http://wiki.sommarmatte.se/wikis/forberedandematte/index.php?title=L%C3%B6sning_4.1.1.e&action=history Versionshistorik för denna sida på wikin sv MediaWiki 1.11.1 Mon, 24 Aug 2026 03:59:56 GMT http://wiki.sommarmatte.se/wikis/forberedandematte/index.php?title=L%C3%B6sning_4.1.1.e&diff=687&oldid=prev <p></p> <table style="background-color: white; color:black;"> <col class='diff-marker' /> <col class='diff-content' /> <col class='diff-marker' /> <col class='diff-content' /> <tr> <td colspan='2' style="background-color: white; color:black;">← Äldre version</td> <td colspan='2' style="background-color: white; color:black;">Versionen från 18 juni 2012 kl. 14.16</td> </tr> <tr><td colspan="2" class="diff-lineno">Rad 1:</td> <td colspan="2" class="diff-lineno">Rad 1:</td></tr> <tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Vi ser att &lt;math&gt; \lim_{x\to 0}\frac{2x^3+x^2+1}{x^3+1}= \lim_{x\to 0}\frac{2x^3}{x^3+1}+ \lim_{x\to</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Vi ser att &lt;math&gt; \lim_{x\to 0}\frac{2x^3+x^2+1}{x^3+1}= \lim_{x\to 0}\frac{2x^3}{x^3+1}+ \lim_{x\to</div></td></tr> <tr><td class='diff-marker'>-</td><td style="background: #ffa; color:black; font-size: smaller;"><div>0}\frac{x^2}{x^3+1}+ \lim_{x\to 0}\frac{1}{x^3+1}&lt;/math&gt; <del style="color: red; font-weight: bold; text-decoration: none;">om </del>vi fortsätter vidare:</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>0}\frac{x^2}{x^3+1}+ \lim_{x\to 0}\frac{1}{x^3+1}<ins style="color: red; font-weight: bold; text-decoration: none;">.</ins>&lt;/math&gt; </div></td></tr> <tr><td colspan="2">&nbsp;</td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div><ins style="color: red; font-weight: bold; text-decoration: none;">Om </ins>vi fortsätter vidare:</div></td></tr> <tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>&lt;math&gt; \lim_{x\to 0}\frac{2x^3}{x^3+1}+0 + 0=\lim_{x\to 0}\frac{2(x^3+1)-2}{x^3+1}=</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>&lt;math&gt; \lim_{x\to 0}\frac{2x^3}{x^3+1}+0 + 0=\lim_{x\to 0}\frac{2(x^3+1)-2}{x^3+1}=</div></td></tr> <tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>=2\lim_{x\to 0}\frac{x^3+1}{x^3+1}-\lim_{x\to 0}\frac{2}{x^3+1}=</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>=2\lim_{x\to 0}\frac{x^3+1}{x^3+1}-\lim_{x\to 0}\frac{2}{x^3+1}=</div></td></tr> <tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>=2\cdot1-0 = 2&lt;/math&gt;</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>=2\cdot1-0 = 2&lt;/math&gt;</div></td></tr> </table> Mon, 18 Jun 2012 14:16:22 GMT Samuel http://wiki.sommarmatte.se/wikis/forberedandematte/index.php/Diskussion:L%C3%B6sning_4.1.1.e http://wiki.sommarmatte.se/wikis/forberedandematte/index.php?title=L%C3%B6sning_4.1.1.e&diff=686&oldid=prev <p></p> <table style="background-color: white; color:black;"> <col class='diff-marker' /> <col class='diff-content' /> <col class='diff-marker' /> <col class='diff-content' /> <tr> <td colspan='2' style="background-color: white; color:black;">← Äldre version</td> <td colspan='2' style="background-color: white; color:black;">Versionen från 18 juni 2012 kl. 14.16</td> </tr> <tr><td colspan="2" class="diff-lineno">Rad 1:</td> <td colspan="2" class="diff-lineno">Rad 1:</td></tr> <tr><td class='diff-marker'>-</td><td style="background: #ffa; color:black; font-size: smaller;"><div>Vi ser att &lt;math&gt; \lim_{x\to 0}\frac{2x^3+x^2+1}{x^3+1}<del style="color: red; font-weight: bold; text-decoration: none;">&amp;</del>= \lim_{x\to 0}\frac{2x^3}{x^3+1}+ \lim_{x\to</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>Vi ser att &lt;math&gt; \lim_{x\to 0}\frac{2x^3+x^2+1}{x^3+1}= \lim_{x\to 0}\frac{2x^3}{x^3+1}+ \lim_{x\to</div></td></tr> <tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>0}\frac{x^2}{x^3+1}+ \lim_{x\to 0}\frac{1}{x^3+1}&lt;/math&gt; om vi fortsätter vidare:</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>0}\frac{x^2}{x^3+1}+ \lim_{x\to 0}\frac{1}{x^3+1}&lt;/math&gt; om vi fortsätter vidare:</div></td></tr> <tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>&lt;math&gt; \lim_{x\to 0}\frac{2x^3}{x^3+1}+0 + 0=\lim_{x\to 0}\frac{2(x^3+1)-2}{x^3+1}=</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>&lt;math&gt; \lim_{x\to 0}\frac{2x^3}{x^3+1}+0 + 0=\lim_{x\to 0}\frac{2(x^3+1)-2}{x^3+1}=</div></td></tr> <tr><td class='diff-marker'>-</td><td style="background: #ffa; color:black; font-size: smaller;"><div><del style="color: red; font-weight: bold; text-decoration: none;">&amp;</del>=2\lim_{x\to 0}\frac{x^3+1}{x^3+1}-\lim_{x\to 0}\frac{2}{x^3+1}=</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>=2\lim_{x\to 0}\frac{x^3+1}{x^3+1}-\lim_{x\to 0}\frac{2}{x^3+1}=</div></td></tr> <tr><td class='diff-marker'>-</td><td style="background: #ffa; color:black; font-size: smaller;"><div><del style="color: red; font-weight: bold; text-decoration: none;">&amp;</del>=2\cdot1-0 = 2&lt;/math&gt;</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>=2\cdot1-0 = 2&lt;/math&gt;</div></td></tr> </table> Mon, 18 Jun 2012 14:16:03 GMT Samuel http://wiki.sommarmatte.se/wikis/forberedandematte/index.php/Diskussion:L%C3%B6sning_4.1.1.e http://wiki.sommarmatte.se/wikis/forberedandematte/index.php?title=L%C3%B6sning_4.1.1.e&diff=685&oldid=prev <p>Ny sida: Vi ser att &lt;math&gt; \lim_{x\to 0}\frac{2x^3+x^2+1}{x^3+1}&amp;= \lim_{x\to 0}\frac{2x^3}{x^3+1}+ \lim_{x\to 0}\frac{x^2}{x^3+1}+ \lim_{x\to 0}\frac{1}{x^3+1}&lt;/math&gt; om vi fortsätter vidare: &lt;mat...</p> <p><b>Ny sida</b></p><div>Vi ser att &lt;math&gt; \lim_{x\to 0}\frac{2x^3+x^2+1}{x^3+1}&amp;= \lim_{x\to 0}\frac{2x^3}{x^3+1}+ \lim_{x\to<br /> 0}\frac{x^2}{x^3+1}+ \lim_{x\to 0}\frac{1}{x^3+1}&lt;/math&gt; om vi fortsätter vidare:<br /> &lt;math&gt; \lim_{x\to 0}\frac{2x^3}{x^3+1}+0 + 0=\lim_{x\to 0}\frac{2(x^3+1)-2}{x^3+1}=<br /> &amp;=2\lim_{x\to 0}\frac{x^3+1}{x^3+1}-\lim_{x\to 0}\frac{2}{x^3+1}=<br /> &amp;=2\cdot1-0 = 2&lt;/math&gt;</div> Mon, 18 Jun 2012 14:15:47 GMT Samuel http://wiki.sommarmatte.se/wikis/forberedandematte/index.php/Diskussion:L%C3%B6sning_4.1.1.e