Find the supremum of the following set and justify why: $$ A:= \left\{\frac{n^{2}+6}{n+8} \ \Bigg|\ n\in\mathbb{N}\right\} . $$
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What have you tried? Don't dump your homework here, please. – Oct 27 '16 at 15:32
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Note that $\frac{n^{2}+6}{n+8}\to\infty$ as $n\to\infty$.
ervx
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Ok thank you, and how would you prove formally that the set has no upper bound or would the above observation be sufficient? Thanks! – Fkins Oct 27 '16 at 15:22
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Well, given any $M<\infty$, by the above observation, there is an $n\in\mathbb{N}$ such that $\frac{n^{2}+6}{n+8}>M$. Thus, $A$ can have no finite upper bound. – ervx Oct 27 '16 at 15:24
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Is there any upper bound to $\frac{n^2+6}{n+8}$? Can you find any number that is larger than $\frac{n^2+6}{n+8}$ for all $n$?
What if instead of $\frac{n^2+6}{n+8}$ you just had $\frac{n^2}{n} = n$? Do the additional constants change the asymptotic behavior for large $n$? What does that tell you about the possibility of any finite supremum?
SarthakC
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