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

1. IrishBoy123

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

2. anonymous

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))...

3. anonymous

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

4. IrishBoy123

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....

5. anonymous

0

6. IrishBoy123

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.

7. anonymous

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

8. perl

now plug in infinity

9. anonymous

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

10. anonymous

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

11. perl

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

12. anonymous

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

13. IrishBoy123

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!

14. anonymous

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..

15. anonymous

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

16. IrishBoy123

well done!