# Thread: lim n->infinity

1. ## lim n->infinity

Hi!,

I have a problem solving this.. I am half way through it but somehow cannot finish it. Any help is really appreciated

lim [ (n ^ ln n)/ (a ^ n)]
n->00

Cheers!
Ramya

2. Originally Posted by Ramya
Hi!,

I have a problem solving this.. I am half way through it but somehow cannot finish it. Any help is really appreciated

lim [ (n ^ ln n)/ (a ^ n)]
n->00

Cheers!
Ramya
Consider the case $\displaystyle 1 < a < + \infty$

$\displaystyle \mathop {\lim }\limits_{n \to \infty } \frac{{{n^{\ln n}}}} {{{a^n}}} = \mathop {\lim }\limits_{n \to \infty } \exp \ln \frac{{{n^{\ln n}}}}{{{a^n}}} = \mathop {\lim }\limits_{n \to \infty } \exp \left\{ {\ln {n^{\ln n}} - \ln {a^n}} \right\} =$

$\displaystyle = \mathop {\lim }\limits_{n \to \infty } \exp \left\{ {{{\ln }^2}n - n\ln a} \right\} = \mathop {\lim }\limits_{n \to \infty } \exp \frac{{{{\ln }^4}n - {n^2}{{\ln }^2}a}}{{{{\ln }^2}n + n\ln a}} =$

$\displaystyle = \mathop {\lim }\limits_{n \to \infty } \exp \frac{{{{\left( {\frac{{\ln n}}{{\sqrt n }}} \right)}^4} - {{\ln }^2}a}}{{{{\left( {\frac{{\ln n}}{n}} \right)}^2} + \frac{{\ln a}}{n}}} = \exp \frac{{{{\left( {\mathop{\lim }\limits_{n \to \infty } \frac{{\ln n}}{{\sqrt n }}} \right)}^4} -{{\ln }^2}a}}{{{{\left( {\mathop {\lim }\limits_{n \to \infty } \frac{{\ln n}} {n}} \right)}^2}}}=$

$\displaystyle = \exp \frac{{{{\left( {\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {\ln n} \right)}^\prime }}}{{{{\left( {\sqrt n } \right)}^\prime }}}} \right)}^4} - {{\ln }^2}a}}{{{{\left( {\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {\ln n} \right)}^\prime }}}{{{{\left( n \right)}^\prime }}}} \right)}^2}}} = \exp$$\displaystyle \frac{{{{\left( { - \frac{1}{2}\mathop {\lim }\limits_{n \to \infty } \frac{1}{{\sqrt n }}} \right)}^4} - {{\ln }^2}a}}{{{{\left( {\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}} \right)}^2}}} =$

$\displaystyle = \exp \frac{{0 - {{\ln }^2}a}}{0} = \exp \left\{ { - \infty } \right\} = 0.$

3. Thanks a ton!