Question

find the limit of k^(1/k) as k --> infinity

Answer 1

hint # 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 \(ln \ y = ln(k^{1/k})\)

Answer 2

yes 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))...

Answer 3

I have the lim ln k^(1/k) ==> lim ln^(k) / k ...what's the derivative of ln^k?

Answer 4

on 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....

Answer 5

0

Answer 6

not 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.

Answer 7

\[\lim k ^{1/k} -> \lim \ln k ^{1/k} = \frac{ 1 }{ k }\ln k ->L'H = \lim \frac{ \frac{ 1 }{ k } }{ 1 }=\frac{ 1 }{k }\]

Answer 8

now plug in infinity

Answer 9

as k --> infinity, 1/k --> 0

Answer 10

yes i know but if i plug in infinity to the original equation, k^(1/k) = 1

Answer 11

if you plug in original equation you get an indeterminate form \( \infty ^0 \)

Answer 12

hmm, true. that's why I had to do L'H. Thanks

Answer 13

recap: 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!

Answer 14

oh. 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..

Answer 15

That took me a long time to get...but finally. Thanks everyone! The problem makes senses now.

Answer 16

well done!