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 $\lim_{n \to \infty}(\ln(n))^{\frac{1}{n}} =(\ln(\infty))^{\frac{1}{\infty}} =(\ln(\infty))^{0}=1$

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

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

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

Therefore:

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

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

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

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

$= e^0$

$= 1$.

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

You need to undergo a transformation to get something of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ so you can use L'Hospital's Rule.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}