limit of ( n/(n-4) )^n as n goes to infinity??

- anonymous

limit of ( n/(n-4) )^n as n goes to infinity??

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- anonymous

positive infinitiy

- anonymous

or wait, nvm

- anonymous

1

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## More answers

- anonymous

yea..the calculator says e^4...have no idea how to get there

- anonymous

\[\lim_{n \rightarrow \infty} [\frac{n}{n-4}]^n\]
?

- anonymous

yes thats the prob

- anonymous

are you sure that your answer is e^4?

- anonymous

because I got 1 too lol

- anonymous

ok yea thats wat i was thinking too..thank u!

- anonymous

\[\lim_{n \rightarrow \infty} e^{n \ln \frac{n}{n-4}}\]
first you find the limit of the fraction :
\[\lim_{n \rightarrow \infty} \frac{n}{n-4} = \lim_{n \rightarrow \infty} \frac{n}{n} = 1\]
so now plug this value in the first equation I wrote:
\[\lim_{n \rightarrow \infty} e^0 = 1\]
np :)

- anonymous

I get \[e^4\]

- anonymous

how? .-.

- anonymous

hmm, I prolly missed out a point that I can't seem to find ._.

- nowhereman

No, 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}{n-4}\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

Set the expression equal to y and take the log of both sides. You'll end up with\[\log y = n \log \frac{n}{n-4}=n \log \frac{1}{1-4/n}=-n \log (1-4/n)=-\frac{\log (1-4/n)}{1/n}\]That's now indeterminate, so you can apply L'Hopital's and take the limit. Then undo the log.

- nowhereman

\[\lim_{n→∞}{\left(\frac{n}{n-4}\right)^{n}} = \lim_{m→∞}{\left(\frac{m+4}{m}\right)^{m+4}} = \lim_{m→∞}{\left(1+\frac{4}{m}\right)^{m}} \cdot \lim_{m→∞}{\left(\frac{m+4}{m}\right)^{4}}\]\[ = e^4 \cdot 1\]

- anonymous

when 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

You should get what nowhereman got.

- anonymous

/I got.

- nowhereman

misha: true, but you can only pull the limit inside, if the outer part does not depend on the variable (n here)

- anonymous

then I was wrong, misha follow both of their ways ^_^

- nowhereman

You should probably take a look at http://en.wikipedia.org/wiki/Exponential_function or something similar.

- anonymous

ok thanks for clarifying!

- anonymous

I gave you a medal sstarica

- anonymous

lol, thank you :) but I didn't work for it

- anonymous

for being a good sport :)

- anonymous

^_^ thank you

- anonymous

You can have one too, nowhereman...

- anonymous

lol I gave both of you a medal

- anonymous

loki

- anonymous

I prolly am just seeing it, but just to make sure, is there something wrong b/w you and nowhereman?

- anonymous

Not that I am aware of. Why?

- anonymous

I've never talked to him.

- anonymous

mm, lol nvm, you guys just act weird around each other?

- anonymous

?

- anonymous

lol, I'm prolly just seeing it, it's nothing ^_^

- anonymous

he doesn't 'chat'

- anonymous

noticed :)

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