 Solution

First take the usual positive-parameter limit $ϵ\downarrow 0$. For real $x$, the equation implies $|x^2-1|\le ϵ$, so every root lies near $1$ or $-1$. Near the positive root,and hence $x^2=1-ϵ$ up to an exponentially small correction. Near the negative root, the corresponding limit is minus one, giving $x^2=1+ϵ$ to all algebraic orders. Expanding the square roots givesThere is one root in each neighborhood: the derivative $2x+{\mathrm{sech}}^2(x/ϵ)$ has the sign of $x$ there for sufficiently small positive $ϵ$. The bound above excludes further real roots.

If the source's unqualified $ϵ\to 0$ is read as a two-sided limit, use the exact identityThe two-sided versions therefore replace $ϵ$ by $|ϵ|$ in both displayed first corrections. The two one-sided algebraic expansions otherwise have the same reasoning.