Skillnad mellan versioner av "2.3a Lösning 6c"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 1: | Rad 1: | ||
:<math> \begin{array}{rcl} \lim_{x \to \infty}\,\, {x^2\,-\,2\,x\,+\,3 \over 2\,x^2\,+\,5\,x\,-\,3} & = & \lim_{x \to \infty}\,\, {x^2/x^2\,-\,2\,x/x^2\,+\,3/x^2 \over 2\,x^2/x^2\,+\,5\,x/x^2\,-\,3/x^2} \,=\, \lim_{x \to \infty}\,\, {1\,-\,2/x\,+\,3/x^2 \over 2\,+\,5/x\,-\,3/x^2} \\ | :<math> \begin{array}{rcl} \lim_{x \to \infty}\,\, {x^2\,-\,2\,x\,+\,3 \over 2\,x^2\,+\,5\,x\,-\,3} & = & \lim_{x \to \infty}\,\, {x^2/x^2\,-\,2\,x/x^2\,+\,3/x^2 \over 2\,x^2/x^2\,+\,5\,x/x^2\,-\,3/x^2} \,=\, \lim_{x \to \infty}\,\, {1\,-\,2/x\,+\,3/x^2 \over 2\,+\,5/x\,-\,3/x^2} \\ | ||
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& = & {1\,-\,0\,+\,0 \over 2\,+\,0\,-\,0} \,=\, {1 \over 2} \end{array}</math> | & = & {1\,-\,0\,+\,0 \over 2\,+\,0\,-\,0} \,=\, {1 \over 2} \end{array}</math> | ||
Versionen från 28 september 2014 kl. 17.00
\[ \begin{array}{rcl} \lim_{x \to \infty}\,\, {x^2\,-\,2\,x\,+\,3 \over 2\,x^2\,+\,5\,x\,-\,3} & = & \lim_{x \to \infty}\,\, {x^2/x^2\,-\,2\,x/x^2\,+\,3/x^2 \over 2\,x^2/x^2\,+\,5\,x/x^2\,-\,3/x^2} \,=\, \lim_{x \to \infty}\,\, {1\,-\,2/x\,+\,3/x^2 \over 2\,+\,5/x\,-\,3/x^2} \\ \\ & = & {1\,-\,0\,+\,0 \over 2\,+\,0\,-\,0} \,=\, {1 \over 2} \end{array}\]
