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What is the limit of the sequence an=(n^2-n+7)/(2n^3+n^2) ?n tends to infinite

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sapon | Student, Undergraduate | eNoter

Posted July 10, 2011 at 10:47 PM via web

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What is the limit of the sequence an=(n^2-n+7)/(2n^3+n^2) ?

n tends to infinite

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giorgiana1976 | College Teacher | Valedictorian

Posted July 10, 2011 at 10:52 PM (Answer #1)

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We'll force the factor n^2 to numerator and we'll get:

lim an = lim n^2*(1 - 1/n + 7/n^2)/(2n^3+n^2)

We'll force the factor n^3 to denominator and we'll get:

lim an = lim n^2*(1 - 1/n + 7/n^2)/n^3*(2 + 1/n)

We'll simplify:

lim an = lim (1 - 1/n + 7/n^2)/n*(2 + 1/n)

The limits of the sequences 1/n , 7/n^2  converge to zero, therefore we'll get:

lim an = 1/2*infinite

lim an = 1/infinite = 0

If n approaches to infinite, the given sequence (an) converges to zero.

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