= Solution
The limiting <polynomial> has a <double root> at $x=-1$, so two roots require a <Puiseux series>. For positive $\epsilon$, put $\delta=\sqrt\epsilon$ and $x=-1+a\delta+b\delta^2+\cdots$. Substitution and <dominant balance> give
$$
0=(a^2-1)\delta^2+(2ab+3a)\delta^3+O(\delta^4).
$$
Thus $a=\pm1$ and $b=-3/2$. The two finite roots have the three-term <asymptotic expansions>
$$
\boxed{x_\pm=-1\pm\epsilon^{1/2}-\frac32\epsilon+O(\epsilon^{3/2}).}
$$
This is <square-root splitting of a double polynomial root>. For negative $\epsilon$, the same formulas apply with a chosen complex square root and describe a <complex conjugate> pair.
The third root is a <divergent algebraic root in a singular perturbation>. Set $x=X/\epsilon$: the equation for $X$ is $X^3+X^2+2\epsilon X+\epsilon^2=0$. Its nonzero limiting root is $X=-1$. Substituting $X=-1+A\epsilon+B\epsilon^2+\cdots$ yields $A=2$ and $B=3$, hence
$$
\boxed{x_L=-\epsilon^{-1}+2+3\epsilon+O(\epsilon^2).}
$$
As an independent check, <Vieta formulas> give the sum of roots as $-1/\epsilon$: the two finite-root expansions sum to $-2-3\epsilon+O(\epsilon^2)$, giving the same large-root expansion.
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