We wish to find \(\delta\) satisfying
\begin{equation*}
0< \abs{x-1}< \delta \implies \abs{f(x)-(-2)}< \frac{1}{2}\text{.}
\end{equation*}
Proceeding as in
ProcedureΒ 4.4, we first solve the inequality
\(\abs{f(x)-(-2)}< \frac{1}{2}\text{:}\)
\begin{align*}
\abs{f(x)-(-2)}< \frac{1}{2} \amp \iff \abs{(1-3x^2)+2}< \frac{1}{2} \\
\amp \iff \abs{3-3x^2}< \frac{1}{2}\\
\amp \iff \abs{(-3)(x^2-1)}< \frac{1}{2}\\
\amp \iff 3\abs{x^2-1}< \frac{1}{2}\\
\amp \iff \abs{x^2-1}< \frac{1}{6}\\
\amp\iff -\frac{1}{6}< x^2-1< \frac{1}{6}\\
\amp \iff\frac{5}{6}< x^2< \frac{7}{6}\\
\amp \iff\sqrt{\frac{5}{6}}< \abs{x}< \sqrt{\frac{7}{6}} \amp \text{(rad. ineq. prop.)}\\
\amp \iff -\sqrt{\frac{7}{6}}< x < -\sqrt{\frac{5}{6}} \text{ or } \sqrt{\frac{5}{6}}< x < \sqrt{\frac{7}{6}}\text{.}
\end{align*}
Notice that in this case, the set of solutions \(S\) to the inequality is a union of two intervals:
\begin{equation*}
S=\left(-\sqrt{\frac{7}{6}},-\sqrt{\frac{5}{6}}\right)\cup \left(\sqrt{\frac{5}{6}},\sqrt{\frac{7}{6}}\right)\text{.}
\end{equation*}
Notice further that our limit point \(1\) is an element of the second interval, and that we have
\begin{equation*}
\left(\sqrt{\frac{5}{6}},\sqrt{\frac{7}{6}}\right)=(1-\delta_1,1+\delta_2) \text{,}
\end{equation*}
where
\begin{align*}
\delta_1 \amp = 1-\sqrt{\frac{5}{6}} \\
\delta_2 \amp = \sqrt{\frac{7}{6}}-1\text{.}
\end{align*}
Thus, setting \(\delta=\min\{\delta_1,\delta_2\}\text{,}\) we have
\begin{equation*}
(1-\delta, 1+\delta)\subseteq \left(\sqrt{\frac{5}{6}},\sqrt{\frac{7}{6}}\right)\subseteq S\text{.}
\end{equation*}
It follows that
\begin{equation*}
0< \abs{x-1}< \delta \implies \abs{f(x)-(-2)}< \frac{1}{2}\text{,}
\end{equation*}
as desired.