# How to solve this asymptotic equality?

1. Jun 5, 2015

### japplepie

How do I solve for x in the relationship below:

nx ~ n ln(n), as n -> infinity

The answer that I'm getting is x=1, but that must be wrong since 1 ~ ln(n) as n-> infinity is wrong.

2. Jun 5, 2015

### pasmith

There is no solution: $$\frac{\ln n}{n^k} \to \begin{cases} 0 & k > 0 \\ \infty & k \leq 0 \end{cases}$$

3. Jun 5, 2015

oh boy