= Solution
The positive <polynomial root> is $x_*=\varepsilon+\sqrt{1+\varepsilon^2}$. Its <Taylor expansion> near zero is
$$
x_*=1+\varepsilon+\frac12\varepsilon^2-\frac18\varepsilon^4+O(\varepsilon^6).
$$
The starting error is therefore $x_0-x_*=-\varepsilon-\tfrac12\varepsilon^2+O(\varepsilon^4)$. Consequently
$$
\boxed{|x_0-x_*|\sim|\varepsilon|,\qquad n_0=1.}
$$
For signed $\varepsilon$, the magnitude scaling should be written with $|\varepsilon|$; this is the precise interpretation of the proportionality in the question.
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