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OpenStudy (anonymous):

find lim x->infinity (1-2x^2)/(3x^3+x)

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OpenStudy (anonymous):

use l'Hôpital's rule: http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule until you get -4/18x which gives you lim (x->infinity)=0

OpenStudy (mathteacher1729):

I don't know how much work they require you to show or what they accept in terms of work shown, but you can note that the power in the denominator is greater than the power in the numerator so ignoring constants you essentially have lim x goes to infinity x^2 / x^3 which is really 1 / x ... and we all know lim x goes to infinity of 1 / x = 0. :) Hope this helps.

OpenStudy (anonymous):

Yes,\[ax ^{m}/ax ^{n}\]m<n, lim goes to 0

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