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<li><a href="#idea-behind-lhospitals-rule" id="toc-idea-behind-lhospitals-rule" class="nav-link active" data-scroll-target="#idea-behind-lhospitals-rule"> <span class="header-section-number">32.1</span> Idea behind LHospitals rule</a></li>
<li><a href="#generalizations" id="toc-generalizations" class="nav-link" data-scroll-target="#generalizations"> <span class="header-section-number">32.2</span> Generalizations</a></li>
<li><a href="#other-indeterminate-forms" id="toc-other-indeterminate-forms" class="nav-link" data-scroll-target="#other-indeterminate-forms"> <span class="header-section-number">32.3</span> Other indeterminate forms</a></li>
<li><a href="#the-assumptions-are-necessary" id="toc-the-assumptions-are-necessary" class="nav-link" data-scroll-target="#the-assumptions-are-necessary"> <span class="header-section-number">32.4</span> The assumptions are necessary</a></li>
<li><a href="#questions" id="toc-questions" class="nav-link" data-scroll-target="#questions"> <span class="header-section-number">32.5</span> Questions</a></li>
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<h1 class="title d-none d-lg-block"><span class="chapter-number">32</span>&nbsp; <span class="chapter-title">LHospitals Rule</span></h1>
</div>
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<p>This section uses these add-on packages:</p>
<div class="sourceCode cell-code" id="cb1"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb1-1"><a href="#cb1-1" aria-hidden="true" tabindex="-1"></a><span class="im">using</span> <span class="bu">CalculusWithJulia</span></span>
<span id="cb1-2"><a href="#cb1-2" aria-hidden="true" tabindex="-1"></a><span class="im">using</span> <span class="bu">Plots</span></span>
<span id="cb1-3"><a href="#cb1-3" aria-hidden="true" tabindex="-1"></a><span class="im">using</span> <span class="bu">SymPy</span></span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<hr>
<p>Lets return to limits of the form <span class="math inline">\(\lim_{x \rightarrow c}f(x)/g(x)\)</span> which have an indeterminate form of <span class="math inline">\(0/0\)</span> if both are evaluated at <span class="math inline">\(c\)</span>. The typical example being the limit considered by Euler:</p>
<p><span class="math display">\[
\lim_{x\rightarrow 0} \frac{\sin(x)}{x}.
\]</span></p>
<p>We know this is <span class="math inline">\(1\)</span> using a bound from geometry, but might also guess this is one, as we know from linearization near <span class="math inline">\(0\)</span> that we have <span class="math inline">\(\sin(x) \approx x\)</span> or, more specifically:</p>
<p><span class="math display">\[
\sin(x) = x - \sin(\xi)x^2/2, \quad 0 &lt; \xi &lt; x.
\]</span></p>
<p>This would yield:</p>
<p><span class="math display">\[
\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = \lim_{x\rightarrow 0} \frac{x -\sin(\xi) x^2/2}{x} = \lim_{x\rightarrow 0} 1 + \sin(\xi) \cdot x/2 = 1.
\]</span></p>
<p>This is because we know <span class="math inline">\(\sin(\xi) x/2\)</span> has a limit of <span class="math inline">\(0\)</span>, when <span class="math inline">\(|\xi| \leq |x|\)</span>.</p>
<p>That doesnt look any easier, as we worried about the error term, but if just mentally replaced <span class="math inline">\(\sin(x)\)</span> with <span class="math inline">\(x\)</span> - which it basically is near <span class="math inline">\(0\)</span> - then we can see that the limit should be the same as <span class="math inline">\(x/x\)</span> which we know is <span class="math inline">\(1\)</span> without thinking.</p>
<p>Basically, we found that in terms of limits, if both <span class="math inline">\(f(x)\)</span> and <span class="math inline">\(g(x)\)</span> are <span class="math inline">\(0\)</span> at <span class="math inline">\(c\)</span>, that we <em>might</em> be able to just take this limit: <span class="math inline">\((f(c) + f'(c) \cdot(x-c)) / (g(c) + g'(c) \cdot (x-c))\)</span> which is just <span class="math inline">\(f'(c)/g'(c)\)</span>.</p>
<p>Wouldnt that be nice? We could find difficult limits just by differentiating the top and the bottom at <span class="math inline">\(c\)</span> (and not use the messy quotient rule).</p>
<p>Well, in fact that is more or less true, a fact that dates back to <a href="http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule">LHospital</a> - who wrote the first textbook on differential calculus - though this result is likely due to one of the Bernoulli brothers.</p>
<blockquote class="blockquote">
<p><em>LHospitals rule</em>: Suppose:</p>
<ul>
<li>that <span class="math inline">\(\lim_{x\rightarrow c+} f(c) =0\)</span> and <span class="math inline">\(\lim_{x\rightarrow c+} g(c) =0\)</span>,</li>
<li>that <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span> are differentiable in <span class="math inline">\((c,b)\)</span>, and</li>
<li>that <span class="math inline">\(g(x)\)</span> exists and is non-zero for <em>all</em> <span class="math inline">\(x\)</span> in <span class="math inline">\((c,b)\)</span>,</li>
</ul>
<p>then <strong>if</strong> the following limit exists: <span class="math inline">\(\lim_{x\rightarrow c+}f'(x)/g'(x)=L\)</span> it follows that <span class="math inline">\(\lim_{x \rightarrow c+}f(x)/g(x) = L\)</span>.</p>
</blockquote>
<p>That is <em>if</em> the right limit of <span class="math inline">\(f(x)/g(x)\)</span> is indeterminate of the form <span class="math inline">\(0/0\)</span>, but the right limit of <span class="math inline">\(f'(x)/g'(x)\)</span> is known, possibly by simple continuity, then the right limit of <span class="math inline">\(f(x)/g(x)\)</span> exists and is equal to that of <span class="math inline">\(f'(x)/g'(x)\)</span>.</p>
<p>The rule equally applies to <em>left limits</em> and <em>limits</em> at <span class="math inline">\(c\)</span>. Later it will see there are other generalizations.</p>
<p>To apply this rule to Eulers example, <span class="math inline">\(\sin(x)/x\)</span>, we just need to consider that:</p>
<p><span class="math display">\[
L = 1 = \lim_{x \rightarrow 0}\frac{\cos(x)}{1},
\]</span></p>
<p>So, as well, <span class="math inline">\(\lim_{x \rightarrow 0} \sin(x)/x = 1\)</span>.</p>
<p>This is due to <span class="math inline">\(\cos(x)\)</span> being continuous at <span class="math inline">\(0\)</span>, so this limit is just <span class="math inline">\(\cos(0)/1\)</span>. (More importantly, the tangent line expansion of <span class="math inline">\(\sin(x)\)</span> at <span class="math inline">\(0\)</span> is <span class="math inline">\(\sin(0) + \cos(0)x\)</span>, so that <span class="math inline">\(\cos(0)\)</span> is why this answer is as it is, but we dont need to think in terms of <span class="math inline">\(\cos(0)\)</span>, but rather the tangent-line expansion, which is <span class="math inline">\(\sin(x) \approx x\)</span>, as <span class="math inline">\(\cos(0)\)</span> appears as the coefficient.</p>
<div class="callout-note callout callout-style-default callout-captioned">
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<p>In <a href="http://www.cybertester.com/data/gruntz.pdf">Gruntz</a>, in a reference attributed to Speiss, we learn that LHospital was a French Marquis who was taught in <span class="math inline">\(1692\)</span> the calculus of Leibniz by Johann Bernoulli. They made a contract obliging Bernoulli to leave his mathematical inventions to LHospital in exchange for a regular compensation. This result was discovered in <span class="math inline">\(1694\)</span> and appeared in LHospitals book of <span class="math inline">\(1696\)</span>.</p>
</div>
</div>
<section id="examples" class="level5">
<h5 class="anchored" data-anchor-id="examples">Examples</h5>
<ul>
<li>Consider this limit at <span class="math inline">\(0\)</span>: <span class="math inline">\((a^x - 1)/x\)</span>. We have <span class="math inline">\(f(x) =a^x-1\)</span> has <span class="math inline">\(f(0) = 0\)</span>, so this limit is indeterminate of the form <span class="math inline">\(0/0\)</span>. The derivative of <span class="math inline">\(f(x)\)</span> is <span class="math inline">\(f'(x) = a^x \log(a)\)</span> which has <span class="math inline">\(f'(0) = \log(a)\)</span>. The derivative of the bottom is also <span class="math inline">\(1\)</span> at <span class="math inline">\(0\)</span>, so we have:</li>
</ul>
<p><span class="math display">\[
\log(a) = \frac{\log(a)}{1} = \frac{f'(0)}{g'(0)} = \lim_{x \rightarrow 0}\frac{f'(x)}{g'(x)} = \lim_{x \rightarrow 0}\frac{f(x)}{g(x)}
= \lim_{x \rightarrow 0}\frac{a^x - 1}{x}.
\]</span></p>
<div class="callout-note callout callout-style-default callout-captioned">
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<p>Why rewrite in the “opposite” direction? Because the theorems result <span class="math inline">\(L\)</span> is the limit is only true if the related limit involving the derivative exists. We dont do this in the following, but did so here to emphasize the need for the limit of the ratio of the derivatives to exist.</p>
</div>
</div>
<ul>
<li>Consider this limit:</li>
</ul>
<p><span class="math display">\[
\lim_{x \rightarrow 0} \frac{e^x - e^{-x}}{x}.
\]</span></p>
<p>It too is of the indeterminate form <span class="math inline">\(0/0\)</span>. The derivative of the top is <span class="math inline">\(e^x + e^{-x}\)</span>, which is <span class="math inline">\(2\)</span> when <span class="math inline">\(x=0\)</span>, so the ratio of <span class="math inline">\(f'(0)/g'(0)\)</span> is seen to be <span class="math inline">\(2\)</span> By continuity, the limit of the ratio of the derivatives is <span class="math inline">\(2\)</span>. Then by LHospitals rule, the limit above is <span class="math inline">\(2\)</span>.</p>
<ul>
<li>Sometimes, LHospitals rule must be applied twice. Consider this limit:</li>
</ul>
<p><span class="math display">\[
\lim_{x \rightarrow 0} \frac{\cos(x)}{1 - x^2}
\]</span></p>
<p>By LHospitals rule <em>if</em> this following limit exists, the two will be equal:</p>
<p><span class="math display">\[
\lim_{x \rightarrow 0} \frac{-\sin(x)}{-2x}.
\]</span></p>
<p>But if we didnt guess the answer, we see that this new problem is <em>also</em> indeterminate of the form <span class="math inline">\(0/0\)</span>. So, repeating the process, this new limit will exist and be equal to the following limit, should it exist:</p>
<p><span class="math display">\[
\lim_{x \rightarrow 0} \frac{-\cos(x)}{-2} = 1/2.
\]</span></p>
<p>As <span class="math inline">\(L = 1/2\)</span> for this related limit, it must also be the limit of the original problem, by LHospitals rule.</p>
<ul>
<li>Our “intuitive” limits can bump into issues. Take for example the limit of <span class="math inline">\((\sin(x)-x)/x^2\)</span> as <span class="math inline">\(x\)</span> goes to <span class="math inline">\(0\)</span>. Using <span class="math inline">\(\sin(x) \approx x\)</span> makes this look like <span class="math inline">\(0/x^2\)</span> which is still indeterminate. (Because the difference is higher order than <span class="math inline">\(x\)</span>.) Using LHospitals, says this limit will exist (and be equal) if the following one does:</li>
</ul>
<p><span class="math display">\[
\lim_{x \rightarrow 0} \frac{\cos(x) - 1}{2x}.
\]</span></p>
<p>This particular limit is indeterminate of the form <span class="math inline">\(0/0\)</span>, so we again try LHospitals rule and consider</p>
<p><span class="math display">\[
\lim_{x \rightarrow 0} \frac{-\sin(x)}{2} = 0
\]</span></p>
<p>So as this limit exists, working backwards, the original limit in question will also be <span class="math inline">\(0\)</span>.</p>
<ul>
<li>This example comes from the Wikipedia page. It “proves” a discrete approximation for the second derivative.</li>
</ul>
<p>Show if <span class="math inline">\(f''(x)\)</span> exists at <span class="math inline">\(c\)</span> and is continuous at <span class="math inline">\(c\)</span>, then</p>
<p><span class="math display">\[
f''(c) = \lim_{h \rightarrow 0} \frac{f(c + h) - 2f(c) + f(c-h)}{h^2}.
\]</span></p>
<p>This will follow from two applications of LHospitals rule to the right-hand side. The first says, the limit on the right is equal to this limit, should it exist:</p>
<p><span class="math display">\[
\lim_{h \rightarrow 0} \frac{f'(c+h) - 0 - f'(c-h)}{2h}.
\]</span></p>
<p>We have to be careful, as we differentiate in the <span class="math inline">\(h\)</span> variable, not the <span class="math inline">\(c\)</span> one, so the chain rule brings out the minus sign. But again, as we still have an indeterminate form <span class="math inline">\(0/0\)</span>, this limit will equal the following limit should it exist:</p>
<p><span class="math display">\[
\lim_{h \rightarrow 0} \frac{f''(c+h) - 0 - (-f''(c-h))}{2} =
\lim_{c \rightarrow 0}\frac{f''(c+h) + f''(c-h)}{2} = f''(c).
\]</span></p>
<p>That last equality follows, as it is assumed that <span class="math inline">\(f''(x)\)</span> exists at <span class="math inline">\(c\)</span> and is continuous, that is, <span class="math inline">\(f''(c \pm h) \rightarrow f''(c)\)</span>.</p>
<p>The expression above finds use when second derivatives are numerically approximated. (The middle expression is the basis of the central-finite difference approximation to the derivative.)</p>
<ul>
<li>LHospital himself was interested in this limit for <span class="math inline">\(a &gt; 0\)</span> (<a href="http://mathoverflow.net/questions/51685/how-did-bernoulli-prove-lh%C3%B4pitals-rule">math overflow</a>)</li>
</ul>
<p><span class="math display">\[
\lim_{x \rightarrow a} \frac{\sqrt{2a^3\cdot x-x^4} - a\cdot(a^2\cdot x)^{1/3}}{ a - (a\cdot x^3)^{1/4}}.
\]</span></p>
<p>These derivatives can be done by hand, but to avoid any minor mistakes we utilize <code>SymPy</code> taking care to use rational numbers for the fractional powers, so as not to lose precision through floating point roundoff:</p>
<div class="cell" data-execution_count="4">
<div class="sourceCode cell-code" id="cb2"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb2-1"><a href="#cb2-1" aria-hidden="true" tabindex="-1"></a><span class="pp">@syms</span> a<span class="op">::</span><span class="dt">positive </span>x<span class="op">::</span><span class="dt">positive</span></span>
<span id="cb2-2"><a href="#cb2-2" aria-hidden="true" tabindex="-1"></a><span class="fu">f</span>(x) <span class="op">=</span> <span class="fu">sqrt</span>(<span class="fl">2</span>a<span class="op">^</span><span class="fl">3</span><span class="op">*</span>x <span class="op">-</span> x<span class="op">^</span><span class="fl">4</span>) <span class="op">-</span> a <span class="op">*</span> (a<span class="op">^</span><span class="fl">2</span><span class="op">*</span>x)<span class="op">^</span>(<span class="fl">1</span><span class="op">//</span><span class="fl">3</span>)</span>
<span id="cb2-3"><a href="#cb2-3" aria-hidden="true" tabindex="-1"></a><span class="fu">g</span>(x) <span class="op">=</span> a <span class="op">-</span> (a<span class="op">*</span>x<span class="op">^</span><span class="fl">3</span>)<span class="op">^</span>(<span class="fl">1</span><span class="op">//</span><span class="fl">4</span>)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="5">
<pre><code>g (generic function with 1 method)</code></pre>
</div>
</div>
<p>We can see that at <span class="math inline">\(x=a\)</span> we have the indeterminate form <span class="math inline">\(0/0\)</span>:</p>
<div class="cell" data-execution_count="5">
<div class="sourceCode cell-code" id="cb4"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb4-1"><a href="#cb4-1" aria-hidden="true" tabindex="-1"></a><span class="fu">f</span>(a), <span class="fu">g</span>(a)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="6">
<pre><code>(0, 0)</code></pre>
</div>
</div>
<p>What about the derivatives?</p>
<div class="cell" data-execution_count="6">
<div class="sourceCode cell-code" id="cb6"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb6-1"><a href="#cb6-1" aria-hidden="true" tabindex="-1"></a>fp, gp <span class="op">=</span> <span class="fu">diff</span>(<span class="fu">f</span>(x),x), <span class="fu">diff</span>(<span class="fu">g</span>(x),x)</span>
<span id="cb6-2"><a href="#cb6-2" aria-hidden="true" tabindex="-1"></a><span class="fu">fp</span>(x<span class="op">=&gt;</span>a), <span class="fu">gp</span>(x<span class="op">=&gt;</span>a)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="7">
<pre><code>(-4*a/3, -3/4)</code></pre>
</div>
</div>
<p>Their ratio will not be indeterminate, so the limit in question is just the ratio:</p>
<div class="cell" data-execution_count="7">
<div class="sourceCode cell-code" id="cb8"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb8-1"><a href="#cb8-1" aria-hidden="true" tabindex="-1"></a><span class="fu">fp</span>(x<span class="op">=&gt;</span>a) <span class="op">/</span> <span class="fu">gp</span>(x<span class="op">=&gt;</span>a)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="8">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
\frac{16 a}{9}
\]
</span>
</div>
</div>
<p>Of course, we could have just relied on <code>limit</code>, which knows about LHospitals rule:</p>
<div class="cell" data-execution_count="8">
<div class="sourceCode cell-code" id="cb9"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb9-1"><a href="#cb9-1" aria-hidden="true" tabindex="-1"></a><span class="fu">limit</span>(<span class="fu">f</span>(x)<span class="op">/</span><span class="fu">g</span>(x), x, a)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="9">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
\frac{16 a}{9}
\]
</span>
</div>
</div>
</section>
<section id="idea-behind-lhospitals-rule" class="level2" data-number="32.1">
<h2 data-number="32.1" class="anchored" data-anchor-id="idea-behind-lhospitals-rule"><span class="header-section-number">32.1</span> Idea behind LHospitals rule</h2>
<p>A first proof of LHospitals rule takes advantage of Cauchys <a href="http://en.wikipedia.org/wiki/Mean_value_theorem#Cauchy.27s_mean_value_theorem">generalization</a> of the mean value theorem to two functions. Suppose <span class="math inline">\(f(x)\)</span> and <span class="math inline">\(g(x)\)</span> are continuous on <span class="math inline">\([c,b]\)</span> and differentiable on <span class="math inline">\((c,b)\)</span>. On <span class="math inline">\((c,x)\)</span>, <span class="math inline">\(c &lt; x &lt; b\)</span> there exists a <span class="math inline">\(\xi\)</span> with <span class="math inline">\(f'(\xi) \cdot (f(x) - f(c)) = g'(\xi) \cdot (g(x) - g(c))\)</span>. In our formulation, both <span class="math inline">\(f(c)\)</span> and <span class="math inline">\(g(c)\)</span> are zero, so we have, provided we know that <span class="math inline">\(g(x)\)</span> is non zero, that <span class="math inline">\(f(x)/g(x) = f'(\xi)/g'(\xi)\)</span> for some <span class="math inline">\(\xi\)</span>, <span class="math inline">\(c &lt; \xi &lt; c + x\)</span>. That the right-hand side has a limit as <span class="math inline">\(x \rightarrow c+\)</span> is true by the assumption that the limit of the ratio of the derivatives exists. (The <span class="math inline">\(\xi\)</span> part can be removed by considering it as a composition of a function going to <span class="math inline">\(c\)</span>.) Thus the right limit of the ratio <span class="math inline">\(f/g\)</span> is known.</p>
<hr>
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<div class="cell-output cell-output-display" data-execution_count="10">
<div class="d-flex justify-content-center"> <figure class="figure"> <img 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" class="card-img-top figure-img" alt="A Figure">
<figcaption class="figure-caption"><div class="markdown">:Geometric
<p>interpretation of \(L=\lim_{x \rightarrow 0} x^2 / (\sqrt{1 + x} - 1 - x^2)\). At \(0\) this limit is indeterminate of the form \(0/0\). The value for a fixed \(x\) can be seen as the slope of a secant line of a parametric plot of the two functions, plotted as \((g, f)\). In this figure, the limiting "tangent" line has \(0\) slope, corresponding to the limit \(L\). In general, L'Hospital's rule is nothing more than a statement about slopes of tangent lines.</p>
<p>$</p>
</div> </figcaption>
</figure>
</div>
</div>
</div>
</section>
<section id="generalizations" class="level2" data-number="32.2">
<h2 data-number="32.2" class="anchored" data-anchor-id="generalizations"><span class="header-section-number">32.2</span> Generalizations</h2>
<p>LHospitals rule generalizes to other indeterminate forms, in particular the indeterminate form <span class="math inline">\(\infty/\infty\)</span> can be proved at the same time as <span class="math inline">\(0/0\)</span> with a more careful <a href="http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#General_proof">proof</a>.</p>
<p>The value <span class="math inline">\(c\)</span> in the limit can also be infinite. Consider this case with <span class="math inline">\(c=\infty\)</span>:</p>
<p><span class="math display">\[
\begin{align*}
\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} &amp;=
\lim_{x \rightarrow 0} \frac{f(1/x)}{g(1/x)}
\end{align*}
\]</span></p>
<p>LHospitals limit applies as <span class="math inline">\(x \rightarrow 0\)</span>, so we differentiate to get:</p>
<p><span class="math display">\[
\begin{align*}
\lim_{x \rightarrow 0} \frac{[f(1/x)]'}{[g(1/x)]'}
&amp;= \lim_{x \rightarrow 0} \frac{f'(1/x)\cdot(-1/x^2)}{g'(1/x)\cdot(-1/x^2)}\\
&amp;= \lim_{x \rightarrow 0} \frac{f'(1/x)}{g'(1/x)}\\
&amp;= \lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)},
\end{align*}
\]</span></p>
<p><em>assuming</em> the latter limit exists, LHospitals rule assures the equality</p>
<p><span class="math display">\[
\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} =
\lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)},
\]</span></p>
<section id="examples-1" class="level5">
<h5 class="anchored" data-anchor-id="examples-1">Examples</h5>
<p>For example, consider</p>
<p><span class="math display">\[
\lim_{x \rightarrow \infty} \frac{x}{e^x}.
\]</span></p>
<p>We see it is of the form <span class="math inline">\(\infty/\infty\)</span>. Taking advantage of the fact that LHospitals rule applies to limits at <span class="math inline">\(\infty\)</span>, we have that this limit will exist and be equal to this one, should it exist:</p>
<p><span class="math display">\[
\lim_{x \rightarrow \infty} \frac{1}{e^x}.
\]</span></p>
<p>This limit is, of course, <span class="math inline">\(0\)</span>, as it is of the form <span class="math inline">\(1/\infty\)</span>. It is not hard to build up from here to show that for any integer value of <span class="math inline">\(n&gt;0\)</span> that:</p>
<p><span class="math display">\[
\lim_{x \rightarrow \infty} \frac{x^n}{e^x} = 0.
\]</span></p>
<p>This is an expression of the fact that exponential functions grow faster than polynomial functions.</p>
<p>Similarly, powers grow faster than logarithms, as this limit shows, which is indeterminate of the form <span class="math inline">\(\infty/\infty\)</span>:</p>
<p><span class="math display">\[
\lim_{x \rightarrow \infty} \frac{\log(x)}{x} =
\lim_{x \rightarrow \infty} \frac{1/x}{1} = 0,
\]</span></p>
<p>the first equality by LHospitals rule, as the second limit exists.</p>
</section>
</section>
<section id="other-indeterminate-forms" class="level2" data-number="32.3">
<h2 data-number="32.3" class="anchored" data-anchor-id="other-indeterminate-forms"><span class="header-section-number">32.3</span> Other indeterminate forms</h2>
<p>Indeterminate forms of the type <span class="math inline">\(0 \cdot \infty\)</span>, <span class="math inline">\(0^0\)</span>, <span class="math inline">\(\infty^\infty\)</span>, <span class="math inline">\(\infty - \infty\)</span> can be re-expressed to be in the form <span class="math inline">\(0/0\)</span> or <span class="math inline">\(\infty/\infty\)</span> and then LHospitals theorem can be applied.</p>
<section id="example-rewriting-0-cdot-infty" class="level6">
<h6 class="anchored" data-anchor-id="example-rewriting-0-cdot-infty">Example: rewriting <span class="math inline">\(0 \cdot \infty\)</span></h6>
<p>What is the limit <span class="math inline">\(x \log(x)\)</span> as <span class="math inline">\(x \rightarrow 0+\)</span>? The form is <span class="math inline">\(0\cdot \infty\)</span>, rewriting, we see this is just:</p>
<p><span class="math display">\[
\lim_{x \rightarrow 0+}\frac{\log(x)}{1/x}.
\]</span></p>
<p>LHospitals rule clearly applies to one-sided limits, as well as two (our proof sketch used one-sided limits), so this limit will equal the following, should it exist:</p>
<p><span class="math display">\[
\lim_{x \rightarrow 0+}\frac{1/x}{-1/x^2} = \lim_{x \rightarrow 0+} -x = 0.
\]</span></p>
</section>
<section id="example-rewriting-00" class="level6">
<h6 class="anchored" data-anchor-id="example-rewriting-00">Example: rewriting <span class="math inline">\(0^0\)</span></h6>
<p>What is the limit <span class="math inline">\(x^x\)</span> as <span class="math inline">\(x \rightarrow 0+\)</span>? The expression is of the form <span class="math inline">\(0^0\)</span>, which is indeterminate. (Even though floating point math defines the value as <span class="math inline">\(1\)</span>.) We can rewrite this by taking a log:</p>
<p><span class="math display">\[
x^x = \exp(\log(x^x)) = \exp(x \log(x)) = \exp(\log(x)/(1/x)).
\]</span></p>
<p>Be just saw that <span class="math inline">\(\lim_{x \rightarrow 0+}\log(x)/(1/x) = 0\)</span>. So by the rules for limits of compositions and the fact that <span class="math inline">\(e^x\)</span> is continuous, we see <span class="math inline">\(\lim_{x \rightarrow 0+} x^x = e^0 = 1\)</span>.</p>
</section>
<section id="example-rewriting-infty---infty" class="level5">
<h5 class="anchored" data-anchor-id="example-rewriting-infty---infty">Example: rewriting <span class="math inline">\(\infty - \infty\)</span></h5>
<p>A limit <span class="math inline">\(\lim_{x \rightarrow c} f(x) - g(x)\)</span> of indeterminate form <span class="math inline">\(\infty - \infty\)</span> can be reexpressed to be of the from <span class="math inline">\(0/0\)</span> through the transformation:</p>
<p><span class="math display">\[
\begin{align*}
f(x) - g(x) &amp;= f(x)g(x) \cdot (\frac{1}{g(x)} - \frac{1}{f(x)}) \\
&amp;= \frac{\frac{1}{g(x)} - \frac{1}{f(x)}}{\frac{1}{f(x)g(x)}}.
\end{align*}
\]</span></p>
<p>Applying this to</p>
<p><span class="math display">\[
L = \lim_{x \rightarrow 1} \big(\frac{x}{x-1} - \frac{1}{\log(x)}\big)
\]</span></p>
<p>We get that <span class="math inline">\(L\)</span> is equal to the following limit:</p>
<p><span class="math display">\[
\lim_{x \rightarrow 1} \frac{\log(x) - \frac{x-1}{x}}{\frac{x-1}{x} \log(x)}
=
\lim_{x \rightarrow 1} \frac{x\log(x)-(x-1)}{(x-1)\log(x)}
\]</span></p>
<p>In <code>SymPy</code> we have:</p>
<div class="cell" data-execution_count="10">
<div class="sourceCode cell-code" id="cb10"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb10-1"><a href="#cb10-1" aria-hidden="true" tabindex="-1"></a>𝒇 <span class="op">=</span> <span class="fu">x*log</span>(x) <span class="op">-</span> (x<span class="op">-</span><span class="fl">1</span>)</span>
<span id="cb10-2"><a href="#cb10-2" aria-hidden="true" tabindex="-1"></a>𝒈 <span class="op">=</span> (x<span class="op">-</span><span class="fl">1</span>)<span class="fu">*log</span>(x)</span>
<span id="cb10-3"><a href="#cb10-3" aria-hidden="true" tabindex="-1"></a><span class="fu">𝒇</span>(<span class="fl">1</span>), <span class="fu">𝒈</span>(<span class="fl">1</span>)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="11">
<pre><code>(0, 0)</code></pre>
</div>
</div>
<p>LHospitals rule applies to the form <span class="math inline">\(0/0\)</span>, so we try:</p>
<div class="cell" data-execution_count="11">
<div class="sourceCode cell-code" id="cb12"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb12-1"><a href="#cb12-1" aria-hidden="true" tabindex="-1"></a>𝒇 <span class="op">=</span> <span class="fu">diff</span>(𝒇, x)</span>
<span id="cb12-2"><a href="#cb12-2" aria-hidden="true" tabindex="-1"></a>𝒈 <span class="op">=</span> <span class="fu">diff</span>(𝒈, x)</span>
<span id="cb12-3"><a href="#cb12-3" aria-hidden="true" tabindex="-1"></a><span class="fu">𝒇</span>(<span class="fl">1</span>), <span class="fu">𝒈</span>(<span class="fl">1</span>)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="12">
<pre><code>(0, 0)</code></pre>
</div>
</div>
<p>Again, we get the indeterminate form <span class="math inline">\(0/0\)</span>, so we try again with second derivatives:</p>
<div class="cell" data-execution_count="12">
<div class="sourceCode cell-code" id="cb14"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb14-1"><a href="#cb14-1" aria-hidden="true" tabindex="-1"></a>𝒇 <span class="op">=</span> <span class="fu">diff</span>(𝒇, x, x)</span>
<span id="cb14-2"><a href="#cb14-2" aria-hidden="true" tabindex="-1"></a>𝒈 <span class="op">=</span> <span class="fu">diff</span>(𝒈, x, x)</span>
<span id="cb14-3"><a href="#cb14-3" aria-hidden="true" tabindex="-1"></a><span class="fu">𝒇</span>(<span class="fl">1</span>), <span class="fu">𝒈</span>(<span class="fl">1</span>)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="13">
<pre><code>(1, 2)</code></pre>
</div>
</div>
<p>From this we see the limit is <span class="math inline">\(1/2\)</span>, as could have been done directly:</p>
<div class="cell" data-execution_count="13">
<div class="sourceCode cell-code" id="cb16"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb16-1"><a href="#cb16-1" aria-hidden="true" tabindex="-1"></a><span class="fu">limit</span>(𝒇<span class="op">/</span>𝒈, x<span class="op">=&gt;</span><span class="fl">1</span>)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="14">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
\frac{1}{2}
\]
</span>
</div>
</div>
</section>
</section>
<section id="the-assumptions-are-necessary" class="level2" data-number="32.4">
<h2 data-number="32.4" class="anchored" data-anchor-id="the-assumptions-are-necessary"><span class="header-section-number">32.4</span> The assumptions are necessary</h2>
<section id="example-the-limit-existing-is-necessary" class="level5">
<h5 class="anchored" data-anchor-id="example-the-limit-existing-is-necessary">Example: the limit existing is necessary</h5>
<p>The following limit is <em>easily</em> seen by comparing terms of largest growth:</p>
<p><span class="math display">\[
1 = \lim_{x \rightarrow \infty} \frac{x - \sin(x)}{x}
\]</span></p>
<p>However, the limit of the ratio of the derivatives <em>does</em> not exist:</p>
<p><span class="math display">\[
\lim_{x \rightarrow \infty} \frac{1 - \cos(x)}{1},
\]</span></p>
<p>as the function just oscillates. This shows that LHospitals rule does not apply when the limit of the the ratio of the derivatives does not exist.</p>
</section>
<section id="example-the-assumptions-matter" class="level5">
<h5 class="anchored" data-anchor-id="example-the-assumptions-matter">Example: the assumptions matter</h5>
<p>This example comes from the thesis of Gruntz to highlight possible issues when computer systems do simplifications.</p>
<p>Consider:</p>
<p><span class="math display">\[
\lim_{x \rightarrow \infty} \frac{1/2\sin(2x) +x}{\exp(\sin(x))\cdot(\cos(x)\sin(x)+x)}.
\]</span></p>
<p>If we apply LHospitals rule using simplification we have:</p>
<div class="cell" data-execution_count="14">
<div class="sourceCode cell-code" id="cb17"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb17-1"><a href="#cb17-1" aria-hidden="true" tabindex="-1"></a><span class="fu">u</span>(x) <span class="op">=</span> <span class="fl">1</span><span class="op">//</span><span class="fl">2</span><span class="fu">*sin</span>(<span class="fl">2</span>x) <span class="op">+</span> x</span>
<span id="cb17-2"><a href="#cb17-2" aria-hidden="true" tabindex="-1"></a><span class="fu">v</span>(x) <span class="op">=</span> <span class="fu">exp</span>(<span class="fu">sin</span>(x))<span class="fu">*</span>(<span class="fu">cos</span>(x)<span class="fu">*sin</span>(x) <span class="op">+</span> x)</span>
<span id="cb17-3"><a href="#cb17-3" aria-hidden="true" tabindex="-1"></a>up, vp <span class="op">=</span> <span class="fu">diff</span>(<span class="fu">u</span>(x),x), <span class="fu">diff</span>(<span class="fu">v</span>(x),x)</span>
<span id="cb17-4"><a href="#cb17-4" aria-hidden="true" tabindex="-1"></a><span class="fu">limit</span>(<span class="fu">simplify</span>(up<span class="op">/</span>vp), x <span class="op">=&gt;</span> oo)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="15">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
0
\]
</span>
</div>
</div>
<p>However, this answer is incorrect. The reason being subtle. The simplification cancels a term of <span class="math inline">\(\cos(x)\)</span> that appears in the numerator and denominator. Before cancellation, we have <code>vp</code> will have infinitely many zeros as <span class="math inline">\(x\)</span> approaches <span class="math inline">\(\infty\)</span> so LHospitals wont apply (the limit wont exist, as every <span class="math inline">\(2\pi\)</span> the ratio is undefined so the function is never eventually close to some <span class="math inline">\(L\)</span>).</p>
<p>This ratio has no limit, as it oscillates, as confirmed by <code>SymPy</code>:</p>
<div class="cell" data-execution_count="15">
<div class="sourceCode cell-code" id="cb18"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb18-1"><a href="#cb18-1" aria-hidden="true" tabindex="-1"></a><span class="fu">limit</span>(<span class="fu">u</span>(x)<span class="op">/</span><span class="fu">v</span>(x), x<span class="op">=&gt;</span> oo)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="16">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
\left\langle e^{-1}, e\right\rangle
\]
</span>
</div>
</div>
</section>
</section>
<section id="questions" class="level2" data-number="32.5">
<h2 data-number="32.5" class="anchored" data-anchor-id="questions"><span class="header-section-number">32.5</span> Questions</h2>
<section id="question" class="level6">
<h6 class="anchored" data-anchor-id="question">Question</h6>
<p>This function <span class="math inline">\(f(x) = \sin(5x)/x\)</span> is <em>indeterminate</em> at <span class="math inline">\(x=0\)</span>. What type?</p>
<div class="cell" data-hold="true" data-execution_count="17">
<div class="cell-output cell-output-display" data-execution_count="18">
<form class="mx-2 my-3 mw-100" name="WeaveQuestion" data-id="12508984653168870405" data-controltype="">
<div class="form-group ">
<div class="controls">
<div class="form" id="controls_12508984653168870405">
<div style="padding-top: 5px">
<div class="form-check">
<label class="form-check-label" for="radio_12508984653168870405_1">
<input class="form-check-input" type="radio" name="radio_12508984653168870405" id="radio_12508984653168870405_1" value="1">
<span class="label-body px-1">
\(0/0\)
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_12508984653168870405_2">
<input class="form-check-input" type="radio" name="radio_12508984653168870405" id="radio_12508984653168870405_2" value="2">
<span class="label-body px-1">
\(\infty/\infty\)
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_12508984653168870405_3">
<input class="form-check-input" type="radio" name="radio_12508984653168870405" id="radio_12508984653168870405_3" value="3">
<span class="label-body px-1">
\(0^0\)
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_12508984653168870405_4">
<input class="form-check-input" type="radio" name="radio_12508984653168870405" id="radio_12508984653168870405_4" value="4">
<span class="label-body px-1">
\(\infty - \infty\)
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_12508984653168870405_5">
<input class="form-check-input" type="radio" name="radio_12508984653168870405" id="radio_12508984653168870405_5" value="5">
<span class="label-body px-1">
\(0 \cdot \infty\)
</span>
</label>
</div>
</div>
</div>
<div id="12508984653168870405_message" style="padding-bottom: 15px"></div>
</div>
</div>
</form>
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</section>
<section id="question-1" class="level6">
<h6 class="anchored" data-anchor-id="question-1">Question</h6>
<p>This function <span class="math inline">\(f(x) = \sin(x)^{\sin(x)}\)</span> is <em>indeterminate</em> at <span class="math inline">\(x=0\)</span>. What type?</p>
<div class="cell" data-hold="true" data-execution_count="18">
<div class="cell-output cell-output-display" data-execution_count="19">
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\(0/0\)
</span>
</label>
</div>
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<label class="form-check-label" for="radio_11524216952751549626_2">
<input class="form-check-input" type="radio" name="radio_11524216952751549626" id="radio_11524216952751549626_2" value="2">
<span class="label-body px-1">
\(\infty/\infty\)
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_11524216952751549626_3">
<input class="form-check-input" type="radio" name="radio_11524216952751549626" id="radio_11524216952751549626_3" value="3">
<span class="label-body px-1">
\(0^0\)
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_11524216952751549626_4">
<input class="form-check-input" type="radio" name="radio_11524216952751549626" id="radio_11524216952751549626_4" value="4">
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\(\infty - \infty\)
</span>
</label>
</div>
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<label class="form-check-label" for="radio_11524216952751549626_5">
<input class="form-check-input" type="radio" name="radio_11524216952751549626" id="radio_11524216952751549626_5" value="5">
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\(0 \cdot \infty\)
</span>
</label>
</div>
</div>
</div>
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</div>
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</div>
</section>
<section id="question-2" class="level6">
<h6 class="anchored" data-anchor-id="question-2">Question</h6>
<p>This function <span class="math inline">\(f(x) = (x-2)/(x^2 - 4)\)</span> is <em>indeterminate</em> at <span class="math inline">\(x=2\)</span>. What type?</p>
<div class="cell" data-hold="true" data-execution_count="19">
<div class="cell-output cell-output-display" data-execution_count="20">
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\(0/0\)
</span>
</label>
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<input class="form-check-input" type="radio" name="radio_9233145049184527033" id="radio_9233145049184527033_2" value="2">
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\(\infty/\infty\)
</span>
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\(0^0\)
</span>
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\(\infty - \infty\)
</span>
</label>
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\(0 \cdot \infty\)
</span>
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</section>
<section id="question-3" class="level6">
<h6 class="anchored" data-anchor-id="question-3">Question</h6>
<p>This function <span class="math inline">\(f(x) = (g(x+h) - g(x-h)) / (2h)\)</span> (<span class="math inline">\(g\)</span> is continuous) is <em>indeterminate</em> at <span class="math inline">\(h=0\)</span>. What type?</p>
<div class="cell" data-hold="true" data-execution_count="20">
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<span class="label-body px-1">
\(0/0\)
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_13103547777178788342_2">
<input class="form-check-input" type="radio" name="radio_13103547777178788342" id="radio_13103547777178788342_2" value="2">
<span class="label-body px-1">
\(\infty/\infty\)
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_13103547777178788342_3">
<input class="form-check-input" type="radio" name="radio_13103547777178788342" id="radio_13103547777178788342_3" value="3">
<span class="label-body px-1">
\(0^0\)
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_13103547777178788342_4">
<input class="form-check-input" type="radio" name="radio_13103547777178788342" id="radio_13103547777178788342_4" value="4">
<span class="label-body px-1">
\(\infty - \infty\)
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_13103547777178788342_5">
<input class="form-check-input" type="radio" name="radio_13103547777178788342" id="radio_13103547777178788342_5" value="5">
<span class="label-body px-1">
\(0 \cdot \infty\)
</span>
</label>
</div>
</div>
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</div>
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</section>
<section id="question-4" class="level6">
<h6 class="anchored" data-anchor-id="question-4">Question</h6>
<p>This function <span class="math inline">\(f(x) = x \log(x)\)</span> is <em>indeterminate</em> at <span class="math inline">\(x=0\)</span>. What type?</p>
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<div class="cell-output cell-output-display" data-execution_count="22">
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<div class="form-check">
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<input class="form-check-input" type="radio" name="radio_7709586706107341691" id="radio_7709586706107341691_1" value="1">
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\(0/0\)
</span>
</label>
</div>
<div class="form-check">
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<input class="form-check-input" type="radio" name="radio_7709586706107341691" id="radio_7709586706107341691_2" value="2">
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\(\infty/\infty\)
</span>
</label>
</div>
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<input class="form-check-input" type="radio" name="radio_7709586706107341691" id="radio_7709586706107341691_3" value="3">
<span class="label-body px-1">
\(0^0\)
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_7709586706107341691_4">
<input class="form-check-input" type="radio" name="radio_7709586706107341691" id="radio_7709586706107341691_4" value="4">
<span class="label-body px-1">
\(\infty - \infty\)
</span>
</label>
</div>
<div class="form-check">
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<input class="form-check-input" type="radio" name="radio_7709586706107341691" id="radio_7709586706107341691_5" value="5">
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\(0 \cdot \infty\)
</span>
</label>
</div>
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</div>
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</section>
<section id="question-5" class="level6">
<h6 class="anchored" data-anchor-id="question-5">Question</h6>
<p>Does LHospitals rule apply to this limit:</p>
<p><span class="math display">\[
\lim_{x \rightarrow \pi} \frac{\sin(\pi x)}{\pi x}.
\]</span></p>
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No. It is not indeterminate
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Yes. It is of the form \(0/0\)
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</section>
<section id="question-6" class="level6">
<h6 class="anchored" data-anchor-id="question-6">Question</h6>
<p>Use LHospitals rule to find the limit</p>
<p><span class="math display">\[
L = \lim_{x \rightarrow 0} \frac{4x - \sin(x)}{x}.
\]</span></p>
<p>What is <span class="math inline">\(L\)</span>?</p>
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</section>
<section id="question-7" class="level6">
<h6 class="anchored" data-anchor-id="question-7">Question</h6>
<p>Use LHospitals rule to find the limit</p>
<p><span class="math display">\[
L = \lim_{x \rightarrow 0} \frac{\sqrt{1+x} - 1}{x}.
\]</span></p>
<p>What is <span class="math inline">\(L\)</span>?</p>
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</section>
<section id="question-8" class="level6">
<h6 class="anchored" data-anchor-id="question-8">Question</h6>
<p>Use LHospitals rule <em>one</em> or more times to find the limit</p>
<p><span class="math display">\[
L = \lim_{x \rightarrow 0} \frac{x - \sin(x)}{x^3}.
\]</span></p>
<p>What is <span class="math inline">\(L\)</span>?</p>
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</div>
</div>
</section>
<section id="question-9" class="level6">
<h6 class="anchored" data-anchor-id="question-9">Question</h6>
<p>Use LHospitals rule <em>one</em> or more times to find the limit</p>
<p><span class="math display">\[
L = \lim_{x \rightarrow 0} \frac{1 - x^2/2 - \cos(x)}{x^3}.
\]</span></p>
<p>What is <span class="math inline">\(L\)</span>?</p>
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</div>
</section>
<section id="question-10" class="level6">
<h6 class="anchored" data-anchor-id="question-10">Question</h6>
<p>Use LHospitals rule <em>one</em> or more times to find the limit</p>
<p><span class="math display">\[
L = \lim_{x \rightarrow \infty} \frac{\log(\log(x))}{\log(x)}.
\]</span></p>
<p>What is <span class="math inline">\(L\)</span>?</p>
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</div>
</section>
<section id="question-11" class="level6">
<h6 class="anchored" data-anchor-id="question-11">Question</h6>
<p>By using a common denominator to rewrite this expression, use LHospitals rule to find the limit</p>
<p><span class="math display">\[
L = \lim_{x \rightarrow 0} \frac{1}{x} - \frac{1}{\sin(x)}.
\]</span></p>
<p>What is <span class="math inline">\(L\)</span>?</p>
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</section>
<section id="question-12" class="level5">
<h5 class="anchored" data-anchor-id="question-12">Question</h5>
<p>Use LHospitals rule to find the limit</p>
<p><span class="math display">\[
L = \lim_{x \rightarrow \infty} \log(x)/x
\]</span></p>
<p>What is <span class="math inline">\(L\)</span>?</p>
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</section>
<section id="question-13" class="level5">
<h5 class="anchored" data-anchor-id="question-13">Question</h5>
<p>Using LHospitals rule, does</p>
<p><span class="math display">\[
\lim_{x \rightarrow 0+} x^{\log(x)}
\]</span></p>
<p>exist?</p>
<p>Consider <span class="math inline">\(x^{\log(x)} = e^{\log(x)\log(x)}\)</span>.</p>
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<section id="question-14" class="level5">
<h5 class="anchored" data-anchor-id="question-14">Question</h5>
<p>Using LHospitals rule, find the limit of</p>
<p><span class="math display">\[
\lim_{x \rightarrow 1} (2-x)^{\tan(\pi/2 \cdot x)}.
\]</span></p>
<p>(Hint, express as <span class="math inline">\(\exp^{\tan(\pi/2 \cdot x) \cdot \log(2-x)}\)</span> and take the limit of the resulting exponent.)</p>
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It does not exist
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\(e^{2/\pi}\)
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\(1\)
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\(0\)
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\({2\pi}\)
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