{"id":106077,"date":"2021-12-09T09:45:45","date_gmt":"2021-12-09T15:45:45","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=106077"},"modified":"2026-03-28T11:47:14","modified_gmt":"2026-03-28T16:47:14","slug":"limit-of-a-sequence","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/limit-of-a-sequence\/","title":{"rendered":"Limit of a Sequence"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_2jj6otzpWMM\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_2jj6otzpWMM\" data-source-videoID=\"2jj6otzpWMM\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Limit of a Sequence Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Limit of a Sequence\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_2jj6otzpWMM:hover {cursor:pointer;} img#videoThumbnailImage_2jj6otzpWMM {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/1726-thumb-final-2.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_2jj6otzpWMM\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_2jj6otzpWMM\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Limit of a Sequence\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_2jj6otzpWMM\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_2jj6otzpWMM\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_2jj6otzpWMM\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction s7t_Function() {\n  var x = document.getElementById(\"s7t\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"s7t_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"s7t\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Simple_Sequence_Example\" class=\"smooth-scroll\">Simple Sequence Example<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#More_Complex_Sequence_Example\" class=\"smooth-scroll\">More Complex Sequence Example<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Converging_on_an_Actual_Number\" class=\"smooth-scroll\">Converging on an Actual Number<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Sequence_Rules\" class=\"smooth-scroll\">Sequence Rules<\/a>\n<ul><\/li>\n<li class=\"toc-h3\"><a href=\"#Example_1\" class=\"smooth-scroll\">Example #1<\/a><\/li>\n<li class=\"toc-h3\"><a href=\"#Example_2\" class=\"smooth-scroll\">Example #2<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>\u201cThe limit does not exist!\u201d Cady Heron\u2019s Mathletes answer may have been correct in the classic teen comedy <em>Mean Girls<\/em>, but limits often do exist in math and today we\u2019re going to look at the <strong>limits of sequences<\/strong>. <\/p>\n<h2><span id=\"Simple_Sequence_Example\" class=\"m-toc-anchor\"><\/span>Simple Sequence Example<\/h2>\n<p>\nLet\u2019s look at the simplest sequence we can think of to get started: <\/p>\n<div class=\"examplesentence\">\\({1,2,3,4,5,6,7\u2026}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow this is a very simple sequence! It\u2019s the set of counting numbers. So the first term is 1, the second term is 2, the seventh term is 7, and so on. The \\(n^{ th}\\) term, therefore, must be \\(n\\).<\/p>\n<p>We could write a rule for this sequence as \\(x_{n}=n\\). As the term number rises, so does the value of that place in the sequence. To find the limit of a sequence, we need to think about what will happen as \\(n\\) gets bigger and bigger and approaches infinity.<\/p>\n<p>In this case, it\u2019s obvious. The value of the sequence will also approach infinity. But since infinity isn\u2019t really a number, this sequence does not really have a limit! In math language that means it <strong>diverges<\/strong>. Or we can simply say it is <strong>divergent<\/strong>. This simply means that the sequence does not converge on an actual number.<\/p>\n<h2><span id=\"More_Complex_Sequence_Example\" class=\"m-toc-anchor\"><\/span>More Complex Sequence Example<\/h2>\n<p>\nOther arithmetic sequences will also not converge on an actual number. Here\u2019s a slightly more complicated one: <\/p>\n<div class=\"examplesentence\">\\({7,4,1,-2,-5,-8\u2026}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis is an arithmetic sequence because each consecutive term is the same number apart. In this case, each term goes down by 3. The rule for this sequence is \\(a_{n}=-3n+10\\). We can explore where this one is going by finding the next terms in the sequence \\((-11, -14, -17, -20\u2026)\\) or we could find the thousandth term using our rule. <\/p>\n<div class=\"examplesentence\">\\(a_{1,000}=-3(1,000)+10=-2,990\\)<\/div>\n<p>\n&nbsp;<br \/>\nOr the millionth term: <\/p>\n<div class=\"examplesentence\">\\(a_{1,000,000}=-3(1,000,000)+10=-2,999,990\\)<\/div>\n<p>\n&nbsp;<br \/>\nIt\u2019s pretty clear where this one is going too. It\u2019s heading towards negative infinity, which means that this sequence also diverges. In math notation we would write it like this: <\/p>\n<div class=\"examplesentence\">\\(\\displaystyle\\lim_{n\\rightarrow \\infty }(-3n+10)=-\\infty\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis looks fancy but it\u2019s just denoting that we\u2019re finding the limit of the expression as \\(n\\) approaches infinity. That little bit under the \u201clim\u201d could actually be something else if we were dealing with functions instead of sequences, like \\(\\rightarrow 0\\).<\/p>\n<p>With sequences, we\u2019re dealing with positive values for the term numbers so we use \\(n\\) and try to discover what happens as \\(n\\) gets infinitely big. <\/p>\n<h2><span id=\"Converging_on_an_Actual_Number\" class=\"m-toc-anchor\"><\/span>Converging on an Actual Number<\/h2>\n<p>\nSo what kind of sequence does converge on an actual number? <\/p>\n<p>Let\u2019s find \\(\\displaystyle\\lim_{n\\rightarrow\\infty}\\frac{1}{n}\\).<\/p>\n<p>We can find the first several terms of this sequence by plugging in values for \\(n\\): <\/p>\n<div class=\"examplesentence\">\\(\\left \\{{\\frac{1}{1},\\frac{1}{2},\\frac{1}{3},\\frac{1}{4},\\frac{1}{5},\\frac{1}{6},\\frac{1}{7},\\frac{1}{8}}  \\right \\}\\)<\/div>\n<p>\n&nbsp;<br \/>\nIf we convert those fractions to decimals it looks like this: <\/p>\n<div class=\"examplesentence\">\\(\\left \\{ 1,0.5,0.\\overline3, 0.25,0.2,0.1\\overline6, 0.\\overline1\\overline3\\overline2\\overline8\\overline5\\overline7, 0.125\\right \\}\\)<\/div>\n<p>\n&nbsp;<br \/>\nOur sequence is not heading towards infinity! Or even negative infinity! It\u2019s getting smaller and smaller but staying positive. So what is it approaching? Let\u2019s find the hundredth term. That would be \\(\\frac{1}{100}\\), which is 0.01. Getting smaller still. The thousandth term is \\(\\frac{1}{1,000}\\), which is 0.001.<\/p>\n<p>I think we can see where it\u2019s headed. Zero! But here\u2019s the thing. It will never actually get there. It will just get closer and closer to 0 without ever actually being 0.<\/p>\n<p>We can say it <strong>converges<\/strong> on 0. Now we have our first convergent sequence. And more importantly, we have an important building block. We now know that \\(\\frac{1}{n}\\) has a limit of 0. And as we\u2019re about to see, that\u2019s very useful when finding the limit of other sequences. <\/p>\n<h2><span id=\"Sequence_Rules\" class=\"m-toc-anchor\"><\/span>Sequence Rules<\/h2>\n<p>\nFortunately, there are some rules for sequences that we can use with what we already know to help us find the limit of more complicated sequences. There\u2019s a rule for addition, subtraction, multiplication and division. In math notation they look like this: <\/p>\n<table class=\"ATable\" style=\"margin: auto; width: 70%;\">\n<tbody>\n<tr>\n<td>Addition Rule<\/td>\n<td>\\(\\displaystyle\\lim_{n\\rightarrow\\infty}(a_{n}+b_{n})=\\lim_{n\\rightarrow\\infty}a_{n}+\\lim_{n\\rightarrow\\infty}b_{n}\\)<\/td>\n<\/tr>\n<tr>\n<td>Subtraction Rule<\/td>\n<td>\\(\\displaystyle\\lim_{n\\rightarrow\\infty}(a_{n}-b_{n})=\\lim_{n\\rightarrow\\infty}a_{n}-\\lim_{n\\rightarrow\\infty}b_{n}\\)<\/td>\n<\/tr>\n<tr>\n<td>Multiplication Rule<\/td>\n<td>\\(\\displaystyle\\lim_{n\\rightarrow\\infty}(a_{n}b_{n})=\\lim_{n\\rightarrow\\infty}a_{n}\\lim_{n\\rightarrow\\infty}b_{n}\\)<\/td>\n<\/tr>\n<tr>\n<td>Division Rule<\/td>\n<td>\\(\\displaystyle\\lim_{n\\rightarrow\\infty}\\frac{a_{n}}{b_{n}}=\\frac{\\displaystyle\\lim_{n\\rightarrow\\infty}}{\\displaystyle\\lim_{n\\rightarrow \\infty}}if \\lim_{n\\rightarrow \\infty}b_{n}\\neq 0\\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\n&nbsp;<br \/>\nThese four rules simply mean that we can break apart our sequence rule, find the limits of the parts, and then put it back together. <\/p>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h3>\n<p>\nLet\u2019s apply the addition rule to a sequence to illustrate this idea. <\/p>\n<p>Find \\(\\displaystyle\\lim_{n\\rightarrow\\infty}\\left ( \\frac{1}{n}+2 \\right )\\).<\/p>\n<p>Our sequence has a rule of \\((\\frac{1}{n}+2)\\). To find the limit of the sequence, we can break it apart and find the \\(\\displaystyle\\lim_{n\\rightarrow\\infty}\\frac{1}{n}\\) and then find the \\(\\displaystyle\\lim_{n\\rightarrow\\infty} 2\\) and simply add the limits together.<\/p>\n<div class=\"examplesentence\">\\(\\displaystyle\\lim_{n\\rightarrow\\infty} (\\frac{1}{n}+2)=\\lim_{n\\rightarrow\\infty}\\frac{1}{n}+\\lim_{n\\rightarrow\\infty}2\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe\u2019ve already found the \\(\\displaystyle\\lim_{n\\rightarrow\\infty}\\frac{1}{n}=0\\). So we could say that the limit for the first term of our expression is 0. So \\(\\displaystyle\\lim_{x\\rightarrow \\infty }a_{n}=0\\). But what is the limit of 2? For any constant sequence, the limit is the constant. The sequence represented by \\(b_{n}=2\\) looks like this: <\/p>\n<div class=\"examplesentence\">\\({2,2,2,2,2,2\u2026}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSince every term is 2, the limit is 2. This is kind of obvious since it has nowhere else to go. Anyway, since we know the limit of the two parts, we can add them together. <\/p>\n<div class=\"examplesentence\">\\(\\displaystyle\\lim_{x\\rightarrow \\infty } \\left ( \\frac{1}{n} +2\\right )=0+2=2\\)<\/div>\n<p>\n&nbsp;<\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h3>\n<p>\nNow let\u2019s try one that lets us use the subtraction and multiplication rules. <\/p>\n<p>Find \\(\\displaystyle\\lim_{x\\rightarrow \\infty } \\left (1-3n^{-1} \\right )\\).<\/p>\n<p>Here we have subtraction so we can break up our sequence: <\/p>\n<div class=\"examplesentence\">\\(\\displaystyle\\lim_{n\\rightarrow\\infty}(1-3n^{-1})=\\lim_{n\\rightarrow \\infty }1-\\lim_{n\\rightarrow \\infty }3n^{-1}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe limit of 1 is 1, as we saw in the last problem. Now we need to find the limit of \\(3n^{-1}\\). We can break that part down using the multiplication rule. <\/p>\n<div class=\"examplesentence\">\\(\\displaystyle\\lim_{n\\rightarrow\\infty}(1-3n^{-1})=\\lim_{n\\rightarrow \\infty } 1-(\\lim_{n\\rightarrow \\infty }3\\cdot \\lim_{n\\rightarrow \\infty }n^{-1})\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe \\(\\displaystyle\\lim_{n\\rightarrow \\infty }3=3\\). Once again, we have the limit of a constant. The \\(\\displaystyle\\lim_{n\\rightarrow \\infty }n^{-1}\\) we also know, though this problem hides it a little bit. \\(n_{-1}\\) is \\(\\frac{1}{n}\\), which we know has a limit of 0. So after all that, here\u2019s what we know: <\/p>\n<div class=\"examplesentence\">\\(\\displaystyle\\lim_{n\\rightarrow \\infty }(1-3n^{-1})=1-(3\\cdot 0)=1\\)<\/div>\n<p>\n&nbsp;<br \/>\nAfter finding our three parts we do some simple math to find our limit is 1. Let\u2019s take a look at the first few terms of our sequence to see if this makes sense. <\/p>\n<div class=\"examplesentence\">\\(\\left \\{ -2, -\\frac{1}{2}, 0,\\frac{1}{4},\\frac{2}{5},\\frac{1}{2},\\frac{4}{7},\\frac{5}{8},\\frac{2}{3} \\right \\}\\)<\/div>\n<p>\n&nbsp;<br \/>\nIt does look like it is climbing towards 1. Let\u2019s find the 100th term in the sequence just to make sure it hasn\u2019t gone over the top. <\/p>\n<div class=\"examplesentence\">\\(a_{100}=1-3(100)^{-1}=\\frac{97}{100}=0.97\\)<\/div>\n<p>\n&nbsp;<br \/>\nLooks like it\u2019s still getting closer to 1. If we look at the 1,000th term, we\u2019d get 0.997, so you can see that we\u2019re getting closer and closer to 1 but will never actually reach it.<\/p>\n<hr>\n<p>\nThat\u2019s it for our introduction to the limits of sequences. I hope this video was helpful. Thanks for watching and happy studying!<\/p>\n<ul class=\"citelist\">\n<li><a href=\"https:\/\/www.mathsisfun.com\/algebra\/sequences-series.html\"target=\"_blank\">\u201cSequences.\u201d Mathsisfun.com<\/a><\/li>\n<li><a href=\"https:\/\/www.mathopenref.com\/calcsequence.html\"target=\"_blank\">\u201cCalculus &#8211; Sequences &#8211; Math Open Reference.\u201d n.d.<\/a><\/li>\n<li><a href=\"https:\/\/www.mathsisfun.com\/calculus\/limits-infinity.html\"target=\"_blank\">\u201cLimits to Infinity.\u201d n.d.<\/a><\/li>\n<li><a href=\"https:\/\/www.youtube.com\/watch?v=XdkoTb8PEG0\"target=\"_blank\">The Organic Chemistry Tutor. \u201cConverging and Diverging Sequences Using Limits &#8211; Practice Problems.\u201d YouTube<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/\">Mometrix Academy &#8211; Home<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Mometrix Academy &#8211; Home<\/p>\n","protected":false},"author":22,"featured_media":106083,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-106077","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-calculus-videos","7":"page_category-video-pages-for-study-course-sidebar-ad","8":"page_type-video","9":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/106077","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/22"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=106077"}],"version-history":[{"count":5,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/106077\/revisions"}],"predecessor-version":[{"id":261283,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/106077\/revisions\/261283"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/106083"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=106077"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}