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anonymous
 one year ago
find the limit of k^(1/k) as k > infinity
anonymous
 one year ago
find the limit of k^(1/k) as k > infinity

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3hint # 1 : stuff some numbers (let k = 100, 100, 10,000] into your calculator and see what you get. it is not \(\infty\) hint #2: take logs and see if you can use L'Hopital. i think you can. [taking logs will get rid of that rather ugly exponent.] so run with \(ln \ y = ln(k^{1/k})\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes i just plugged in some #s and the lim is definitely not infinity... I'll have to use L'H but I'm stuck with ln(K^(1/k)) now... do I take the e^ ln(y) and e^ln(k^(1/k))...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have the lim ln k^(1/k) ==> lim ln^(k) / k ...what's the derivative of ln^k?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3on RHS you should have \(ln(k^{1/k}) = \frac{1}{k} ln \ k\ = \frac{ln \ k}{k} \). that is l'Hopital territory. incidentally, after plugging in some numbers, what do you think the limit is? it is much easier when you know that....

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3not sure which question you are now answering :p if it is as \(k \rightarrow \ \infty, \ k^{(1/k)} = 0\), you need to try again. e.g.: \(100^{(0.01)}\) = ???? for \(\frac{ln \ k}{k}\), you should see that as \(k \rightarrow \infty\), you have \(\frac{ln \ \infty}{\infty} = \frac{\infty}{\infty}\), so you can use L'Hopital.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim k ^{1/k} > \lim \ln k ^{1/k} = \frac{ 1 }{ k }\ln k >L'H = \lim \frac{ \frac{ 1 }{ k } }{ 1 }=\frac{ 1 }{k }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0as k > infinity, 1/k > 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes i know but if i plug in infinity to the original equation, k^(1/k) = 1

perl
 one year ago
Best ResponseYou've already chosen the best response.0if you plug in original equation you get an indeterminate form \( \infty ^0 \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm, true. that's why I had to do L'H. Thanks

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3recap: we made the LHS ln(y) and we want to know what **y** is as \(k \rightarrow \infty\) you had this licked when you said: "as k > infinity, 1/k > 0". that was L'Hopital applied to the RHS. which means the LHS is what?!?!? IOW, problem solved!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh. to get rid of ln, I need to raise e^(ln(y)) which means I need to do e^(1/k) on the RHS..which is e^0 = 1..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That took me a long time to get...but finally. Thanks everyone! The problem makes senses now.
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