{"id":178,"date":"2019-08-27T17:38:35","date_gmt":"2019-08-27T17:38:35","guid":{"rendered":"http:\/\/104.42.120.246.xip.io\/calculus-tutorials\/?page_id=178"},"modified":"2019-12-02T21:32:52","modified_gmt":"2019-12-02T21:32:52","slug":"lhopitals-rule","status":"publish","type":"page","link":"https:\/\/math.hmc.edu\/calculus\/hmc-mathematics-calculus-online-tutorials\/single-variable-calculus\/lhopitals-rule\/","title":{"rendered":"L&#8217;Hopital&#8217;s Rule"},"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>L&#8217;H\u00f4pital&#8217;s Rule &#8211; HMC Calculus Tutorial<\/title>\n\n\n\n<p>\n<!------------------------>\n\nConsider the limit\n\\[\\lim_{x\\to a}\\, \\frac{f(x)}{g(x)}.\\]\nIf both the numerator and the denominator are finite at $a$ and\n$g(a)\\neq 0$, then \n\\[\\lim_{x\\to a}\\, \\frac{f(x)}{g(x)}=\\frac{f(a)}{g(a)}.\\]\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\n$\\displaystyle\\lim_{x\\to 3}\\,\n\\frac{x^2+1}{x+2}=\\frac{10}{5}=2$.  \n\n<\/p>\n\n\n\n<p>\nBut what happens if both the numerator and the denominator tend to\n$0$?  It is not clear what the limit is.  In fact, depending on what\nfunctions $f(x)$ and $g(x)$ are, the limit can be anything at all!\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\n$\\displaystyle\\begin{array}{l@{\\qquad\\qquad}l}\n\t\\displaystyle\\lim_{x\\to 0}\\, \\frac{x^3}{x^2}=\\lim_{x\\to 0} x=0. &amp;\n\t\\displaystyle\\lim_{x\\to 0}\\, \n\t\\frac{-x}{x^3}=\\lim_{x\\to 0} \\frac{-1}{x^2}=-\\infty.\\\\  \n\t\\displaystyle\\lim_{x\\to 0}\\, \\frac{x}{x^2}=\\lim_{x\\to 0} \\frac{1}{x}=\\infty. &amp; \n\t\\displaystyle\\lim_{x\\to 0}\\, \\frac{kx}{x}=\\lim_{x\\to 0} k=k.\n\\end{array}$\n\n<\/p>\n\n\n\n<p>\nThese limits are examples of <i>indeterminate forms<\/i> of type\n$\\frac{0}{0}$.  L&#8217;H\u00f4pital&#8217;s Rule provides a method\nfor evaluating such limits.  We will denote $\\displaystyle\\lim_{x\\to a}, \n\\lim_{x\\to a^+}, \\lim_{x\\to a^-}, \\lim_{x\\to \\infty}, {\\small\\textrm{ and }} \\lim_{x\\to \n-\\infty}$\ngenerically by $\\lim$ in what follows.\n\n<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">L&#8217;H\u00f4pital&#8217;s Rule for $\\displaystyle\\frac{0}{0}$<\/h4>\n\n\n\n<p>\n\nSuppose $\\lim f(x)=\\lim g(x)=0$.  Then\n<\/p>\n\n\n\n<ol class=\"wp-block-list\"><li> If $\\displaystyle \\lim\\, \\frac{f'(x)}{g'(x)}=L$, then $\\displaystyle \\lim\\, \\frac{f(x)}{g(x)}=\\lim \\frac{f'(x)}{g'(x)}=L$. <\/li><li> If $\\displaystyle \\lim\\, \\frac{f'(x)}{g'(x)}$ tends to $+\\infty$ or $-\\infty$ in the limit, then so does $\\displaystyle\\frac{f(x)}{g(x)}$.<\/li><\/ol>\n\n\n<style>.kt-accordion-id_22c309-f2 .kt-accordion-inner-wrap{column-gap:var(--global-kb-gap-md, 2rem);row-gap:0px;}.kt-accordion-id_22c309-f2 .kt-accordion-panel-inner{border-top-width:0px;border-right-width:0px;border-bottom-width:0px;border-left-width:0px;padding-top:var(--global-kb-spacing-sm, 1.5rem);padding-right:var(--global-kb-spacing-sm, 1.5rem);padding-bottom:var(--global-kb-spacing-sm, 1.5rem);padding-left:var(--global-kb-spacing-sm, 1.5rem);}.kt-accordion-id_22c309-f2 > .kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > 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.kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active{color:#ffffff;background:#444444;border-top-color:#444444;border-right-color:#444444;border-bottom-color:#444444;border-left-color:#444444;}.kt-accordion-id_22c309-f2:not( .kt-accodion-icon-style-basiccircle ):not( .kt-accodion-icon-style-xclosecircle ):not( .kt-accodion-icon-style-arrowcircle )  > .kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:after, .kt-accordion-id_22c309-f2:not( .kt-accodion-icon-style-basiccircle ):not( .kt-accodion-icon-style-xclosecircle ):not( .kt-accodion-icon-style-arrowcircle )  > .kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:before{background:#ffffff;}.kt-accordion-id_22c309-f2:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger{background:#ffffff;}.kt-accordion-id_22c309-f2:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:after, .kt-accordion-id_22c309-f2:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:before{background:#444444;}@media all and (max-width: 767px){.kt-accordion-id_22c309-f2 .kt-accordion-inner-wrap{display:block;}.kt-accordion-id_22c309-f2 .kt-accordion-inner-wrap .kt-accordion-pane:not(:first-child){margin-top:0px;}}<\/style>\n<div class=\"wp-block-kadence-accordion alignnone\"><div class=\"kt-accordion-wrap kt-accordion-wrap kt-accordion-id_22c309-f2 kt-accordion-has-2-panes kt-active-pane-0 kt-accordion-block kt-pane-header-alignment-left kt-accodion-icon-style-basic kt-accodion-icon-side-right\" style=\"max-width:none\"><div class=\"kt-accordion-inner-wrap\" data-allow-multiple-open=\"true\" data-start-open=\"none\">\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-1 kt-pane_2588d1-ab\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><div class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">Geometric Interpretation<\/span><\/div><div class=\"kt-blocks-accordion-icon-trigger\"><\/div><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<div class=\"wp-block-media-text alignwide has-media-on-the-right\" style=\"grid-template-columns:auto 47%\"><figure class=\"wp-block-media-text__media\"><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" width=\"344\" height=\"388\" src=\"https:\/\/i0.wp.com\/math.hmc.edu\/calculus-tutorials\/wp-content\/uploads\/sites\/3\/2019\/08\/lhopital_geometric.gif?resize=344%2C388&#038;ssl=1\" alt=\"Follow the image link for a complete description of the image\" class=\"wp-image-356\"\/><\/figure><div class=\"wp-block-media-text__content\">\n<p>\n\nConsider $\\displaystyle \\lim_{t\\to a^+}\\, \\frac{y(t)}{x(t)}$, where $x(a)=y(a)=0$.  At time $t$, the secant line through $(x(t), y(t))$ and $(0,0)$ has slope  \\[\\frac{y(t)-0}{x(t)-0}=\\frac{y(t)}{x(t)}.\\] As $t\\to a^+$, $x(t)\\to 0$ and $y(t)\\to 0$, and we expect the secant line to approximate the tangent line at $(0,0)$ better and better.  In the limit as $t\\to a^+$,  \\[{\\small\\textrm{slope of tangent at $(0,0)$}} = \\lim_{t\\to a^+}\\, \\frac{y(t)}{x(t)}.\\]  But we can also calculate the slope of the tangent line at $(0,0)$ as \n\n<\/p>\n<\/div><\/div>\n\n\n\n<p>\n\n \\[{\\small\\textrm{slope of tangent at $(0,0)$}} = \\left.\\frac{dy}{dx}\\right|_{x=0}=\\frac{dy\/dt|_{y=0}}{dx\/dt|_{x=0}}=\\lim_{t\\to a^+} \\frac{y'(t)}{x'(t)}.\\]  Thus,  \\[\\lim_{t\\to a^+}\\frac{y(t)}{x(t)}=\\lim_{t\\to a^+}\\frac{y'(t)}{x'(t)}.\\] This is an informal geometrical interpretation, and certainly not a proof, of L&#8217;H\u00f4pital&#8217;s Rule.  However, it does give us insight into the formal statement of the rule.  \n\n<\/p>\n<\/div><\/div><\/div>\n\n\n\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-2 kt-pane_1d247a-f7\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><div class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">Sketch of the Proof of L&#8217;H\u00f4pital&#8217;s Rule $\\displaystyle \\left(\\frac{0}{0} {\\small\\bf\\textrm{ Case}}\\right)$<\/span><\/div><div class=\"kt-blocks-accordion-icon-trigger\"><\/div><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<p>\n<!------------------------>\n\nWe will use an extension of the Mean Value Theorem:\n\n<\/p>\n\n\n\n<p>\n<b>Extended (Cauchy) Mean Value Theorem<\/b>\n<\/p>\n\n\n\n<p>\n\nLet $f$ and $g$ be differentiable on $(a,b)$ and continuous on\n$[a,b]$.  Suppose that $g'(x)\\neq 0$ in $(a,b)$.  Then there is at\nleast one point $c$ in $(a,b)$ such that\n\\[\\frac{f'(c)}{g'(c)}=\\frac{f(b)-f(a)}{g(b)-g(a)}.\\]\nThe proof of this theorem is fairly simple and can be found in most\ncalculus texts.\n\n<\/p>\n\n\n\n<p>\nWe will now sketch the proof of L&#8217;H\u00f4pital&#8217;s Rule for the\n$\\frac{0}{0}$ case in the limit as $x\\to c^+$, where $c$ is finite.\nThe case $x\\to c^-$ can be proven in a similar manner, and these two\ncases together can be used to prove L&#8217;H\u00f4pital&#8217;s Rule for a\ntwo-sided limit.  This proof is taken from Salas and Hille&#8217;s\n<i>Calculus:  One Variable<\/i>.\n\n<\/p>\n\n\n\n<p>\nLet $f$ and $g$ be defined on an interval $(c,b)$, where $f(x)\\to 0$\nand $g(x)\\to 0$ as $x\\to c^+$ but $\\displaystyle \\frac{f'(x)}{g'(x)}$\ntends to a finite limit $L$.  Then $f&#8217;$ and $g&#8217;$ exist on some set\n$(c, c+g]$ and $g&#8217;\\neq 0$ on $(c, c+h]$.  Also, $f$ and $g$ are\ncontinuous on $[c,c+h]$, where we define $f(c)=0$ and $g(c)=0$.\n\n<\/p>\n\n\n\n<p>\nBy the Extended Mean Value Theorem, there exists $c_h\\in (c,c+h)$ such\nthat \n\\[\\frac{f'(c_h)}{g'(c_h)}=\\frac{f(c+h)-f(c)}{g(c+h)-g(c)}=\\frac{f(c+h)}\n{g(c+h)}\\]\nsince $f(c)=g(c)=0$.  Letting $h\\to 0^+$, $\\displaystyle \\lim_{h\\to 0^+}\\,\n\\frac{f'(c_h)}{g'(c_h)}=\\lim_{x\\to c^+}\\, \\frac{f'(x)}{g'(x)}$ while\n$\\displaystyle  \\lim_{h\\to 0^+}\\,\n\\frac{f(c+h)}{g(c+h)}=\\lim_{x\\to c^+}\\, \\frac{f(x)}{g(x)}$.  Thus,\n\\[\\lim_{x\\to c^+}\\frac{f(x)}{g(x)}=\\lim_{x\\to c^+} \\frac{f'(x)}{g'(x)}.\\]   \n\n<!------------------------>\n\n\n<\/p>\n<\/div><\/div><\/div>\n<\/div><\/div><\/div>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<ul class=\"wp-block-list\"><li> $\\displaystyle \\lim_{x\\to 0}\\, \\frac{\\sin x}{x}=\\lim_{x\\to\n\t\t0}\\, \\frac{\\frac{d}{dx}(\\sin x)}{\\frac{d}{dx}(x)}=\\lim_{x\\to 0}\\,\n\t\t\\frac{\\cos x}{1}=1.$\n\t<\/li><li> $\\displaystyle \\lim_{x\\to 1}\\, \\frac{2\\ln x}{x-1}=\\lim_{x\\to\n\t\t1}\\, \\frac{\\frac{d}{dx}(2\\ln x)}{\\frac{d}{dx}(x-1)}=\\lim_{x\\to 1}\\,\n\t\t\\frac{~\\frac{2}{x}~}{1}=2.$\n\t<\/li><li> $\\displaystyle \\lim_{x\\to 0}\\, \\frac{e^x-1}{x^2}=\\lim_{x\\to\n\t\t0}\\, \\frac{\\frac{d}{dx}(e^x-1)}{\\frac{d}{dx}(x^2)}=\\lim_{x\\to 0}\\,\n\t\t\\frac{e^x}{2x}=\\text{does not exist}.$\n<\/li><\/ul>\n\n\n\n<p>\nIf the numerator and the denominator both tend to $\\infty$ or\n$-\\infty$, L&#8217;H\u00f4pital&#8217;s Rule still applies.\n\n<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">L&#8217;H\u00f4pital&#8217;s Rule for $\\displaystyle\\frac{\\infty}{\\infty}$<\/h4>\n\n\n\n<p>\n\nSuppose $\\lim f(x)$ and $\\lim g(x)$ are both infinite.  Then\n<\/p>\n\n\n\n<ol class=\"wp-block-list\"><li> If $\\displaystyle \\lim\\, \\frac{f'(x)}{g'(x)}=L$, then\n\t\t$\\displaystyle \\lim\\, \\frac{f(x)}{g(x)}=\\lim \\frac{f'(x)}{g'(x)}=L$.\n\t<\/li><li> If $\\displaystyle \\lim\\, \\frac{f'(x)}{g'(x)}$ tends to $+\\infty$\n\t\tor $-\\infty$ in the limit, then so does $\\displaystyle\\frac{f(x)}{g(x)}$.  \n<\/li><\/ol>\n\n\n\n<p>\nThe proof of this form of L&#8217;H\u00f4pital&#8217;s Rule requires more advanced\nanalysis.\n\n<\/p>\n\n\n\n<p>\nHere are some examples of indeterminate forms of type\n$\\displaystyle\\frac{\\infty}{\\infty}$.\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\n$\\displaystyle\\lim_{x\\to\\infty}\n\\frac{e^x}{x}=\\lim_{x\\to\\infty}  \\frac{e^x}{1}=\\infty.$\n\n<\/p>\n\n\n\n<p>\nSometimes it is necessary to use L&#8217;H\u00f4pital&#8217;s Rule several times in\nthe same problem:\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n$\\displaystyle\\lim_{x\\to 0} \\frac{1-\\cos\nx}{x^2}=\\lim_{x\\to 0}\\frac{\\sin \nx}{2x}=\\lim_{x\\to 0}\\frac{\\cos x}{2}=\\frac{1}{2}.$\n\n<\/p>\n\n\n\n<p>\nOccasionally, a limit can be re-written in order to apply\nL&#8217;H\u00f4pital&#8217;s Rule:\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n$\\displaystyle\\lim_{x\\to 0}\\, x\\ln\nx=\\lim_{x\\to 0}\\frac{\\ln x}{\\frac{1}{x}}=\\lim_{x\\to 0}\\,\n\\frac{~\\frac{1}{x}~}{-\\frac{1}{x^2}}=\\lim_{x\\to 0}\\, (-x)=0.$\n\n<\/p>\n\n\n\n<p>\nWe can use other tricks to apply L&#8217;H\u00f4pital&#8217;s Rule.  In the next\nexample, we use L&#8217;H\u00f4pital&#8217;s Rule to evaluate an indeterminate form\nof type $0^0$:\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\nTo evaluate $\\displaystyle \\lim_{x\\to 0^+}\\, x^x$, we will first\nevaluate $\\displaystyle \\lim_{x\\to 0^+}\\, \\ln (x^x)$.\n\\[\\lim_{x\\to 0^+}\\, \\ln (x^x)=\\lim_{x\\to 0^+}\\, x\\ln (x)=0,\\quad\n{\\small\\textrm{ by the previous example}}.\\]\nThen since $\\displaystyle\\lim_{x\\to 0^+}\\, \\ln (x^x)\\to 0$ as $x\\to\n0^+$ and $\\ln (u)=0$ if and only if $u=1$,\n\\[x^x\\to 1 \\quad\\textrm{as}\\quad x\\to 0^+.\\]\nThus, \n\\[\\lim_{x\\to 0^+}\\, x^x=1.\\]\n\n<\/p>\n\n\n\n<p>\nNotice that L&#8217;H\u00f4pital&#8217;s Rule only applies to indeterminate forms.\nFor the limit in the first example of this tutorial, L&#8217;H\u00f4pital&#8217;s\nRule does not apply and would give an incorrect result of 6.\nL&#8217;H\u00f4pital&#8217;s Rule is powerful and remarkably easy to use to\nevaluate indeterminate forms of type $\\frac{0}{0}$ and \n$\\frac{\\infty}{\\infty}$. \n\n<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<center>\n<h4>Key Concepts<\/h4>\n<\/center>\n\n\n\n<b>L&#8217;H\u00f4pital&#8217;s Rule for $\\frac{0}{0}$<\/b>\n\n\n\n<p>\nSuppose $\\lim f(x) = \\lim g(x) = 0$. Then \n<\/p>\n\n\n\n<ol class=\"wp-block-list\"><li> If $\\displaystyle\\lim \\frac{f'(x)}{g'(x)} = L,$ then $\\displaystyle\\lim \n\t\t\\frac{f(x)}{g(x)} = \\displaystyle\\lim \\frac{f'(x)}{g'(x)} = L$.\n\t<\/li><li> If $\\displaystyle\\lim \\frac{f'(x)}{g'(x)}$ tends to $+\\infty$ or \n\t\t$-\\infty$ in the limit, then so does $\\frac{f(x)}{g(x)}$.\n<\/li><\/ol>\n\n\n\n<p>\n<b>L&#8217;H\u00f4pital&#8217;s Rule for $\\frac{\\infty}{\\infty}$<\/b> <br>\nSuppose $\\lim f(x)$ and $\\lim g(x)$ are both infinite. Then\n<\/p>\n\n\n\n<ol class=\"wp-block-list\"><li> If $\\displaystyle\\lim \\frac{f'(x)}{g'(x)} = L$, then $\\displaystyle\\lim \n\t\t\\frac{f(x)}{g(x)} = \\displaystyle\\lim \\frac{f'(x)}{g'(x)} = L$.\n\t<\/li><li> If $\\displaystyle\\lim \\frac{f'(x)}{g'(x)}$ tends to $+\\infty$ or \n\t\t$-\\infty$ in the limit, then so does $\\frac{f(x)}{g(x)}$.\n<\/li><\/ol>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p>\n\n[<a href=\"https:\/\/physics.hmc.edu\/ct\/quiz\/QZ2210\/\">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>L&#8217;H\u00f4pital&#8217;s Rule &#8211; HMC Calculus Tutorial Consider the limit \\[\\lim_{x\\to a}\\, \\frac{f(x)}{g(x)}.\\] If both the numerator and the denominator are finite at $a$ and $g(a)\\neq 0$, then \\[\\lim_{x\\to a}\\, \\frac{f(x)}{g(x)}=\\frac{f(a)}{g(a)}.\\] Example $\\displaystyle\\lim_{x\\to 3}\\, \\frac{x^2+1}{x+2}=\\frac{10}{5}=2$. But what happens if both the numerator and the denominator tend to $0$? It is not clear what the limit is.&hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"parent":57,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[],"class_list":["post-178","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/178","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=178"}],"version-history":[{"count":6,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/178\/revisions"}],"predecessor-version":[{"id":1105,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/178\/revisions\/1105"}],"up":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/57"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/media?parent=178"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/tags?post=178"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}