon[\/latex].<\/p>\n<p>Therefore, [latex]\\underset{x\\to 2}{\\lim}3x-2=4[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170571571830\">Find [latex]\\delta[\/latex] corresponding to [latex]\\varepsilon >0[\/latex] for a proof that [latex]\\underset{x\\to 9}{\\lim}\\sqrt{x}=3[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572512097\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572512097\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572512097\">Choose [latex]\\delta =\\text{min}\\{9-(3-\\varepsilon)^2,(3+\\varepsilon)^2-9\\}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572243120\">Complete the proof that [latex]\\underset{x\\to 1}{\\lim}x^2=1[\/latex].<\/p>\n<p id=\"fs-id1170572334818\">Let [latex]\\varepsilon >0[\/latex]; choose [latex]\\delta =\\text{min}\\{1,\\frac{\\varepsilon}{3}\\}[\/latex]; assume [latex]0<|x-1|<\\delta[\/latex].<\/p>\n<p id=\"fs-id1170571543218\">Since [latex]|x-1|<1[\/latex], we may conclude that [latex]-1<x-1<1[\/latex]. Thus, [latex]1<x+1<3[\/latex]. Hence, [latex]|x+1|<3[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571712301\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170571712301\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571712301\">[latex]|x^2-1|=|x-1| \\cdot |x+1|<\\varepsilon\/3 \\cdot 3=\\varepsilon[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2>Advanced Applications of the Epsilon-Delta Definition: Proofs, Non-Existence, and Algebraic Calculations<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Triangle Inequality:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">For any real numbers [latex]a[\/latex] and [latex]b[\/latex], [latex]|a+b| \\leq |a| + |b|[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Key tool in epsilon-delta proofs<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Proving Limit Laws:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Use epsilon-delta definition to prove properties of limits<\/li>\n<li class=\"whitespace-normal break-words\">Example: Sum of limits is limit of sum<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Non-Existence of Limits:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">A limit doesn&#8217;t exist if no real number satisfies the epsilon-delta definition<\/li>\n<li class=\"whitespace-normal break-words\">Requires finding an [latex]\\varepsilon > 0[\/latex] that works for all [latex]\\delta > 0[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Algebraic Approach to Finding Deltas:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Solve [latex]|f(x) - L| < \\varepsilon[\/latex] for [latex]x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Find [latex]\\delta[\/latex] that ensures [latex]|x - x_0| < \\delta[\/latex] implies [latex]|f(x) - L| < \\varepsilon[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572559647\">Find an open interval about [latex]x_0[\/latex] on which the inequality [latex]|f(x)-L| < 0[\/latex] holds. Then give the largest value [latex]\\delta &gt; 0[\/latex] such that for all [latex]x[\/latex] satisfying [latex]0 < |x-x_0| < \\delta[\/latex] the inequality [latex]|f(x)-L| < \\varepsilon[\/latex] holds.<\/p>\n<div style=\"text-align: center;\">[latex]f(x)=\\sqrt{x+4}, \\,\\, L=3, \\,\\, x_0=5, \\,\\, \\varepsilon=1[\/latex]<\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572601145\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572601145\" class=\"hidden-answer\" style=\"display: none\">First we will need to start with the inequality [latex]|f(x)-L| < \\varepsilon[\/latex] and plug in our numbers. Then we will solve for [latex]x[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]|\\sqrt(x+4) - 3| < \\varepsilon[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]|\\sqrt{x+4} - 3| < 1[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]-1 < \\sqrt{x+4} - 3 < 1[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]2 < \\sqrt{x+4} < 4[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]4 < x + 4 < 16[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]0 < x + 4 < 12[\/latex]<\/div>\n<p>Therefore, the interval is [latex](0,12)[\/latex]. For the second answer, we will start with\u00a0[latex]0 < |x-x_0| < \\delta[\/latex].\u00a0 We will plug in our value and solve:<\/p>\n<div style=\"text-align: center;\">[latex]|x-5| &lt; \\delta[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]-\\delta &lt; x-5 &lt; \\delta[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]5-\\delta &lt x &lt; 5+\\delta[\/latex]<\/div>\n<p>Now we will set each piece equal to the endpoints we found above.<\/p>\n<div style=\"text-align: center;\">[latex]5-\\delta=0[\/latex] and [latex]5+\\delta=12[\/latex]<\/div>\n<p>After solving we will get [latex]\\delta=5 \\text{ and }\\delta=7[\/latex]. Since the question is asking for the smallest interval, we choose the smaller number.\u00a0 Therefore the answer is [latex]\\delta=5[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">One-Sided and Infinite Limits<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">One-Sided Limits:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Limit from the right: [latex]\\lim_{x \\to a^+} f(x) = L[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Limit from the left: [latex]\\lim_{x \\to a^-} f(x) = L[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Modifications to standard epsilon-delta definition<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Infinite Limits:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Positive infinite limit: [latex]\\lim_{x \\to a} f(x) = +\\infty[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Negative infinite limit: [latex]\\lim_{x \\to a} f(x) = -\\infty[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Replace [latex]\\varepsilon[\/latex] with [latex]M[\/latex] for arbitrarily large values<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p class=\"font-bold\"><strong>Formal Definitions<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Limit from the Right: For every [latex]\\varepsilon > 0[\/latex], there exists [latex]\\delta > 0[\/latex] such that: If [latex]0 < x - a < \\delta[\/latex], then [latex]|f(x) - L| < \\varepsilon[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Limit from the Left: For every [latex]\\varepsilon > 0[\/latex], there exists [latex]\\delta > 0[\/latex] such that: If [latex]0 < a - x < \\delta[\/latex], then [latex]|f(x) - L| < \\varepsilon[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Positive Infinite Limit: For every [latex]M > 0[\/latex], there exists [latex]\\delta > 0[\/latex] such that: If [latex]0 < |x - a| < \\delta[\/latex], then [latex]f(x) > M[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Negative Infinite Limit: For every [latex]M > 0[\/latex], there exists [latex]\\delta > 0[\/latex] such that: If [latex]0 < |x - a| < \\delta[\/latex], then [latex]f(x) < -M[\/latex]<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1170571711163\">Find [latex]\\delta[\/latex] corresponding to [latex]\\varepsilon[\/latex]\u00a0for a proof that [latex]\\underset{x\\to 1^-}{\\lim}\\sqrt{1-x}=0[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q142584\">Show Solution<\/button><\/p>\n<div id=\"q142584\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\delta =\\varepsilon^2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Prove that [latex]\\lim_{x \\to 4^+} \\sqrt{x-4} = 0[\/latex].<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q789575\">Show Answer<\/button><\/p>\n<div id=\"q789575\" class=\"hidden-answer\" style=\"display: none\">\n<p class=\"whitespace-pre-wrap break-words\">Let [latex]\\varepsilon > 0[\/latex]. Choose [latex]\\delta = \\varepsilon^2[\/latex]. Assume [latex]0 < x - 4 < \\delta[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Then:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\begin{array}{rcl} 0 < x - 4 < \\delta &=& \\varepsilon^2 \\ 0 < \\sqrt{x-4} < \\varepsilon \\ |\\sqrt{x-4} - 0| < \\varepsilon \\end{array}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Therefore, [latex]\\lim_{x \\to 4^+} \\sqrt{x-4} = 0[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p>&nbsp;<\/p>\n","protected":false},"author":15,"menu_order":20,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":185,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2464"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2464\/revisions"}],"predecessor-version":[{"id":4716,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2464\/revisions\/4716"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/185"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2464\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=2464"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=2464"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=2464"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=2464"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}