Timeline for Compare the growth of $2^{n}$ to $n^{a}$ for all $a \in \mathbb{N}$?
Current License: CC BY-SA 3.0
6 events
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| Apr 13, 2017 at 12:48 | history | edited | CommunityBot |
replaced http://cs.stackexchange.com/ with https://cs.stackexchange.com/
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| Dec 14, 2015 at 1:05 | comment | added | D.W.♦ | Ahh, my mistake: I missed the title. (I've edited the question to make sure it appears in the body, so others don't miss it.) So, that addresses Raphael's point a), and your edit to the answer addresses Raphael's point 3). Thank you for the nice improvements! Seems like a nice answer to me. | |
| Dec 13, 2015 at 20:16 | history | edited | Renato Sanhueza | CC BY-SA 3.0 |
Add L'Hopital's rule's conditions.
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| Dec 5, 2015 at 14:57 | comment | added | Renato Sanhueza | Hi. The OP asked for $a\in \mathbb{N}$. What do you mean about inserting $\infty$ as a symbolic value?. Finally L'Hopital's conditions meet so I can't see the problem. | |
| Dec 5, 2015 at 9:23 | comment | added | Raphael | a) What if $a \not\in\mathbb{N}$? b) Your limit calculation is wrong. You don't get to insert $\infty$ as a symbolic value. 3) L'Hospital has certain conditions you have to check. | |
| Dec 5, 2015 at 4:42 | history | answered | Renato Sanhueza | CC BY-SA 3.0 |