<?xml version="1.0" encoding="UTF-8"?><rss version="2.0"
	xmlns:content="http://purl.org/rss/1.0/modules/content/"
	xmlns:wfw="http://wellformedweb.org/CommentAPI/"
	xmlns:dc="http://purl.org/dc/elements/1.1/"
	xmlns:atom="http://www.w3.org/2005/Atom"
	xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"
	xmlns:slash="http://purl.org/rss/1.0/modules/slash/"
	>

<channel>
	<title>Gina &#8211; Die Kluge Eule</title>
	<atom:link href="https://dieklugeeule.com/author/gina/feed/" rel="self" type="application/rss+xml" />
	<link>https://dieklugeeule.com</link>
	<description></description>
	<lastBuildDate>Wed, 18 Mar 2020 18:02:51 +0000</lastBuildDate>
	<language>de-DE</language>
	<sy:updatePeriod>
	hourly	</sy:updatePeriod>
	<sy:updateFrequency>
	1	</sy:updateFrequency>
	<generator>https://wordpress.org/?v=6.0.2</generator>

<image>
	<url>https://dieklugeeule.com/wp-content/uploads/2022/04/cropped-logo-smal-2-32x32.jpg</url>
	<title>Gina &#8211; Die Kluge Eule</title>
	<link>https://dieklugeeule.com</link>
	<width>32</width>
	<height>32</height>
</image> 
	<item>
		<title>Wie finden Sie das Limit von # (1-cosx) / x #, wenn sich x 0 nähert?</title>
		<link>https://dieklugeeule.com/wie-finden-sie-das-limit-von-1-cosx-x-wenn-sich-x-0-nahert/</link>
		
		<dc:creator><![CDATA[Gina]]></dc:creator>
		<pubDate>Wed, 18 Mar 2020 18:02:51 +0000</pubDate>
				<category><![CDATA[Infinitesimalrechnung]]></category>
		<guid isPermaLink="false">https://dieklugeeule.com/?p=5373</guid>

					<description><![CDATA[Wie finden Sie das Limit von # (1-cosx) / x #, wenn sich x 0 nähert? Antworten: #0# Erläuterung: #1-cosx=2sin^2(x/2)# so #(1-cos x)/x=(x/4) (sin(x/2)/(x/2))^2# dann #lim_(x-&#62;0)(1-cos x)/x equiv lim_(x-&#62;0)(x/4) (sin(x/2)/(x/2))^2 = 0 cdot 1 = 0#]]></description>
										<content:encoded><![CDATA[<h1 class="questionTitle">Wie finden Sie das Limit von # (1-cosx) / x #, wenn sich x 0 nähert?</h1>
<div class="answerContainer clearfix">
<div class='answerText'>
<div class="answerSummary">
<h4 class="answerHeader">Antworten:</h4>
<div>
<div class='markdown'>
<p>#0#</p>
</div></div>
</p></div>
<div class="answerDescription">
<h4 class="answerHeader">Erläuterung:</h4>
<div>
<div class='markdown'>
<p>#1-cosx=2sin^2(x/2)#  so</p>
<p>#(1-cos x)/x=(x/4) (sin(x/2)/(x/2))^2# dann</p>
<p>#lim_(x-&gt;0)(1-cos x)/x equiv lim_(x-&gt;0)(x/4) (sin(x/2)/(x/2))^2 = 0 cdot 1 = 0#</p>
</div></div>
</p></div>
</p></div>
</p></div>
]]></content:encoded>
					
		
		
			</item>
	</channel>
</rss>
