1. ## Another Limit Problem

Just wondering if somebody can help out with the workings for the following:- lim n-> inf (ln n)^1/n

2. I would say $\displaystyle \lim_{n \to \infty}(\ln(n))^{\frac{1}{n}} =(\ln(\infty))^{\frac{1}{\infty}} =(\ln(\infty))^{0}=1$

3. Thanks for that. So bloody obvious.

4. Originally Posted by p75213
Just wondering if somebody can help out with the workings for the following:- lim n-> inf (ln n)^1/n
$\displaystyle (\ln{n})^{\frac{1}{n}} = e^{\ln{[(\ln{n})^{\frac{1}{n}}]}}$

$\displaystyle = e^{\frac{1}{n}\ln{(\ln{n})}}$

$\displaystyle = e^{\frac{\ln{(\ln{n})}}{n}}$.

Therefore:

$\displaystyle \lim_{n \to \infty}(\ln{n})^{\frac{1}{n}} = \lim_{n \to \infty} e^{\frac{\ln{(\ln{n})}}{n}}$

$\displaystyle = e^{\lim_{n \to \infty}\frac{\ln{(\ln{n})}}{n}}$

$\displaystyle = e^{\lim_{n \to \infty}\frac{\frac{1}{n\ln{n}}}{1}}$ by L'Hospital's Rule

$\displaystyle = e^{\lim_{n \to \infty}\frac{1}{n\ln{n}}}$

$\displaystyle = e^0$

$\displaystyle = 1$.

5. Originally Posted by pickslides
I would say $\displaystyle \lim_{n \to \infty}(\ln(n))^{\frac{1}{n}} =(\ln(\infty))^{\frac{1}{\infty}} =(\ln(\infty))^{0}=1$
Actually $\displaystyle \infty^0$ is an indeterminate form.

You need to undergo a transformation to get something of the form $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$ so you can use L'Hospital's Rule.