= Solution
{c}
First take the usual positive-parameter limit $\epsilon\downarrow0$. For real $x$, the equation implies $|x^2-1|\leq\epsilon$, so every root lies near $1$ or $-1$. Near the positive root,
$$
\tanh(x/\epsilon)=1+O(e^{-2x/\epsilon}),
$$
and hence $x^2=1-\epsilon$ up to an exponentially small correction. Near the negative root, the corresponding limit is minus one, giving $x^2=1+\epsilon$ to all algebraic orders. Expanding the square roots gives
$$
\boxed{x_+=1-\frac{\epsilon}{2}+O(\epsilon^2),
\qquad x_-=-1-\frac{\epsilon}{2}+O(\epsilon^2).}
$$
There is one root in each neighborhood: the <derivative> $2x+\operatorname{sech}^2(x/\epsilon)$ has the sign of $x$ there for sufficiently small positive $\epsilon$. The bound above excludes further real roots.
If the source's unqualified $\epsilon\to0$ is read as a two-sided limit, use the exact identity
$$
\epsilon\tanh(x/\epsilon)
=|\epsilon|\tanh(x/|\epsilon|).
$$
The two-sided versions therefore replace $\epsilon$ by $|\epsilon|$ in both displayed first corrections. The two one-sided algebraic expansions otherwise have the same reasoning.
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