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<!-- #BeginEditable "Body" -->
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<h1 class="center">L'Hôpital's Rule</h1>
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<p><span class="center">L'Hôpital's Rule</span> can help us calculate a <a href="limits.html">limit</a> that may otherwise be hard or impossible.</p>
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<div class="words">
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<p><span class="center">L'Hôpital is pronounced "lopital"</span>. He was a French mathematician from the 1600s.</p>
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</div>
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<p> </p>
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<p>It says that the <b>limit</b> when we divide one function by another is the same after we take the <a href="derivatives-introduction.html">derivative</a> of each function (with some special conditions shown later).</p>
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<p>In symbols we can write:</p>
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<p class="center large"><span class="lim"><em>lim</em><strong>x→c</strong></span><span class="intbl"><em>f(x)</em><strong>g(x)</strong></span> = <span class="lim"><em>lim</em><strong>x→c</strong></span><span class="intbl"><em>f’(x)</em><strong>g’(x)</strong></span></p>
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<!-- limx->c f(x)/g(x) = limx->c f'(x)/g'(x) -->
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<p class="center"><i>The limit as x approaches c of "f-of−x over g-of−x" equals the<br>
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the limit as x approaches c of "f-dash-of−x over g-dash-of−x"</i></p>
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<p>All we did is add that little dash mark <span class="hilite"> ’ </span> on each function, which means to take the derivative.</p>
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<div class="example">
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<h3>Example:
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<div class="center large"><span class="lim"><em>lim</em><strong>x→2</strong></span><span class="intbl"><em>x<sup>2</sup>+x−6</em><strong>x<sup>2</sup>−4</strong></span></div>
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<!-- limx->2 x^2~+x−6/x^2~−4 --></h3>
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<p>At <b>x=2</b> we would normally get:</p>
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<p class="center large"><span class="intbl"><em>2<sup>2</sup>+2−6</em><strong>2<sup>2</sup>−4</strong></span> = <span class="intbl"><em>0</em><strong>0</strong></span></p>
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<!---− 2^2~+2−6/2^2~−4 = 0/0 ---->
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<p>Which is <a href="../numbers/dividing-by-zero.html">indeterminate</a>, so we are stuck. Or are we?</p>
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<p>Let's try <span class="center">L'Hôpita</span>l!</p>
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<p>Differentiate both top and bottom (see <a href="derivatives-rules.html">Derivative Rules</a>):</p>
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<p class="center large"><span class="lim"><em>lim</em><strong>x→2</strong></span><span class="intbl"><em>x<sup>2</sup>+x−6</em><strong>x<sup>2</sup>−4</strong></span> = <span class="lim"><em>lim</em><strong>x→2</strong></span><span class="intbl"><em>2x+1−0</em><strong>2x−0</strong></span></p>
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<!-- limx->2 x^2~+x−6/x^2~−4 = limx->2 2x+1-0/2x-0 -->
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<p>Now we just substitute <b>x=2</b> to get our answer:</p>
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<p class="center large"><span class="lim"><em>lim</em><strong>x→2</strong></span><span class="intbl"><em>2x+1−0</em><strong>2x−0</strong></span> = <span class="intbl"><em>5</em><strong>4</strong></span></p>
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<!-- limx->2 2x+1-0/2x-0 = 5/4 -->
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<p>Here is the graph, notice the "hole" at x=2:</p>
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<p class="center"><img src="images/x2pxm6etc.svg" alt="(x^2+x-6)/(x^2-4)" height="330" width="570"></p>
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<p><i>Note: we can also get this answer by factoring, see <a href="limits-evaluating.html">Evaluating Limits</a></i>.</p>
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</div>
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<!-- limx->2 x^2~+x−6/x^2~−4 = limx->2 2x+1-0/2x-0 = 5/4 -->
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<div class="example">
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<h3>Example:
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<div class="center large"><span class="lim"><em>lim</em><strong>x→∞</strong></span><span class="intbl"><em>e<sup>x</sup></em><strong>x<sup>2</sup></strong></span></div>
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<!-- limx->INF e^x/x^2 --></h3>
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<p>Normally this is the result:</p>
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<p class="center large"><span class="lim"><em>lim</em><strong>x→∞</strong></span><span class="intbl"><em>e<sup>x</sup></em><strong>x<sup>2</sup></strong></span> = <span class="intbl"><em>∞</em><strong>∞</strong></span></p>
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<!-- limx->INF e^x/x^2 = INF/INF -->
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<p>Both head to infinity. Which is indeterminate.</p>
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<p>But let's differentiate both top and bottom (note that the derivative of e<sup>x</sup> is e<sup>x</sup>):</p>
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<p class="center large"><span class="lim"><em>lim</em><strong>x→∞</strong></span><span class="intbl"><em>e<sup>x</sup></em><strong>x<sup>2</sup></strong></span> = <span class="lim"><em>lim</em><strong>x→∞</strong></span><span class="intbl"><em>e<sup>x</sup></em><strong>2x</strong></span></p>
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<!-- limx->INF e^x/x^2 = limx->INF e^x/2x -->
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<p>Hmmm, still not solved, both tending towards infinity. But we can use it again:</p>
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<p class="center large"><span class="lim"><em>lim</em><strong>x→∞</strong></span><span class="intbl"><em>e<sup>x</sup></em><strong>x<sup>2</sup></strong></span> = <span class="lim"><em>lim</em><strong>x→∞</strong></span><span class="intbl"><em>e<sup>x</sup></em><strong>2x</strong></span> = <span class="lim"><em>lim</em><strong>x→∞</strong></span><span class="intbl"><em>e<sup>x</sup></em><strong>2</strong></span></p>
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<p><!-- limx->INF e^x/x^2 = limx->INF e^x/2x = limx->INF e^x/2 -->
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Now we have:</p>
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<p class="center large"><span class="lim"><em>lim</em><strong>x→∞</strong></span><span class="intbl"><em>e<sup>x</sup></em><strong>2</strong></span> = ∞</p>
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<!-- limx->INF e^x/2 = INF/0 = INF -->
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<p>It has shown us that e<sup>x</sup> grows much faster than x<sup>2</sup>.</p>
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</div>
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<h2>Cases</h2>
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<p>We have already seen a <span class="intbl"><em>0</em><strong>0</strong></span> and <span class="intbl"><em>∞</em><strong>∞</strong></span> example. Here are all the indeterminate forms that <span class="center">L'Hopital's Rule</span> may be able to help with:</p>
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<p class="center large"><span class="intbl"><em>0</em><strong>0</strong></span> <span class="intbl"><em>∞</em><strong>∞</strong></span> 0×∞ 1<sup>∞</sup> 0<sup>0</sup> ∞<sup>0</sup> ∞−∞</p>
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<!-- 0/0 INF/INF 0*INF 1^INF 0^0 INF^0 INF-INF -->
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<h2 id="cond">Conditions</h2>
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<h3>Differentiable</h3>
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<p>For a limit approaching c, the original functions must be differentiable either side of c, but not necessarily at c.</p>
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<p>Likewise g’(x) is not equal to zero either side of c.</p>
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<h3>The Limit Must Exist</h3>
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This limit must exist:
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<p class="center large"><span class="lim"><em>lim</em><strong>x→c</strong></span><span class="intbl"><em>f’(x)</em><strong>g’(x)</strong></span></p>
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<!-- limx->c f'(x)/g'(x) -->
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<p>Why? Well a good example is functions that never settle to a value.</p>
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<div class="example">
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<h3>Example:
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<div class="center large"><span class="lim"><em>lim</em><strong>x→∞</strong></span><span class="intbl"><em>x+cos(x)</em><strong>x</strong></span></div>
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<!-- limx->INF x+cos(x)/x --></h3>
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<p>Which is a <span class="intbl"><em>∞</em><strong>∞</strong></span> case. Let's differentiate top and bottom:</p>
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<p class="center large"><span class="lim"><em>lim</em><strong>x→∞</strong></span><span class="intbl"><em>1−sin(x)</em><strong>1</strong></span></p>
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<!-- limx->INF 1-sin(x)/1 -->
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<p>And because it just wiggles up and down it never approaches any value.</p>
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<p>So that new limit does not exist!</p>
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<p><b>And so <span class="center">L'Hôpita</span>l's Rule is not usable in this case.</b></p>
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<p>BUT we can do this:</p>
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<p class="center large"><span class="lim"><em>lim</em><strong>x→∞</strong></span><span class="intbl"><em>x+cos(x)</em><strong>x</strong></span> = <span class="lim"><em>lim</em><strong>x→∞</strong></span>(1 + <span class="intbl"><em>cos(x)</em><strong>x</strong></span>)</p>
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<!-- limx->INF x+cos(x)/x = limx->INF (1 + cos(x)/x ) = 1 -->
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<p>As x goes to infinity then <span class="intbl"><em>cos(x)</em><strong>x</strong></span> tends to between <span class="intbl"><em>−1</em><strong>∞</strong></span> and <span class="intbl"><em>+1</em><strong>∞</strong></span>, and both tend to zero.</p>
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<p>And we are left with just the "1", so:</p>
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<p class="center large"><span class="lim"><em>lim</em><strong>x→∞</strong></span><span class="intbl"><em>x+cos(x)</em><strong>x</strong></span> = <span class="lim"><em>lim</em><strong>x→∞</strong></span>(1 + <span class="intbl"><em>cos(x)</em><strong>x</strong></span>) = 1</p>
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<!-- limx->INF x+cos(x)/x = limx->INF (1 + cos(x)/x ) = 1 -->
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</div>
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<p> </p>
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<div class="related">
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<a href="limits.html">Limits (An Introduction)</a>
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<a href="index.html">Calculus Index</a>
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