{"id":152,"date":"2019-08-15T17:08:18","date_gmt":"2019-08-15T17:08:18","guid":{"rendered":"http:\/\/104.42.120.246.xip.io\/calculus-tutorials\/?page_id=152"},"modified":"2020-06-17T18:42:56","modified_gmt":"2020-06-17T18:42:56","slug":"computing-limits","status":"publish","type":"page","link":"https:\/\/math.hmc.edu\/calculus\/hmc-mathematics-calculus-online-tutorials\/precalculus\/computing-limits\/","title":{"rendered":"Computing Limits"},"content":{"rendered":"\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n\n\n\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<meta http-equiv=\"X-UA-Compatible\" content=\"IE=EmulateIE7\">\n\n\n\n<title>Computing Limits &#8211; HMC Calculus Tutorial<\/title>\n\n\n\n<p>  Intuitively, we say that $\\displaystyle \\lim_{x\\to c} f(x)=L$ if $f$ is defined near, but not necessarily at, $c$ and $f(x)$ approaches $L$ as $x$ approaches $c$.   <\/p>\n\n\n\n<p>\nIf we let $x$ approach $c$ from the left side only, we write\n$\\displaystyle \\lim_{x\\to c^-} f(x)$ since $x$ is approaching $c$ from\nsmaller values.  Similarly, for $x$ approaching $c$ from the right, we\nwrite $\\displaystyle \\lim_{x\\to c^+} f(x)$.  The <i>two-sided<\/i>\nlimit $\\displaystyle \\lim_{x\\to c} f(x)$ exists if and only if both of\nthese <i>one-sided<\/i> limits exist and are equal.\n\n<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">An Intuitive Example<\/h4>\n\n\n\n<p>\n\nConsider the graph of a function $f(x)$ shown below.\n<br>\n<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" width=\"375\" height=\"246\" src=\"https:\/\/i0.wp.com\/math.hmc.edu\/calculus\/wp-content\/uploads\/sites\/3\/2020\/06\/lim_graph.gif?resize=375%2C246&#038;ssl=1\" alt=\"Follow the link for a description of the image.\" class=\"wp-image-1197\"\/><\/figure><\/div>\n\n\n\n<p>\nEvaluate each of the following. Click each one to check your reasoning. \n\n\n<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Definition of the Limit<\/h4>\n\n\n\n<p>\n\nMore rigorously, let $f$ be defined at all $x$ in an open interval\ncontaining $c$, except possibly at $c$ itself.\n\n<\/p>\n\n\n\n<p> Then  \\[\\lim_{x\\to c} f(x)=L\\] if and only if for each $\\varepsilon &gt;0$, there exists a $\\delta &gt;0$ such that \\[{\\small\\textrm{if }} 0&lt;|x-c|&lt;\\delta {\\small\\textrm{ then }}|f(x)-L|&lt;\\varepsilon.\\]  In words, $\\displaystyle \\lim_{x\\to c} f(x)=L$ if and only if by taking $x$ close enough to $c$ we can get $f(x)$ arbitrarily close to $L$. <\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Properties of the Limit<\/h4>\n\n\n\n<p>\n\nEach of the following properties is proven using the rigorous\ndefinition of the limit.  Let $\\lim$ stand for $\\displaystyle\n\\lim_{x\\to c}$, $\\displaystyle \\lim_{x\\to c^+}$, or $\\displaystyle\n\\lim_{x\\to c^-}$.  Assume $\\lim f(x)$ and $\\lim g(x)$ both exist.\n<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li> <b>Uniqueness<\/b>&nbsp; If $\\lim f(x)=L_1$ and $\\lim f(x)=L_2$, then $L_1=L_2$. <br><br> <\/li><li> <b>Addition<\/b>&nbsp; $\\lim [f(x)+g(x)]=\\lim f(x) +\\lim g(x)$. <br><br> <\/li><li> <b>Scalar multiplication<\/b>&nbsp; $\\lim [\\alpha f(x)]=\\alpha \\lim f(x)$. <br><br> <\/li><li> <b>Multiplication<\/b>&nbsp; $\\lim [f(x)g(x)]=\\lim f(x) \\cdot \\lim g(x)$. <br><br> <\/li><li> <b>Division<\/b>&nbsp;  $\\displaystyle \\lim \\frac{f(x)}{g(x)}=\\frac{\\lim f(x)}{\\lim g(x)}$, provided $\\lim g(x)\\neq 0$. <br><br> <\/li><li> <b>Powers<\/b>&nbsp; $\\displaystyle \\lim [f(x)]^n=[\\lim f(x)]^n$ for any positive integer $n$. <\/li><\/ul>\n\n\n\n<p>\n\nIn practice, much of the time we can &#8220;reason out&#8221; the value of a\nlimit without explicitly using the $\\varepsilon$&#8211;$\\delta$ definition.\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<ul class=\"wp-block-list\"><li> $\\displaystyle\\lim_{x\\to 2} \\sqrt{x^2+12}=4$ since the function\n\t\t$f(x)=\\sqrt{x^2+12}$ is continuous at $x=2$ and $f(2)=4$.\n\n\t<\/li><li> $\\displaystyle\\lim_{x\\to \\infty} \\frac{1}{x}=0$ since as $x$\n\t\tincreases, $\\displaystyle \\frac{1}{x}$ gets arbitrarily close to $0$.\n\n\t<\/li><li> $\\displaystyle\\lim_{x\\to 0^+} \\ln |x|$ tends to $-\\infty$ and so\n\t\tdoes not exist since as $x$ decreases to $0$, $\\ln |x|$ gets\n\t\tarbitrarily large in magnitude and negative.\n\n\t<\/li><li> $\\displaystyle\\lim_{x\\to 3} \\frac{x^2-9}{x-3}=6$ even though\n\t\t$\\displaystyle f(x)=\\frac{x^2-9}{x-3}$ is undefined at $3$ since\n\t\t$\\displaystyle \\frac{x^2-9}{x-3}=x+3$ and $\\displaystyle \\lim_{x\\to 3}\n\t\tx+3 =6$.\n<\/li><\/ul>\n\n\n\n<p>\n\nWhat about something like $\\displaystyle\\lim_{x\\to 0} \\frac{\\sin\nx}{x}$?  When we cannot easily &#8220;reason out&#8221; the value of a limit, we\ncan often use numerical methods or L&#8217;H\u00f4pital&#8217;s Rule to determine\nthe value of the limit.  Can you convince yourself that \n$\\displaystyle\\lim_{x\\to 0} \\frac{\\sin x}{x}=1$?\n\n<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<center>\n<h3>Key Concepts<\/h3>\n<\/center>\n\n\n\n<p>\n\nLet $f$ be defined at all $x$ in an open interval\ncontaining $c$, except possibly at $c$ itself.\n\n<\/p>\n\n\n\n<p>\nThen \n\\[\\lim_{x\\to c} f(x)=L\\]\nif and only if for each $\\varepsilon &gt;0$, there exists a $\\delta &gt;0$\nsuch that\n\\[{\\small\\textrm{if }} 0&lt;|x-c|&lt;\\delta {\\small\\textrm{ then }}|f(x)-L|&lt;\\varepsilon.\\] \nIn words, $\\displaystyle \\lim_{x\\to c} f(x)=L$ if and only if by\ntaking $x$ close enough to $c$ we can get $f(x)$ arbitrarily close to $L$.\n\n<!------------------------>\n\n\n<br>\n\n<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p>\n\n[<a href=\"https:\/\/physics.hmc.edu\/ct\/quiz\/QZ2710\/\">I&#8217;m ready to take the quiz.<\/a>]\n[<a href=\"#top\">I need to review more.<\/a>]<br>\n\n\n<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Computing Limits &#8211; HMC Calculus Tutorial Intuitively, we say that $\\displaystyle \\lim_{x\\to c} f(x)=L$ if $f$ is defined near, but not necessarily at, $c$ and $f(x)$ approaches $L$ as $x$ approaches $c$. If we let $x$ approach $c$ from the left side only, we write $\\displaystyle \\lim_{x\\to c^-} f(x)$ since $x$ is approaching $c$ from&hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"parent":55,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[],"class_list":["post-152","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/152","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/comments?post=152"}],"version-history":[{"count":10,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/152\/revisions"}],"predecessor-version":[{"id":1198,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/152\/revisions\/1198"}],"up":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/55"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/media?parent=152"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/tags?post=152"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}