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Let $x(n)$ be the solution to the following equation $$ x=-\frac{\log(x)}{n} \quad \quad \quad \quad (1) $$ as a function of $n,$ where $n \in \mathbb N.$ How would you find the asymptotic behaviour of the solution, i.e. a function $f$ of $n$ such that there exist constants $A,B$ and $n_0\in\mathbb N$ so that it holds $$Af(n) \leq x(n) \leq Bf(n)$$ for all $n > n_0$ ?

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Call $u_n:t\mapsto t\mathrm e^{nt}$, then $x(n)$ solves $u_n(x(n))=1$. For every $a$, introduce $$ x_a(n)=\frac{\log n}n-a\frac{\log\log n}n. $$ Simple computations show that, for every fixed $a$, $u_n(x_a(n))\cdot(\log n)^{a-1}\to1$ when $n\to\infty$. Thus, for every $a\gt1$, there exists some finite index $n(a)$ such that $x(n)\geqslant x_a(n)$ for every $n\geqslant n(a)$, and, for every $a\lt1$, there exists some finite index $n'(a)$ such that $x(n)\leqslant x_a(n)$ for every $n\geqslant n'(a)$. Finally, when $n\to\infty$, $$ nx(n)=\log n-\log\log n+o(\log\log n). $$ The assertion in your post holds with $f(n)=(\log n)/n$, $n_0=\max\{n(A),n'(B)\}$, $B=1$ and every $A\lt1$.

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The solution can be written as $x_n=W(n)/n$, where $W$ is Lambert's function. –  Julián Aguirre Mar 8 '12 at 9:21
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