# help finding a limit

• May 18th 2008, 05:35 PM
cowboys111
help finding a limit
Hi, can someone show me how to find this limit i cant figure it out. Lim as k goes to infinity (k^(1/k))/5 ?
Thank you
• May 18th 2008, 05:41 PM
icemanfan
Quote:

Originally Posted by cowboys111
Hi, can someone show me how to find this limit i cant figure it out. Lim as k goes to infinity (k^(1/k))/5 ?
Thank you

I'm not sure how to prove it aside from just plugging in really large numbers, but $\lim_{k \to \infty} k^{1/k} = 1$.
• May 18th 2008, 05:48 PM
Mathstud28
Quote:

Originally Posted by cowboys111
Hi, can someone show me how to find this limit i cant figure it out. Lim as k goes to infinity (k^(1/k))/5 ?
Thank you

$L=\lim_{k\to\infty}\frac{\sqrt[k]{k}}{5}$

Taking the natural log of both sides gives

$\ln(L)=\lim_{k\to\infty}\frac{\ln(k)}{k}-\ln(5)$

Since this yields an indeterminate form we apply L'hopitals rule go get

$\ln(L)=\lim_{k\to\infty}\frac{\frac{1}{k}}{1}-\ln(5)=-\ln(5)$

So $L=e^{-\ln(5)}=\frac{1}{5}$
• May 18th 2008, 06:12 PM
Mathstud28
Quote:

Originally Posted by icemanfan
I'm not sure how to prove it aside from just plugging in really large numbers, but $\lim_{k \to \infty} k^{1/k} = 1$.

Try considering

$\forall{x}>1,1\leq{x^{\frac{1}{x}}}\leq\bigg(1+\fr ac{1}{\sqrt{x}}\bigg)$
• May 18th 2008, 06:18 PM
cowboys111
thank you