Skillnad mellan versioner av "2.3a Lösning 6d"

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:<math> \lim_{x \to \infty}\,\, {x\,+\,1 \over x^2\,+\,1} & = & \lim_{x \to \infty}\,\, {x/x^2\,+\,1/x^2 \over 2\,x^2/x^2\,+\,1/x^2} \,=\, \lim_{x \to \infty}\,\, {1\,-\,2/x\,+\,3/x^2 \over 2\,+\,5/x\,-\,3/x^2}  </math>
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:<math> \lim_{x \to \infty}\,\, {x\,+\,1 \over x^2\,+\,1} \, = \, \lim_{x \to \infty}\,\, {x/x^2\,+\,1/x^2 \over 2\,x^2/x^2\,+\,1/x^2} \,=\, \lim_{x \to \infty}\,\, {1\,-\,2/x\,+\,3/x^2 \over 2\,+\,5/x\,-\,3/x^2}  </math>

Versionen från 28 september 2014 kl. 17.12

\[ \lim_{x \to \infty}\,\, {x\,+\,1 \over x^2\,+\,1} \, = \, \lim_{x \to \infty}\,\, {x/x^2\,+\,1/x^2 \over 2\,x^2/x^2\,+\,1/x^2} \,=\, \lim_{x \to \infty}\,\, {1\,-\,2/x\,+\,3/x^2 \over 2\,+\,5/x\,-\,3/x^2} \]