A question regarding the order of an asymptotic estimate

Question

Suppose that $m, n \in \mathbb{N}$ such that \begin{equation} m \cdot \log m = n, \end{equation} where the logarithm is in the natural base. How can we estimate the solution $m = m(n)$ asymptotically? Especially, can we find a function $f(n)$ such that \begin{equation} \lim_{n \rightarrow \infty} \frac{m}{f(n)} = 1. \end{equation} Thanks very much.

Answer 1

Let $f(n)=n/\log(n)$. Then, $$ \frac{m}{f(n)}=\frac{m}{n/\log(n)}=\frac{m\log(n)}{n}. $$ Since $n=m\log m$, we can substitute to get $$ \frac{m}{f(n)}=\frac{m\log(n)}{m\log(m)}=\frac{\log(n)}{\log(m)}. $$ Since $n=m\log m$, $\log(n)=\log(m)+\log(\log(m))$. Therefore, $$ \frac{m}{f(n)}=\frac{\log(n)}{\log(m)}=\frac{\log(m)+\log(\log(m))}{\log(m)}=1+\frac{\log(\log(m))}{\log(m)}. $$ As $m$ grows, the second term vanishes.