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For real $x$, the function $F(x)=x^2{e}^{-x}$ has derivative $F^{\prime }(x)=x(2-x)e^{-x}$. It decreases from infinity to zero on the negative half-line, increases from zero to $4e^{-2}$ on $(0,2)$, and then decreases to zero. Hence sufficiently small positive $ϵ$ gives exactly three real roots: two near zero and one tending to positive infinity.

Set $\delta =\sqrt{ϵ}$ and write either small root as $x=s\delta +A{\delta }^2+O({\delta }^3)$, with $s=\pm 1$. Expanding $x^2{e}^{-x}$ gives ${\delta }^2+s(2A-1){\delta }^3+O({\delta }^4)$. Matching its value to ${\delta }^2$ fixes $A=1/2$. Thus the two asymptotic expansions areFor the large root, put $L=\log (1/ϵ)$. Taking logarithms gives $x-2\log x=L$, so $x/L\to 1$. Writing $x=L+r$ then gives $r=2\log L+2\log (1+r/L)$. Iteration yieldsIts two asymptotic expansion terms are of different logarithmic orders; a regular power series in $ϵ$ cannot describe it.

As an exact check, the real branches of Lambert W give $x_{-}=-2W_0(\sqrt{ϵ}/2)$, $x_{+}=-2W_0(-\sqrt{ϵ}/2)$ and $x_{\mathrm{large}}=-2W_{-1}(-\sqrt{ϵ}/2)$. If complex roots are intended instead, there are infinitely many, given by $-2W_j(\pm \sqrt{ϵ}/2)$. For each fixed nonprincipal branch their two logarithmic terms follow from $W_j(z)=L_j-\log L_j+O(\log |L_j|/|L_j|)$, where $L_j=\log z+2\pi ij$ and logarithms are continued on that branch. The three real roots are the ones selected above.

There are two positive roots for $u>1$: the derivative of $x{e}^{1/x}$ is $e^{1/x}(1-1/x)$, so the function decreases to its minimum $e$ at $x=1$ and then increases. Negative $x$ cannot solve the equation. For the larger root, expansion of the exponential function in $1/x$ givesso series reversion yieldsFor the smaller root, set $y=1/x$ and take a logarithm. Then $y-\log y=u$, so successive substitution gives $y=u+\log u+O(\log u/u)$ andThese are the first two nonzero terms of the two asymptotic expansions. Equivalently, the roots are $-1/W_0(-e^{-u})$ and $-1/W_{-1}(-e^{-u})$, from the real branches of Lambert W.