{"id":72517,"date":"2021-05-06T08:48:51","date_gmt":"2021-05-06T13:48:51","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=72517"},"modified":"2026-03-28T11:17:51","modified_gmt":"2026-03-28T16:17:51","slug":"lhopitals-rule","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/lhopitals-rule\/","title":{"rendered":"L&#8217;Hopital&#8217;s Rule"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_pHhRsUAzPTY\" 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_pHhRsUAzPTY\" data-source-videoID=\"pHhRsUAzPTY\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"L&#8217;Hopital&#8217;s Rule Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"L&#8217;Hopital&#8217;s Rule\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_pHhRsUAzPTY:hover {cursor:pointer;} img#videoThumbnailImage_pHhRsUAzPTY {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/1751-l-hopital-s-rule-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_pHhRsUAzPTY\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_pHhRsUAzPTY\" 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\",\"L&#8217;Hopital&#8217;s Rule\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_pHhRsUAzPTY\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_pHhRsUAzPTY\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_pHhRsUAzPTY\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction TaY_Function() {\n  var x = document.getElementById(\"TaY\");\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=\"TaY_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=\"TaY\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#L%E2%80%99Hopital%E2%80%99s_Rule_and_Its_Significance\" class=\"smooth-scroll\">L\u2019H\u00f4pital\u2019s Rule and Its Significance<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Indeterminate_Forms_and_Limits\" class=\"smooth-scroll\">Indeterminate Forms and Limits<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Handling_Indeterminate_Forms\" class=\"smooth-scroll\">Handling Indeterminate Forms<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Examples\" class=\"smooth-scroll\">Examples<\/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>Hello and welcome to this video about l\u2019H\u00f4pital\u2019s rule! When taking certain types of limits, you\u2019ll find this 300-year-old rule can come in extremely handy.<\/p>\n<h2>L&#8217;H\u00f4pital&#8217;s Rule and Its Significance<\/h2>\n<p>\nGuillaume Fran\u00e7ois Antoine de l&#8217;H\u00f4pital was a French mathematician in the late 1600s who rubbed elbows with the likes of the Bernoulli brothers and one of the fathers of calculus himself, Gottfried Leibniz.<\/p>\n<p>In 1696, when he was about 35 years old, he published his book <em>Infinitesimal Calculus with Applications to Curved Lines<\/em>, the first textbook on infinitesimal calculus, and the first book to contain the famous rule. Though the rule was given to him by Johann Bernoulli two years prior, stirring some controversy, it bears l\u2019H\u00f4pital\u2019s name and is still in use today.<\/p>\n<h2><span id=\"Indeterminate_Forms_and_Limits\" class=\"m-toc-anchor\"><\/span>Indeterminate Forms and Limits<\/h2>\n<p>\nLet\u2019s start by taking some limits and establishing where l\u2019H\u00f4pital\u2019s rule comes into play and why it is so important.<\/p>\n<p>First, consider \\(\\lim_{x\\rightarrow 5}\\frac{x^2-25}{x-5}\\). Substituting 5 into the expression yields the indeterminate form \\(\\frac{0}{0}\\), but factoring the numerator results in \\(\\frac{(x+5)(x-5)}{(x-5)}\\), which reduces to \\(x+5\\). Substitute 5 again to find a limit of 10.<\/p>\n<p>Recall that an indeterminate form means that we cannot yet tell if a limit exists. The idea is that if we substitute the value the input approaches and the result is indeterminate, do some algebra and substitute again. If we get a limit (or there for sure isn\u2019t one), great. If the result is indeterminate, do some more algebra and try again. Repeat as needed.<\/p>\n<p>Take a look at this limit: \\(\\lim_{x\\rightarrow \u221e}\\frac{5x-2x^3}{3x^3+1}\\). As \\(x\\) approaches \\(\u221e\\), the function approaches \\(\\frac{\u221e}{\u221e}\\), indeterminate once again. We are only concerned with the cubed terms in the numerator and denominator because they\u2019ll grow the fastest and determine the overall limit.<\/p>\n<p>So, only considering those, the limit becomes \\(\\lim_{x\\rightarrow \u221e}\\frac{-2x^3}{3x^3}\\), which is \\(-\\frac{2}{3}\\).<\/p>\n<p>Unfortunately, these methods of finding limits do not always work.<\/p>\n<h2><span id=\"Handling_Indeterminate_Forms\" class=\"m-toc-anchor\"><\/span>Handling Indeterminate Forms<\/h2>\n<p>\nSuppose we were going to evaluate this limit: <\/p>\n<div class=\"equation-container\">\\(\\lim_{x\\rightarrow 0}\\frac{sin x}{x}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSubstituting 0, we\u2019ll get \\(\\frac{0}{0}\\). No problem, right? We got an indeterminate form, so all we need to do is a little bit of algebra. But now we have a serious problem. That quotient won\u2019t simplify algebraically.<\/p>\n<p>So we can\u2019t yet tell if there\u2019s a limit, but we have no algebraic way of finding one. Or do we?<\/p>\n<p>It\u2019s easy to see on the graph that this limit is 1 (note that the function is undefined at \\(x=0\\)), but of course graphs are not always practical.<\/p>\n<p>L&#8217;Hopital&#8217;s rule says we can do this:<\/p>\n<div class=\"equation-container\" style=\"text-align:center;overflow-y:hidden;overflow-x:auto;\">\\(\\lim_{x\\rightarrow 0}\\frac{sinx}{x}=\\lim_{x\\rightarrow 0}\\frac{\\frac{d}{dx}(sin x)}{\\frac{d}{dx}(x)}=\\lim_{x\\rightarrow 0}\\frac{cos x}{1}=cos 0=1\\)<\/div>\n<p>\n&nbsp;<br \/>\nIt\u2019s not a coincidence that we find a limit of 1, just as the graph showed.<\/p>\n<p>The idea is not exactly intuitive at first, but with a \\(\\frac{0}{0}\\) limit, it\u2019s an application of local linearity. Really, really close to \\(x=0\\), \\(sin x\\) looks like the line tangent to \\(sin x\\) there.<\/p>\n<p>In other words, since the derivative of \\(sin x\\) is \\(cos x\\), the graph of \\(sin x\\) close to 0 looks like its tangent line at 0, which has slope \\(cos(0)\\), or 1. Of course, the function in the denominator, \\(x\\), looks exactly the same.<\/p>\n<p>Because of local linearity, the ratio of function derivatives is usually a very good approximation of the ratio of function values.<\/p>\n<h2><span id=\"Examples\" class=\"m-toc-anchor\"><\/span>Examples<\/h2>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h3>\n<p>\nTake a look at this one. When we go to take the limit<\/p>\n<div class=\"equation-container\">\\(\\lim_{x\\rightarrow \u221e}\\frac{e^x}{2x^3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nwe end up with the indeterminate form \\(\\frac{\\infty}{\\infty}\\) with no way to simplify algebraically. In the case of \\(\\frac{0}{0}\\), we \u201czoom in\u201d really close to where we need to look.<\/p>\n<p>In an \\(\\frac{\\infty}{\\infty}\\) case like this, we\u2019ll use the same method, but look out at one of the ends. Since both functions approach infinity as \\(x\\) approaches infinity, the question is: which one goes to infinity faster?<\/p>\n<p>Using l&#8217;Hopital&#8217;s rule, we can do this: <\/p>\n<div class=\"equation-container\" style=\"text-align:center;overflow-y:hidden;overflow-x:auto;\">\\(\\lim_{x\\rightarrow \u221e}\\frac{ex^2}{x^2}=\\lim_{x\\rightarrow \u221e}\\frac{\\frac{d}{dx}(e^x)}{\\frac{d}{dx}(2x^2)}=\\lim_{x\\rightarrow \u221e}\\frac{e^x}{4x}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSince both functions still approach infinity, apply the rule again!<\/p>\n<div class=\"equation-container\" style=\"text-align:center;overflow-y:hidden;overflow-x:auto;\">\\(\\lim_{x\\rightarrow \u221e}\\frac{e^x}{4x}=\\lim_{x\\rightarrow \u221e}\\frac{\\frac{d}{dx}(e^x)}{\\frac{d}{dx}(4x)}=\\lim_{x\\rightarrow \u221e}\\frac{e^x}{4}=\u221e \\)<\/div>\n<p>\n&nbsp;<br \/>\nAfter the second application, it\u2019s clear we have a limit of infinity.<\/p>\n<p>Like many mathematical rules, l&#8217;Hopital&#8217;s rule won\u2019t always work 100% of the time. It is a powerful technique to learn because it will get you out of many tight spots!<\/p>\n<p>Here is the rule formally stated:<\/p>\n<div class=\"equation-container\" style=\"text-align:center;overflow-y:hidden;overflow-x:auto;\">\n \\(\\text{If }\\lim_{x\\rightarrow a}\\frac{f(x)}{g(x)}=\\frac{0}{0} \\text{ or } \\lim_{x\\rightarrow a}\\frac{f(x)}{g(x)}=\\frac{\u00b1\u221e}{\u00b1\u221e}, \\text{ then } \\lim_{x\\rightarrow a}\\frac{f(x)}{g(x)}=\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}\\)<\/div>\n<p>\n&nbsp;<\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h3>\n<p>\nLet\u2019s try a few more!<\/p>\n<div class=\"equation-container\">\\(\\lim_{x\\rightarrow 2}\\frac{x^3-7xT2+10x}{x^2-x-2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nOn the first attempt, the result is \\(\\frac{0}{0}\\). Time for l&#8217;Hopital&#8217;s rule!<\/p>\n<div class=\"equation-container\" style=\"text-align:center;overflow-y:hidden;overflow-x:auto;\">\\(\\lim_{x\\rightarrow 2}\\frac{x^3-7x^2+10x}{x^2-x-2}=\\lim_{x\\rightarrow 2}\\frac{3x^2-14x+10}{2x-1}=-\\frac{6}{3}=-2\\)<\/div>\n<p>\n&nbsp;<\/p>\n<h3><span id=\"Example_3\" class=\"m-toc-anchor\"><\/span>Example #3<\/h3>\n<p>\nOr what about:<\/p>\n<div class=\"equation-container\">\\(\\lim_{x\\rightarrow -3}\\frac{sin\u2061(\u03c0x)}{x^2-9}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe first substitution yields \\(\\frac{0}{0}\\).<\/p>\n<div class=\"equation-container\" style=\"text-align:center;overflow-y:hidden;overflow-x:auto;\">\\(\\lim_{x\\rightarrow -3}\\frac{sin\u2061(\u03c0x)}{x^2-9}=\\lim_{x\\rightarrow -3}\\frac{\u03c0cos (\u03c0x)}{2x}=\\frac{\u03c0}{6}\\)<\/div>\n<p>\n&nbsp;<br \/>\nLet\u2019s try a natural log:<\/p>\n<div class=\"equation-container\">\\(\\lim_{t\\rightarrow \u221e}\\frac{ln\u2061(5t)}{t^2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nBoth functions approach infinity as x approaches infinity. But which one approaches infinity faster?<\/p>\n<div class=\"equation-container\" style=\"text-align:center;overflow-y:hidden;overflow-x:auto;\">\\(\\lim_{t\\rightarrow \u221e}\\frac{ln\u2061(5t)}{t^2}=\\lim_{t\\rightarrow \u221e}\\frac{\\frac{1}{t}}{2t}=\\lim_{t\\rightarrow \u221e}\\frac{1}{2t^2}=0\\)<\/div>\n<p>\n&nbsp;<\/p>\n<hr>\n<p>\nSo that\u2019s l&#8217;Hopital&#8217;s rule, a valuable addition to any limit-taking repertoire. I hope that this video helped you understand the rule, when to use it, and how it works!<\/p>\n<p>Thanks for watching, and happy studying!<\/p>\n<\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/calculus\/\">Return to Calculus Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Calculus Videos<\/p>\n","protected":false},"author":1,"featured_media":100768,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-72517","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\/72517","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\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=72517"}],"version-history":[{"count":5,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/72517\/revisions"}],"predecessor-version":[{"id":280904,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/72517\/revisions\/280904"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/100768"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=72517"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}