 Solution

Write the equation as $ϵx^3-(x-1{)}^2=0$. For $ϵ\to 0^{+}$, two roots of a polynomial approach the double root one. Their splitting follows a fractional-power dominant balance, rather than an ordinary expansion in integer powers.

Put $\delta =\sqrt{ϵ}$ and $x=1+a\delta +b{\delta }^2+\cdots$. Substitution givesThus $a=\pm 1$ and $b=3/2$. The two bounded roots have the Puiseux series expansionsThe first two nonzero terms are $1\pm \sqrt{ϵ}$; the displayed correction also resolves their common order-$ϵ$ shift.

The third root escapes to infinity because the leading polynomial degree drops when $ϵ=0$. Balance $ϵx^3$ against $x^2$, giving $x=O({ϵ}^{-1})$. Set $X=ϵx$; the equation becomesFor $X=1+Aϵ+B{ϵ}^2+\cdots$, the first two coefficient equations are $A+2=0$ and $B+2A^2+2A-1=0$. HenceIts first two nonzero terms are ${ϵ}^{-1}-2$. These three distinct branches account for all roots of the cubic for sufficiently small positive $ϵ$. The Vieta formulas provide a consistency check: their sum is ${ϵ}^{-1}$, with the constant and order-$ϵ$ terms cancelling.

If $ϵ$ approaches zero through negative values, the bounded pair is complex: $x_{\pm }=1\pm i\sqrt{|ϵ|}+\frac{3}{2}ϵ+O(|ϵ{|}^{3/2})$. The large-root expansion is unchanged as a real asymptotic formula. The positive-parameter regime is used for the boundary-layer problem below.