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anonymous
 5 years ago
limit of ( n/(n4) )^n as n goes to infinity??
anonymous
 5 years ago
limit of ( n/(n4) )^n as n goes to infinity??

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea..the calculator says e^4...have no idea how to get there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} [\frac{n}{n4}]^n\] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you sure that your answer is e^4?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because I got 1 too lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok yea thats wat i was thinking too..thank u!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} e^{n \ln \frac{n}{n4}}\] first you find the limit of the fraction : \[\lim_{n \rightarrow \infty} \frac{n}{n4} = \lim_{n \rightarrow \infty} \frac{n}{n} = 1\] so now plug this value in the first equation I wrote: \[\lim_{n \rightarrow \infty} e^0 = 1\] np :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm, I prolly missed out a point that I can't seem to find ._.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.1No, that is not how it works. This way there wouldn't be any constant e. The problem with your approach is, that when the ln goes to zero, at the same time n goes to infinity and so you can't say that their product goes to zero! But you can do the following substitution: \[\lim_{n→∞}{\left(\frac{n}{n4}\right)^{n}} = \lim_{m→∞}{\left(\frac{m+4}{m}\right)^{m+4}} = \lim_{m→∞}{\left(1+\frac{4}{m}\right)^{m}} \cdot \lim_{m→∞}{\left(\frac{m+4}{m}\right)^{4}} = e^4 \cdot 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Set the expression equal to y and take the log of both sides. You'll end up with\[\log y = n \log \frac{n}{n4}=n \log \frac{1}{14/n}=n \log (14/n)=\frac{\log (14/n)}{1/n}\]That's now indeterminate, so you can apply L'Hopital's and take the limit. Then undo the log.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{n→∞}{\left(\frac{n}{n4}\right)^{n}} = \lim_{m→∞}{\left(\frac{m+4}{m}\right)^{m+4}} = \lim_{m→∞}{\left(1+\frac{4}{m}\right)^{m}} \cdot \lim_{m→∞}{\left(\frac{m+4}{m}\right)^{4}}\]\[ = e^4 \cdot 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when the degree is same on the top and bottom, dont you jus look at the coeffecients of the highest degree which is 1/1= 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should get what nowhereman got.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.1misha: true, but you can only pull the limit inside, if the outer part does not depend on the variable (n here)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then I was wrong, misha follow both of their ways ^_^

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.1You should probably take a look at http://en.wikipedia.org/wiki/Exponential_function or something similar.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok thanks for clarifying!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I gave you a medal sstarica

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, thank you :) but I didn't work for it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for being a good sport :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can have one too, nowhereman...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol I gave both of you a medal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I prolly am just seeing it, but just to make sure, is there something wrong b/w you and nowhereman?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Not that I am aware of. Why?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've never talked to him.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mm, lol nvm, you guys just act weird around each other?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, I'm prolly just seeing it, it's nothing ^_^
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