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Take $ϵ\to 0^{+}$. A root on the smaller scale has $x=ϵu$, giving $u+2+ϵu^3=0$. Put $u=u_0+ϵu_1+O({ϵ}^2)$. Successive coefficients give $u_0=-2$ and $u_1+u_0^3=0$, hence $u_1=8$. The real root is thereforeThe other dominant balance is between the cubic and linear terms, requiring $x={ϵ}^{1/2}z$. Set $t={ϵ}^{1/2}$; the equation becomes $z^3+z+2t=0$. At $t=0$ the two nonzero roots are $z_0=\pm i$. With $z=z_0+t{z}_1+O(t^2)$, the first correction satisfies $(3z_0^2+1)z_1+2=0$, so $-2z_1+2=0$ and $z_1=1$. ThusThe limiting roots in both rescalings are simple, so the implicit function theorem justifies these regular expansions in their respective small parameters. The fractional powers in the original variable are a fractional-power root expansion. The real derivative $3x^2+ϵ$ is strictly positive; the cubic therefore has exactly one real root. Its other roots are the displayed complex conjugate pair, accounting for all three roots. Their leading real corrections sum to $-2ϵ+ϵ+ϵ=0$, as required by the absence of a quadratic coefficient.

Take $ϵ\to 0^{+}$. There are two real branches, on different scales. For a root tending to zero, the Taylor series givesThe dominant balance is $x^3\sim -6ϵ$, so a fractional-power root expansion yieldsMore precisely $x_{-}=-(6ϵ{)}^{1/3}+O(ϵ)$. An ordinary power series in integer powers of $ϵ$ would miss this scale.

There is also a root escaping to positive infinity. Boundedness of the sine term makes the leading dominant balance $x-ϵx^2=0$. The nonzero branch isIndeed the equation rearranges to $ϵx_{+}^2-x_{+}=ϵ-\sin x_{+}$, which yields $x_{+}={ϵ}^{-1}+O(1)$ on this branch. The bounded oscillatory correction need not have a constant limiting value.

To check that these are all the real solutions, solve for the parameter:Positive values occur only on $(-1,0)$ and $(1,\infty )$; $x=\pm 1$ are not roots. A useful exact derivative isIt vanishes only at isolated points, so $E$ is strictly decreasing on each of these intervals. Its endpoint limits are $+\infty$ and zero on each interval. This monotone parametrization of sine perturbation roots gives exactly one root in each interval for every $ϵ>0$, proving that both asymptotic branches, and no others, have been included. If negative small parameters are allowed, the transformation $(x,ϵ)\mapsto (-x,-ϵ)$ reflects these two branches and gives all real solutions for that sign as well.

For $ϵ>0$, dominant balance for algebraic roots gives two roots on the scale $\sqrt{ϵ}$ and a third on the scale ${ϵ}^{2/3}$. On the first scale put $t=\sqrt{ϵ}s$. The equation becomes $s^5-s^3+\sqrt{ϵ}=0$. The derivative of $s^5-s^3$ at either $s=1$ or $s=-1$ is two, so the first correction is $-\sqrt{ϵ}/2$. On the second scale put $t={ϵ}^{2/3}v$; then $1-v^3+{ϵ}^{1/3}{v}^5=0$. Writing $v=1+a{ϵ}^{1/3}+\cdots$ gives $a=1/3$. Thus all three real-root expansions for positive $ϵ$ areThese are fractional-power root expansions. To check that none is missing, the derivative of the polynomial is $t^2(5t^2-3ϵ)$. Its nonzero critical points are $\pm \sqrt{3ϵ/5}$; for sufficiently small positive $ϵ$ the polynomial is positive at the negative one and negative at the positive one. Its three monotone ranges therefore contain exactly these three real roots. The remaining two small roots correspond to the nonreal cube roots of unity in the second balance.

The printed limit does not specify the sign of $ϵ$. If a two-sided real limit is intended, also put $ϵ=-\eta$ with $\eta >0$. The polynomial $t^5+\eta t^3-{\eta }^3$ is strictly increasing and has just one real root. The same dominant balance for algebraic roots gives the negative-parameter branch

A Puiseux series permits fractional exponents with a fixed common denominator: ${\sum }_{n\ge n_0}{a}_n{z}^{n/q}$. These are natural local expansions of branches of algebraic functions. A fractional-power root expansion uses such powers to resolve roots that coalesce as a parameter tends to zero.