Asymptotic expansion, solving roots to an equation with dominant balance, what went wrong in my approach? So I wanted to compute the asymptotic expansion of the roots to, as $\epsilon \to 0$,
$$\epsilon x^3-x^2+2x-1=0$$
Now when I tried to find $x\sim x_0+\epsilon x_1+\epsilon^2x_2+...$ I ran into trouble as at $O(\epsilon)$, I had $0=2x_1-2x_1+1.$ Now I just wanted to know why did I reach such contradiction? Is it because I implicitly balanced $x$ with $1$? In other words, is it because if I assume $x=O(1)$ then $x^2$ would also have order $1$ and thus the two terms I balanced are not dominant enough?
Moreover, I applied dominant balance to $\epsilon x^3$ and $x^2$ to reach one of the root, how could I reach the other two roots via dominant balance? This is because if I balance any other two terms in the above expression, there is always another term that is larger or have the same order as $\epsilon \to 0.$
Many thank in advance!

Another way to obtain the expansion is by using series inversion. We have
\begin{align*}
\varepsilon  = \frac{{(x - 1)^2 }}{{x^3 }} & = \frac{{(x - 1)^2 }}{{1 + 3(x - 1) + 3(x - 1)^2  + (x - 1)^3 }} \\ & = (x - 1)^2 (1 - 3(x - 1) + 6(x - 1)^2  -  \cdots ),
\end{align*}
and thus
\begin{align*}
 \pm \sqrt \varepsilon &  = (x - 1)\left( {1 - \frac{3}{2}(x - 1) + \frac{{15}}{8}(x - 1)^2  -  \cdots } \right) \\ & = (x - 1) - \frac{3}{2}(x - 1)^2  + \frac{{15}}{8}(x - 1)^3  -  \cdots \,.
\end{align*}
By series inversion
$$
x - 1 =  \pm \sqrt \varepsilon   + \frac{3}{2}\varepsilon  \pm \frac{{57}}{8}\varepsilon ^{3/2}  +  \cdots ,
$$
i.e.,
$$
x = 1 \pm \sqrt \varepsilon   + \frac{3}{2}\varepsilon \pm \frac{{57}}{8}\varepsilon ^{3/2}  +  \cdots \,.
$$
Addendum. Following David's comment, I add the expansion for the third root. By Viète's formula and simple series manipulation, we find
\begin{align*}
x & = \frac{1}{\varepsilon }\frac{1}{{1 + \sqrt \varepsilon   + \frac{3}{2}\varepsilon  + \frac{{57}}{8}\varepsilon ^{3/2}  +  \cdots }}\frac{1}{{1 - \sqrt \varepsilon   + \frac{3}{2}\varepsilon  - \frac{{57}}{8}\varepsilon ^{3/2}  +  \cdots }}
\\ & = \frac{1}{\varepsilon }\frac{1}{{1 + 2\varepsilon  +  \cdots }} = \frac{1}{\varepsilon }(1 - 2\varepsilon  +  \cdots ) = \frac{1}{\varepsilon } - 2 +  \cdots \,.
\end{align*}
