Let's call $u\_n$ the solution (it is unique) of $f\_n(x)=x^5+nx-1=0$. I'd like to find an asymptotic expansion of $u\_n$ of order two. I found $$u\_n=\frac{1}{n}-\frac{1}{n^6}+o(\frac{1}{n^6})$$

The method I used:

* We have $0 < u\_n < 1$, hence $0 < n u\_n =1-u\_n^5 < 1$ and $\lim\limits\_{n \to +\infty} u\_n = 0$.
* Therefore $\lim\limits\_{n \to +\infty} u\_n^5 = 0$ and $\lim\limits\_{n \to +\infty} nu\_n = 1$, which provides the first term of the development.
* I followed with $v\_n=u\_n-\frac{1}{n}$ to get $(v\_n+\frac{1}{n})^5+n v\_n=0$. I proved that $0 < 1+ n v\_n <1$ and followed to get the second term.

Somehow tedious...

**I there a more systematic approach using calculus?**