{"id":862,"date":"2025-06-20T17:18:40","date_gmt":"2025-06-20T17:18:40","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=862"},"modified":"2025-07-29T15:47:14","modified_gmt":"2025-07-29T15:47:14","slug":"basics-of-differential-equations-learn-it-4-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/basics-of-differential-equations-learn-it-4-2\/","title":{"raw":"Sequences and Their Properties: Learn It 4","rendered":"Sequences and Their Properties: Learn It 4"},"content":{"raw":"<h2>Advanced Techniques for Sequence Limits<\/h2>\r\nIn the previous section, we learned the basic tools for finding sequence limits. Now we'll explore more sophisticated techniques that handle challenging cases where basic limit laws aren't enough.\r\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">L'H\u00f4pital's Rule for Sequences<\/h3>\r\n<p class=\"whitespace-normal break-words\">Sometimes you'll encounter sequences where both the numerator and denominator grow without bound, making it unclear what happens to their ratio. These indeterminate forms require special techniques.<\/p>\r\n<p class=\"whitespace-normal break-words\">We often need to analyze sequences featuring ratios where both parts increase without bound and it's not immediately clear what the limit will be. Fortunately, we can use L'H\u00f4pital's Rule from our study of functions.<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>L'H\u00f4pital's Rule<\/strong>\r\n<p class=\"whitespace-normal break-words\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval [latex](a, \\infty)[\/latex]. If either:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\underset{x\\to \\infty}{\\lim}f(x) = 0[\/latex] and [latex]\\underset{x\\to \\infty}{\\lim}g(x) = 0[\/latex]\r\n<strong>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 or<\/strong><\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\underset{x\\to \\infty}{\\lim}f(x) = \\infty[\/latex] (or [latex]-\\infty[\/latex]) and [latex]\\underset{x\\to \\infty}{\\lim}g(x) = \\infty[\/latex] (or [latex]-\\infty[\/latex])<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Then: [latex]\\underset{x\\to \\infty}{\\lim}\\frac{f(x)}{g(x)} = \\underset{x\\to \\infty}{\\lim}\\frac{f'(x)}{g'(x)}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">(assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]-\\infty[\/latex])<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169739012244\" data-type=\"problem\">\r\n<p id=\"fs-id1169739012246\">Consider the sequence [latex]\\{\\frac{(5{n}^{2}+1)}{{e}^{n}}\\}[\/latex]. Determine whether or not the sequence converges. If it converges, find its limit.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558891\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n<div id=\"fs-id1169736602009\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169736602015\">Use L\u2019H\u00f4pital\u2019s rule.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1169736601996\" data-type=\"solution\">\r\n<p id=\"fs-id1169736601998\">The sequence converges, and its limit is [latex]0[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6Wri3AvC6xg?controls=0&amp;start=400&amp;end=442&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n<p class=\"p1\">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>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.1.2_400to442_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.1.2\" here (opens in new window)<\/a>.\r\n\r\n<\/section>\r\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Limits with Continuous Functions<\/h3>\r\n<p class=\"whitespace-normal break-words\">Recall that if [latex]f[\/latex] is a continuous function at a value [latex]L[\/latex], then [latex]f(x)\\to f(L)[\/latex] as [latex]x\\to L[\/latex]. This idea applies to sequences as well.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\">Consider the sequence [latex]{\\sqrt{5-\\frac{3}{n^2}}}[\/latex]. We know that [latex]5-\\frac{3}{n^2} \\to 5[\/latex]. Since [latex]\\sqrt{x}[\/latex] is continuous at [latex]x = 5[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{n\\to \\infty}{\\lim}\\sqrt{5-\\frac{3}{n^2}} = \\sqrt{\\underset{n\\to \\infty}{\\lim}(5-\\frac{3}{n^2})} = \\sqrt{5}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>continuous functions and convergent sequences<\/h3>\r\n<p class=\"whitespace-normal break-words\">If sequence [latex]{a_n}[\/latex] converges to [latex]L[\/latex] and function [latex]f[\/latex] is continuous at [latex]L[\/latex], then:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{n\\to \\infty}f(a_n) = f(\\underset{n\\to \\infty}a_n) = f(L)[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169739375460\">Let [latex] &gt;0[\/latex]. Since [latex]f[\/latex] is continuous at [latex]L[\/latex], there exists [latex]\\delta &gt;0[\/latex] such that [latex]|f(x)-f(L)|&lt;\\epsilon [\/latex] if [latex]|x-L|&lt;\\delta [\/latex]. Since the sequence [latex]\\{{a}_{n}\\}[\/latex] converges to [latex]L[\/latex], there exists [latex]N[\/latex] such that [latex]|{a}_{n}-L|&lt;\\delta [\/latex] for all [latex]n\\ge N[\/latex]. Therefore, for all [latex]n\\ge N[\/latex], [latex]|{a}_{n}-L|&lt;\\delta [\/latex], which implies [latex]|f({a}_{n})\\text{-}f(L)|&lt;\\epsilon [\/latex]. We conclude that the sequence [latex]\\{f({a}_{n})\\}[\/latex] converges to [latex]f(L)[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>[caption id=\"\" align=\"aligncenter\" width=\"379\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234250\/CNX_Calc_Figure_09_01_006.jpg\" alt=\"A graph in quadrant 1 with points (a_1, f(a_1)), (a_3, f(a_3)), (L, f(L)), (a_4, f(a_4)), and (a_2, f(a_2)) connected by smooth curves.\" width=\"379\" height=\"269\" data-media-type=\"image\/jpeg\" \/> Figure 4. Because [latex]f[\/latex] is a continuous function as the inputs [latex]{a}_{1},{a}_{2},{a}_{3}\\text{,}\\ldots[\/latex] approach [latex]L[\/latex], the outputs [latex]f\\left({a}_{1}\\right),f\\left({a}_{2}\\right),f\\left({a}_{3}\\right)\\text{,}\\ldots[\/latex] approach [latex]f\\left(L\\right)[\/latex].[\/caption]<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169739062266\" data-type=\"problem\">\r\n<p id=\"fs-id1169739062272\">Determine whether the sequence [latex]\\{\\cos(\\frac{3}{{n}^{2}})\\}[\/latex] converges. If it converges, find its limit.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1169739062302\" data-type=\"solution\">\r\n<p id=\"fs-id1169739062304\">Since the sequence [latex]\\{\\frac{3}{{n}^{2}}\\}[\/latex] converges to [latex]0[\/latex] and [latex]\\cos{x}[\/latex] is continuous at [latex]x=0[\/latex], we can conclude that the sequence [latex]\\{\\cos(\\frac{3}{{n}^{2})\\}[\/latex] converges and<\/p>\r\n\r\n<div id=\"fs-id1169739062373\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\cos(\\frac{3}{{n}^{2}})=\\cos(0)=1[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">The Squeeze Theorem for Sequences<\/h3>\r\n<p class=\"whitespace-normal break-words\">Another powerful technique extends the Squeeze Theorem you learned for function limits. This method is particularly useful when you can't find a sequence's limit directly, but you can \"sandwich\" it between two sequences whose limits you do know.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>squeeze theorem for sequences<\/h3>\r\n<p id=\"fs-id1169736790732\">Consider sequences [latex]\\{{a}_{n}\\}[\/latex], [latex]\\{{b}_{n}\\}[\/latex], and [latex]\\{{c}_{n}\\}[\/latex]. Suppose there exists an integer [latex]N[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1169736790784\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}\\le {b}_{n}\\le {c}_{n}\\text{for all}n\\ge N[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736790825\">If there exists a real number [latex]L[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1169739209622\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=L=\\underset{n\\to \\infty }{\\text{lim}}{c}_{n}[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739209673\">then [latex]\\{{b}_{n}\\}[\/latex] converges and [latex]\\underset{n\\to \\infty }{\\text{lim}}{b}_{n}=L[\/latex] (Figure 5).<\/p>\r\n\r\n<\/section>&nbsp;\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"437\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234253\/CNX_Calc_Figure_09_01_007.jpg\" alt=\"A graph in quadrant 1 with the line y = L and the x-axis labeled as the n axis. Points are plotted above and below the line, converging to L as n goes to infinity. Points a_n, b_n, and c_n are plotted at the same n-value. A_n and b_n are above y = L, and c_n is below it.\" width=\"437\" height=\"266\" data-media-type=\"image\/jpeg\" \/> Figure 5. Each term [latex]{b}_{n}[\/latex] satisfies [latex]{a}_{n}\\le {b}_{n}\\le {c}_{n}[\/latex] and the sequences [latex]\\left\\{{a}_{n}\\right\\}[\/latex] and [latex]\\left\\{{c}_{n}\\right\\}[\/latex] converge to the same limit, so the sequence [latex]\\left\\{{b}_{n}\\right\\}[\/latex] must converge to the same limit as well.[\/caption]<section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169739209725\">Let [latex]\\epsilon &gt;0[\/latex]. Since the sequence [latex]\\{{a}_{n}\\}[\/latex] converges to [latex]L[\/latex], there exists an integer [latex]{N}_{1}[\/latex] such that [latex]|{a}_{n}-L|&lt;\\epsilon [\/latex] for all [latex]n\\ge {N}_{1}[\/latex]. Similarly, since [latex]\\{{c}_{n}\\}[\/latex] converges to [latex]L[\/latex], there exists an integer [latex]{N}_{2}[\/latex] such that [latex]|{c}_{n}-L|&lt;\\epsilon [\/latex] for all [latex]n\\ge {N}_{2}[\/latex]. By assumption, there exists an integer [latex]N[\/latex] such that [latex]{a}_{n}\\le {b}_{n}\\le {c}_{n}[\/latex] for all [latex]n\\ge N[\/latex]. Let [latex]M[\/latex] be the largest of [latex]{N}_{1},{N}_{2}[\/latex], and [latex]N[\/latex]. We must show that [latex]|{b}_{n}-L|&lt;\\epsilon [\/latex] for all [latex]n\\ge M[\/latex]. For all [latex]n\\ge M[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1169736707656\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{-}\\epsilon &lt;\\text{-}|{a}_{n}-L|\\le {a}_{n}-L\\le {b}_{n}-L\\le {c}_{n}-L\\le |{c}_{n}-L|&lt;\\epsilon [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739171825\">Therefore, [latex]\\text{-}\\epsilon &lt;{b}_{n}-L&lt;\\epsilon [\/latex], and we conclude that [latex]|{b}_{n}-L|&lt;\\epsilon [\/latex] for all [latex]n\\ge M[\/latex], and we conclude that the sequence [latex]\\{{b}_{n}\\}[\/latex] converges to [latex]L[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">\r\n<p class=\"whitespace-normal break-words\"><strong>When to Use the Squeeze Theorem<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Look for the Squeeze Theorem when:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">The sequence involves trigonometric functions (since [latex]-1 \\leq \\sin x, \\cos x \\leq 1[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">You have a sequence that oscillates but is bounded<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Direct limit calculation seems difficult, but you can find upper and lower bounds<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-pre-wrap break-words\">The key is identifying good \"squeezing\" sequences that are easier to analyze.<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169736700632\" data-type=\"problem\">\r\n<p id=\"fs-id1169736700637\">Use the Squeeze Theorem to find the limit of each of the following sequences.<\/p>\r\n\r\n<ol id=\"fs-id1169736700640\" type=\"a\">\r\n \t<li>[latex]\\{\\frac{\\cos{n}}{{n}^{2}}\\}[\/latex]<\/li>\r\n \t<li>[latex]\\{{(-\\frac{1}{2})}^{n}\\}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558869\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n<div id=\"fs-id1169736700699\" data-type=\"solution\">\r\n<ol id=\"fs-id1169736700701\" type=\"a\">\r\n \t<li>Since [latex]-1\\le \\cos{n}\\le 1[\/latex] for all integers [latex]n[\/latex], we have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736851869\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]-\\frac{1}{{n}^{2}}\\le \\frac{\\cos{n}}{{n}^{2}}\\le \\frac{1}{{n}^{2}}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]\\frac{-1}{{n}^{2}}\\to 0[\/latex] and [latex]\\frac{1}{{n}^{2}}\\to 0[\/latex], we conclude that [latex]\\frac{\\cos{n}}{{n}^{2}}\\to 0[\/latex] as well.<\/li>\r\n \t<li>Since<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736851981\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]-\\frac{1}{{2}^{n}}\\le {(-\\frac{1}{2})}^{n}\\le \\frac{1}{{2}^{n}}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor all positive integers [latex]n[\/latex], [latex]\\frac{-1}{{2}^{n}}\\to 0[\/latex] and [latex]\\frac{1}{{2}^{n}}\\to 0[\/latex], we can conclude that [latex]{(\\frac{-1}{2})}^{n}\\to 0[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<h2>Advanced Techniques for Sequence Limits<\/h2>\n<p>In the previous section, we learned the basic tools for finding sequence limits. Now we&#8217;ll explore more sophisticated techniques that handle challenging cases where basic limit laws aren&#8217;t enough.<\/p>\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">L&#8217;H\u00f4pital&#8217;s Rule for Sequences<\/h3>\n<p class=\"whitespace-normal break-words\">Sometimes you&#8217;ll encounter sequences where both the numerator and denominator grow without bound, making it unclear what happens to their ratio. These indeterminate forms require special techniques.<\/p>\n<p class=\"whitespace-normal break-words\">We often need to analyze sequences featuring ratios where both parts increase without bound and it&#8217;s not immediately clear what the limit will be. Fortunately, we can use L&#8217;H\u00f4pital&#8217;s Rule from our study of functions.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>L&#8217;H\u00f4pital&#8217;s Rule<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval [latex](a, \\infty)[\/latex]. If either:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\underset{x\\to \\infty}{\\lim}f(x) = 0[\/latex] and [latex]\\underset{x\\to \\infty}{\\lim}g(x) = 0[\/latex]<br \/>\n<strong>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 or<\/strong><\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\underset{x\\to \\infty}{\\lim}f(x) = \\infty[\/latex] (or [latex]-\\infty[\/latex]) and [latex]\\underset{x\\to \\infty}{\\lim}g(x) = \\infty[\/latex] (or [latex]-\\infty[\/latex])<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Then: [latex]\\underset{x\\to \\infty}{\\lim}\\frac{f(x)}{g(x)} = \\underset{x\\to \\infty}{\\lim}\\frac{f'(x)}{g'(x)}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">(assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]-\\infty[\/latex])<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169739012244\" data-type=\"problem\">\n<p id=\"fs-id1169739012246\">Consider the sequence [latex]\\{\\frac{(5{n}^{2}+1)}{{e}^{n}}\\}[\/latex]. Determine whether or not the sequence converges. If it converges, find its limit.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558891\">Hint<\/button><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169736602009\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169736602015\">Use L\u2019H\u00f4pital\u2019s rule.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558892\">Show Solution<\/button><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169736601996\" data-type=\"solution\">\n<p id=\"fs-id1169736601998\">The sequence converges, and its limit is [latex]0[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6Wri3AvC6xg?controls=0&amp;start=400&amp;end=442&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p class=\"p1\">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+II\/Transcripts\/5.1.2_400to442_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.1.2&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Limits with Continuous Functions<\/h3>\n<p class=\"whitespace-normal break-words\">Recall that if [latex]f[\/latex] is a continuous function at a value [latex]L[\/latex], then [latex]f(x)\\to f(L)[\/latex] as [latex]x\\to L[\/latex]. This idea applies to sequences as well.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\">Consider the sequence [latex]{\\sqrt{5-\\frac{3}{n^2}}}[\/latex]. We know that [latex]5-\\frac{3}{n^2} \\to 5[\/latex]. Since [latex]\\sqrt{x}[\/latex] is continuous at [latex]x = 5[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{n\\to \\infty}{\\lim}\\sqrt{5-\\frac{3}{n^2}} = \\sqrt{\\underset{n\\to \\infty}{\\lim}(5-\\frac{3}{n^2})} = \\sqrt{5}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>continuous functions and convergent sequences<\/h3>\n<p class=\"whitespace-normal break-words\">If sequence [latex]{a_n}[\/latex] converges to [latex]L[\/latex] and function [latex]f[\/latex] is continuous at [latex]L[\/latex], then:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{n\\to \\infty}f(a_n) = f(\\underset{n\\to \\infty}a_n) = f(L)[\/latex]<\/p>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<p id=\"fs-id1169739375460\">Let [latex]>0[\/latex]. Since [latex]f[\/latex] is continuous at [latex]L[\/latex], there exists [latex]\\delta >0[\/latex] such that [latex]|f(x)-f(L)|<\\epsilon[\/latex] if [latex]|x-L|<\\delta[\/latex]. Since the sequence [latex]\\{{a}_{n}\\}[\/latex] converges to [latex]L[\/latex], there exists [latex]N[\/latex] such that [latex]|{a}_{n}-L|<\\delta[\/latex] for all [latex]n\\ge N[\/latex]. Therefore, for all [latex]n\\ge N[\/latex], [latex]|{a}_{n}-L|<\\delta[\/latex], which implies [latex]|f({a}_{n})\\text{-}f(L)|<\\epsilon[\/latex]. We conclude that the sequence [latex]\\{f({a}_{n})\\}[\/latex] converges to [latex]f(L)[\/latex].<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<figure style=\"width: 379px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234250\/CNX_Calc_Figure_09_01_006.jpg\" alt=\"A graph in quadrant 1 with points (a_1, f(a_1)), (a_3, f(a_3)), (L, f(L)), (a_4, f(a_4)), and (a_2, f(a_2)) connected by smooth curves.\" width=\"379\" height=\"269\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 4. Because [latex]f[\/latex] is a continuous function as the inputs [latex]{a}_{1},{a}_{2},{a}_{3}\\text{,}\\ldots[\/latex] approach [latex]L[\/latex], the outputs [latex]f\\left({a}_{1}\\right),f\\left({a}_{2}\\right),f\\left({a}_{3}\\right)\\text{,}\\ldots[\/latex] approach [latex]f\\left(L\\right)[\/latex].<\/figcaption><\/figure>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169739062266\" data-type=\"problem\">\n<p id=\"fs-id1169739062272\">Determine whether the sequence [latex]\\{\\cos(\\frac{3}{{n}^{2}})\\}[\/latex] converges. If it converges, find its limit.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558890\">Show Solution<\/button><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739062302\" data-type=\"solution\">\n<p id=\"fs-id1169739062304\">Since the sequence [latex]\\{\\frac{3}{{n}^{2}}\\}[\/latex] converges to [latex]0[\/latex] and [latex]\\cos{x}[\/latex] is continuous at [latex]x=0[\/latex], we can conclude that the sequence [latex]\\{\\cos(\\frac{3}{{n}^{2})\\}[\/latex] converges and<\/p>\n<div id=\"fs-id1169739062373\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\cos(\\frac{3}{{n}^{2}})=\\cos(0)=1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">The Squeeze Theorem for Sequences<\/h3>\n<p class=\"whitespace-normal break-words\">Another powerful technique extends the Squeeze Theorem you learned for function limits. This method is particularly useful when you can&#8217;t find a sequence&#8217;s limit directly, but you can &#8220;sandwich&#8221; it between two sequences whose limits you do know.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>squeeze theorem for sequences<\/h3>\n<p id=\"fs-id1169736790732\">Consider sequences [latex]\\{{a}_{n}\\}[\/latex], [latex]\\{{b}_{n}\\}[\/latex], and [latex]\\{{c}_{n}\\}[\/latex]. Suppose there exists an integer [latex]N[\/latex] such that<\/p>\n<div id=\"fs-id1169736790784\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}\\le {b}_{n}\\le {c}_{n}\\text{for all}n\\ge N[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736790825\">If there exists a real number [latex]L[\/latex] such that<\/p>\n<div id=\"fs-id1169739209622\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=L=\\underset{n\\to \\infty }{\\text{lim}}{c}_{n}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739209673\">then [latex]\\{{b}_{n}\\}[\/latex] converges and [latex]\\underset{n\\to \\infty }{\\text{lim}}{b}_{n}=L[\/latex] (Figure 5).<\/p>\n<\/section>\n<p>&nbsp;<\/p>\n<figure style=\"width: 437px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234253\/CNX_Calc_Figure_09_01_007.jpg\" alt=\"A graph in quadrant 1 with the line y = L and the x-axis labeled as the n axis. Points are plotted above and below the line, converging to L as n goes to infinity. Points a_n, b_n, and c_n are plotted at the same n-value. A_n and b_n are above y = L, and c_n is below it.\" width=\"437\" height=\"266\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 5. Each term [latex]{b}_{n}[\/latex] satisfies [latex]{a}_{n}\\le {b}_{n}\\le {c}_{n}[\/latex] and the sequences [latex]\\left\\{{a}_{n}\\right\\}[\/latex] and [latex]\\left\\{{c}_{n}\\right\\}[\/latex] converge to the same limit, so the sequence [latex]\\left\\{{b}_{n}\\right\\}[\/latex] must converge to the same limit as well.<\/figcaption><\/figure>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<p id=\"fs-id1169739209725\">Let [latex]\\epsilon >0[\/latex]. Since the sequence [latex]\\{{a}_{n}\\}[\/latex] converges to [latex]L[\/latex], there exists an integer [latex]{N}_{1}[\/latex] such that [latex]|{a}_{n}-L|<\\epsilon[\/latex] for all [latex]n\\ge {N}_{1}[\/latex]. Similarly, since [latex]\\{{c}_{n}\\}[\/latex] converges to [latex]L[\/latex], there exists an integer [latex]{N}_{2}[\/latex] such that [latex]|{c}_{n}-L|<\\epsilon[\/latex] for all [latex]n\\ge {N}_{2}[\/latex]. By assumption, there exists an integer [latex]N[\/latex] such that [latex]{a}_{n}\\le {b}_{n}\\le {c}_{n}[\/latex] for all [latex]n\\ge N[\/latex]. Let [latex]M[\/latex] be the largest of [latex]{N}_{1},{N}_{2}[\/latex], and [latex]N[\/latex]. We must show that [latex]|{b}_{n}-L|<\\epsilon[\/latex] for all [latex]n\\ge M[\/latex]. For all [latex]n\\ge M[\/latex],<\/p>\n<div id=\"fs-id1169736707656\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{-}\\epsilon <\\text{-}|{a}_{n}-L|\\le {a}_{n}-L\\le {b}_{n}-L\\le {c}_{n}-L\\le |{c}_{n}-L|<\\epsilon[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739171825\">Therefore, [latex]\\text{-}\\epsilon <{b}_{n}-L<\\epsilon[\/latex], and we conclude that [latex]|{b}_{n}-L|<\\epsilon[\/latex] for all [latex]n\\ge M[\/latex], and we conclude that the sequence [latex]\\{{b}_{n}\\}[\/latex] converges to [latex]L[\/latex].<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\n<p class=\"whitespace-normal break-words\"><strong>When to Use the Squeeze Theorem<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Look for the Squeeze Theorem when:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">The sequence involves trigonometric functions (since [latex]-1 \\leq \\sin x, \\cos x \\leq 1[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">You have a sequence that oscillates but is bounded<\/li>\n<li class=\"whitespace-normal break-words\">Direct limit calculation seems difficult, but you can find upper and lower bounds<\/li>\n<\/ul>\n<p class=\"whitespace-pre-wrap break-words\">The key is identifying good &#8220;squeezing&#8221; sequences that are easier to analyze.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169736700632\" data-type=\"problem\">\n<p id=\"fs-id1169736700637\">Use the Squeeze Theorem to find the limit of each of the following sequences.<\/p>\n<ol id=\"fs-id1169736700640\" type=\"a\">\n<li>[latex]\\{\\frac{\\cos{n}}{{n}^{2}}\\}[\/latex]<\/li>\n<li>[latex]\\{{(-\\frac{1}{2})}^{n}\\}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558869\">Show Solution<\/button><\/p>\n<div id=\"q44558869\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169736700699\" data-type=\"solution\">\n<ol id=\"fs-id1169736700701\" type=\"a\">\n<li>Since [latex]-1\\le \\cos{n}\\le 1[\/latex] for all integers [latex]n[\/latex], we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736851869\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]-\\frac{1}{{n}^{2}}\\le \\frac{\\cos{n}}{{n}^{2}}\\le \\frac{1}{{n}^{2}}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]\\frac{-1}{{n}^{2}}\\to 0[\/latex] and [latex]\\frac{1}{{n}^{2}}\\to 0[\/latex], we conclude that [latex]\\frac{\\cos{n}}{{n}^{2}}\\to 0[\/latex] as well.<\/li>\n<li>Since<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736851981\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]-\\frac{1}{{2}^{n}}\\le {(-\\frac{1}{2})}^{n}\\le \\frac{1}{{2}^{n}}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor all positive integers [latex]n[\/latex], [latex]\\frac{-1}{{2}^{n}}\\to 0[\/latex] and [latex]\\frac{1}{{2}^{n}}\\to 0[\/latex], we can conclude that [latex]{(\\frac{-1}{2})}^{n}\\to 0[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":8,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":671,"module-header":"- Select Header -","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\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/862"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":4,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/862\/revisions"}],"predecessor-version":[{"id":1537,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/862\/revisions\/1537"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/671"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/862\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=862"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=862"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=862"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=862"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}