obtain a three-term asymptotic solution In the limit as $\epsilon \to \infty$, obtain a three-term asymptotic solution to the roots of the following equation.
$$\epsilon x^3+x^2-2x+1=0$$
What I've done so far
I've assumed $x=O(\epsilon^r)$ as $\epsilon \to \infty$ therefore,
$\epsilon x^3=O(\epsilon^{1+3r})$
$x^2=O(\epsilon^{2r})$
$x=O(\epsilon^{r})$
$1=O(\epsilon^{0})$
I'm not too sure what needs to be done from here. any help would be much appreciated.
 A: Outline (rather messy: is there a more elegant approach?): 


*

*First, prove that you must have $x=x(\epsilon)\xrightarrow[\epsilon\to\infty]{} 0$. 

*Then, you can get the expansion one term at a time...
Now, the second bullet (up to order two) below -- if I have not screwed up.
$$
1-2x + x^2+\epsilon x^3 = 1+\epsilon x^3 + o(1)
$$
so you need $\epsilon x^3 \operatorname*{\sim}_{\epsilon\to\infty} -1$. I.e., $x = -\frac{1}{\epsilon^{1/3}} + o\left(\frac{1}{\epsilon^{1/3}}\right)$.
Write $x = -\frac{1}{\epsilon^{1/3}} + h$, so that $x^3 = -\frac{1}{\epsilon} + \frac{3h}{\epsilon^{2/3}} + o\left(\frac{1}{\epsilon^{2/3}}\right)$. Plugging it back,
$$
1-2x + x^2+\epsilon x^3 = 1 -1 + 3h\epsilon^{1/3} +2\frac{1}{\epsilon^{1/3}} -2h + x^2 = 3h\epsilon^{1/3} +2\frac{1}{\epsilon^{1/3}} + o(h\epsilon^{1/3})
$$
To get that equal to $0$, this implies in particular $3h\epsilon^{1/3} +2\frac{1}{\epsilon^{1/3}}=o(1)$, so that
$$
h = \frac{-2}{3\epsilon^{2/3}} + o\left(\frac{1}{\epsilon^{2/3}}\right)
$$
i.e.
$$
x = -\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + o\left(\frac{1}{\epsilon^{2/3}}\right)
$$
Now, for the final step, write 
$$
x = -\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + g
$$
with $g=o\left(\frac{1}{\epsilon^{2/3}}\right)$.
$$
1-2x + x^2+\epsilon x^3 = 1 -2(-\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + g)+\left(-\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + g\right)^2+\epsilon\left(-\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + g\right)^3
$$
and expand (possibly with Taylor expansions to the right order if this helps). To satisfy the equation, this must be $o(1)$: with this constraint, you will get $g$.
A: Formally, we are looking for $x(\epsilon) \sim \epsilon^{-1/3}x_1 + \epsilon^{-2/3}x_2 + \epsilon^{-1}x_3 + \epsilon^{-4/3}x_4 + \dots$. This suggests the change of the parameter $s = \epsilon^{-1/3}$, and we now are interested in the behavior of solutions of $x(s)^3 + s^{3}(x(s)^2 - 2x(s) + 1) = 0$ as $s \to 0$. 
Now clearly $x(s)$ cannot go to infinity as $s \to 0$. So the $x(s)$ remain bounded as $s \to 0$. Take any subsequence $s_n \to 0$ such that $\lim_n x(s_n) = c$ exists, then $c^3 + 0\dot(c^2-2c+1) = 0$ and hence $c=0$. Therefore $x(s) \to 0$ as $s \to 0$..
Now substitute the expression for $x(s)$ into the equation, expand, and collect terms. You will obtain
$$ 0 = (1+x_1^3) + s(-2x_1+3x_1^2x_2) + s^2(x_1^2-2x_2+3x_1x_2^2+3x_1^2x_3) + \dots
$$
The constant term must be zero, so $x_1 = -1$. The term with $s$ must also be zero, so $x_2 = -2/3$. And so on. You will get $x_3 = -1/3, x_4 = -10/81, x_5 = -7/243$, using Mathematica.
