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lim as n-->infiniti, of a(sub n) = n/(1+(n^1/2)) i do not know how to approach it
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divide each term by n^1/2
\[\lim_{n \rightarrow \infty}= n / 1 + \sqrt{n} solve for convergence or divergence, i do \not know where \to start\]
oh ty
so it diverges
i got stuck at diviging n(2/2) / n(1/2)
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we get lim n^(.5) /[ 1/ n^.5 + 1 )
so thats infinity / ( 0 + 1)
Or you can realize that the order of the top is n, and the order of the bottom is sqrt(n). Since the order on top is greater, it diverges, so the limit tends to infinity.
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