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Mathematics 9 Online
OpenStudy (anonymous):

lim x→∞ [(n^2+n)/n^3]=0

OpenStudy (anonymous):

What do you need? Check your answer or an explanation on how to get 0?

OpenStudy (anonymous):

An explanation please...

OpenStudy (anonymous):

Help figuring it out, not an answer.

OpenStudy (anonymous):

I have to prove that the limit equals zero.

OpenStudy (anonymous):

\[\lim_{n\to \infty}\frac{n^2+1}{n^3}\] Dividing by the highest power of n in the numerator, you have \[\large\lim_{n\to \infty}\frac{\frac{n^2+1}{n^2}}{\frac{n^3}{n^2}}=\lim_{n\to \infty}\frac{1+\frac{1}{n^2}}{n}\] \[\text{As }n\to\infty, \frac{1}{n^2}\to0, \text{ so you have}\\ \lim (\cdots)= \frac{1+0}{\infty}=0\]

OpenStudy (anonymous):

Wait, a proof? Like epsilon-delta?

OpenStudy (anonymous):

There are no epsilon nor deltas in his study notes, so I don't think so

OpenStudy (anonymous):

I guess just "proving" using limit rules, laws, facts??? His course notes are sparse since it is a "review course"

OpenStudy (anonymous):

The professor's only hint for the problem set is (You can use the fact lim n→∞ (1/n) = 0)

OpenStudy (anonymous):

Oh, well in that case, I hope my explanation was clear enough. I did use that hint/fact towards the end.

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