= Solution
{c}
Take $\epsilon\to0^+$. There are two real branches, on different scales. For a root tending to zero, the <Taylor series> gives
$$
-\frac{x^3}{6}+O(x^5)=\epsilon(1-x^2).
$$
The <dominant balance> is $x^3\sim-6\epsilon$, so a <fractional-power root expansion> yields
$$
\boxed{x_-\sim-(6\epsilon)^{1/3}}.
$$
More precisely $x_-=-(6\epsilon)^{1/3}+O(\epsilon)$. An ordinary <power series> in integer powers of $\epsilon$ would miss this scale.
There is also a root escaping to positive infinity. Boundedness of the <sine> term makes the leading <dominant balance> $x-\epsilon x^2=0$. The nonzero branch is
$$
\boxed{x_+\sim\epsilon^{-1}}.
$$
Indeed the equation rearranges to $\epsilon x_+^2-x_+=\epsilon-\sin x_+$, which yields $x_+=\epsilon^{-1}+O(1)$ 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:
$$
\epsilon=E(x):=\frac{x-\sin x}{x^2-1}.
$$
Positive values occur only on $(-1,0)$ and $(1,\infty)$; $x=\pm1$ are not roots. A useful exact <derivative> is
$$
E'(x)=-\frac{2\{x\cos(x/2)-\sin(x/2)\}^2}{(x^2-1)^2}\leq0.
$$
It 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 $\epsilon>0$, proving that both asymptotic branches, and no others, have been included. If negative small parameters are allowed, the transformation $(x,\epsilon)\mapsto(-x,-\epsilon)$ reflects these two branches and gives all real solutions for that sign as well.
Back to article page
