[tex]{x sqrt{x+1}}{(1-sqrt{2x+3})}/({7-6x+4x^2})[/tex]
Sliter med den algebraiske omformingen her....og LH`s regel skal ikke brukes...
Lim *vanskelig*
Moderators: Vektormannen, espen180, Aleks855, Solar Plexsus, Gustav, Nebuchadnezzar, Janhaa
[tex]{x sqrt{x+1}}{(1-sqrt{2x+3})}/({7-6x+4x^2})[/tex]
[tex]\frac{x\sqrt{x+1}-x\sqrt{2x^2+6x+3}}{4x^2+6x+7[/tex]
[tex]\frac{x^2\sqrt{\frac{1}{x}+\frac{1}{x^2}}-x^2\sqrt{2+\frac{6}{x}+\frac{3}x^2}}{x^2(4+\frac{6}{x}+\frac{7}{x^2})[/tex]
[tex]\frac{\sqrt{\frac{1}{x}+\frac{1}{x^2}}-\sqrt{2+\frac{6}{x}+\frac{3}x^2}}{(4+\frac{6}{x}+\frac{7}{x^2})[/tex]
[tex]\lim_{x\to \infty}\frac{\sqrt{\frac{1}{x}+\frac{1}{x^2}}-\sqrt{2+\frac{6}{x}+\frac{3}x^2}}{(4+\frac{6}{x}+\frac{7}{x^2})}=\frac{\sqrt{0+0}-\sqrt{2+0+0}}{(4+0+0)}=-\frac{\sqrt{2}}{4} = -\frac{1}{2\sqrt{2}[/tex]
[tex]\frac{x\sqrt{x+1}-x\sqrt{2x^2+6x+3}}{4x^2+6x+7[/tex]
[tex]\frac{x^2\sqrt{\frac{1}{x}+\frac{1}{x^2}}-x^2\sqrt{2+\frac{6}{x}+\frac{3}x^2}}{x^2(4+\frac{6}{x}+\frac{7}{x^2})[/tex]
[tex]\frac{\sqrt{\frac{1}{x}+\frac{1}{x^2}}-\sqrt{2+\frac{6}{x}+\frac{3}x^2}}{(4+\frac{6}{x}+\frac{7}{x^2})[/tex]
[tex]\lim_{x\to \infty}\frac{\sqrt{\frac{1}{x}+\frac{1}{x^2}}-\sqrt{2+\frac{6}{x}+\frac{3}x^2}}{(4+\frac{6}{x}+\frac{7}{x^2})}=\frac{\sqrt{0+0}-\sqrt{2+0+0}}{(4+0+0)}=-\frac{\sqrt{2}}{4} = -\frac{1}{2\sqrt{2}[/tex]
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Solar Plexsus
- Over-Guru
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- Joined: 03/10-2005 12:09
Her trengs ingen "avansert" algebraisk omforming. Det holder å dele med x[sup]2[/sup] i teller og nevner. Da får vi
[tex]\frac{\sqrt{1 \:+\: \frac{1}{x}} \; \Big( \frac{1}{\sqrt{x}} \: - \: \sqrt{2 \:+\: \frac{3}{x}} \; \Big)}{4 \:-\: \frac{6}{x} \:+\: \frac{7}{x^2}} \; \rightarrow \; - \, \frac{\sqrt{2}}{4} \;[/tex] når [tex]\; x \rightarrow \infty.[/tex]
[tex]\frac{\sqrt{1 \:+\: \frac{1}{x}} \; \Big( \frac{1}{\sqrt{x}} \: - \: \sqrt{2 \:+\: \frac{3}{x}} \; \Big)}{4 \:-\: \frac{6}{x} \:+\: \frac{7}{x^2}} \; \rightarrow \; - \, \frac{\sqrt{2}}{4} \;[/tex] når [tex]\; x \rightarrow \infty.[/tex]
Jarle10 wrote:[tex]{x sqrt{x+1}}{(1-sqrt{2x+3})}/({7-6x+4x^2})[/tex]
[tex]\frac{x\sqrt{x+1}-x\sqrt{2x^2+6x+3}}{4x^2+6x+7[/tex]
[tex]\frac{x^2\sqrt{\frac{1}{x}+\frac{1}{x^2}}-x^2\sqrt{2+\frac{6}{x}+\frac{3}x^2}}{x^2(4+\frac{6}{x}+\frac{7}{x^2})[/tex]
[tex]\frac{\sqrt{\frac{1}{x}+\frac{1}{x^2}}-\sqrt{2+\frac{6}{x}+\frac{3}x^2}}{(4+\frac{6}{x}+\frac{7}{x^2})[/tex]
[tex]\lim_{x\to \infty}\frac{\sqrt{\frac{1}{x}+\frac{1}{x^2}}-\sqrt{2+\frac{6}{x}+\frac{3}x^2}}{(4+\frac{6}{x}+\frac{7}{x^2})}=\frac{\sqrt{0+0}-\sqrt{2+0+0}}{(4+0+0)}=-\frac{\sqrt{2}}{4} = -\frac{1}{2\sqrt{2}[/tex]
[tex]\frac{x^2\sqrt{\frac{1}{x}+\frac{1}{x^2}}-x^2\sqrt{2+\frac{6}{x}+\frac{3}x^2}}{x^2(4+\frac{6}{x}+\frac{7}{x^2})[/tex]
Denne delen din skjønner jeg ikke...hvor kom 6x fra, og hvordan fikk du x^2 foran rottegnet?
(Algebraen min er litt rusten...)
[tex]x \sqrt{x+1}(1-\sqrt{2x+3}) = 1 \cdot x \sqrt{x+1}-x \sqrt{x+1} \cdot \sqrt{2x+3}=x\sqrt{x+1}-x\sqrt{(x+1)(2x+3)} = x\sqrt{x+1}-x\sqrt{2x^2+5x+3)}[/tex]
Glapp visst inn en liten slurvefeil, det skal være 5x, og ikke 6x, men det hadde ingenting å si for grenseverdien.
Glapp visst inn en liten slurvefeil, det skal være 5x, og ikke 6x, men det hadde ingenting å si for grenseverdien.