http://wiki.sommarmatte.se/wikis/forberedandematte/index.php?action=history&feed=atom&title=L%C3%B6sning_4.1.1.e
Lösning 4.1.1.e - Versionshistorik
2026-08-24T04:03:37Z
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http://wiki.sommarmatte.se/wikis/forberedandematte/index.php?title=L%C3%B6sning_4.1.1.e&diff=687&oldid=prev
Samuel den 18 juni 2012 kl. 14.16
2012-06-18T14:16:22Z
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<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>
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<tr><td colspan="2" class="diff-lineno">Rad 1:</td>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Vi ser att <math> \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 <math> \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}</math> <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></math> </div></td></tr>
<tr><td colspan="2"> </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><math> \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><math> \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</math></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>=2\cdot1-0 = 2</math></div></td></tr>
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Samuel
http://wiki.sommarmatte.se/wikis/forberedandematte/index.php?title=L%C3%B6sning_4.1.1.e&diff=686&oldid=prev
Samuel den 18 juni 2012 kl. 14.16
2012-06-18T14:16:03Z
<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 <math> \lim_{x\to 0}\frac{2x^3+x^2+1}{x^3+1}<del style="color: red; font-weight: bold; text-decoration: none;">&</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 <math> \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}</math> 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}</math> om vi fortsätter vidare:</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div><math> \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><math> \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;">&</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;">&</del>=2\cdot1-0 = 2</math></div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>=2\cdot1-0 = 2</math></div></td></tr>
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Samuel
http://wiki.sommarmatte.se/wikis/forberedandematte/index.php?title=L%C3%B6sning_4.1.1.e&diff=685&oldid=prev
Samuel: Ny sida: Vi ser att <math> \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 0}\frac{x^2}{x^3+1}+ \lim_{x\to 0}\frac{1}{x^3+1}</math> om vi fortsätter vidare: <mat...
2012-06-18T14:15:47Z
<p>Ny sida: Vi ser att <math> \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 0}\frac{x^2}{x^3+1}+ \lim_{x\to 0}\frac{1}{x^3+1}</math> om vi fortsätter vidare: <mat...</p>
<p><b>Ny sida</b></p><div>Vi ser att <math> \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<br />
0}\frac{x^2}{x^3+1}+ \lim_{x\to 0}\frac{1}{x^3+1}</math> om vi fortsätter vidare:<br />
<math> \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 />
&=2\lim_{x\to 0}\frac{x^3+1}{x^3+1}-\lim_{x\to 0}\frac{2}{x^3+1}=<br />
&=2\cdot1-0 = 2</math></div>
Samuel