1. The problem is to find the value of $\delta$ such that if $0 < |x - a| < \delta$ then $|f(x) - L| < \varepsilon$ for the function $f(x) = 8 - x^2$ with $a = 2$, $L = 4$, and $\varepsilon = 0.5$. 2. The definition of limit states: For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$. 3. Start by expressing $|f(x) - L|$: $$|f(x) - L| = |8 - x^2 - 4| = |4 - x^2| = |(2 - x)(2 + x)|$$ 4. We want to find $\delta$ such that: $$|(2 - x)(2 + x)| < 0.5$$ 5. Since $x$ is close to $a = 2$, we can bound $|2 + x|$ by choosing $\delta \leq 1$ so that $|x - 2| < 1$ implies $1 < x < 3$ and thus: $$|2 + x| < |2 + 3| = 5$$ 6. Using this bound: $$|2 - x| \cdot |2 + x| < |2 - x| \cdot 5 < 0.5$$ 7. Therefore: $$|2 - x| < \frac{0.5}{5} = 0.1$$ 8. Since $|2 - x| = |x - 2|$, we choose: $$\delta = \min(1, 0.1) = 0.1$$ 9. Conclusion: If $0 < |x - 2| < 0.1$, then $|f(x) - 4| < 0.5$. This satisfies the limit definition for the given $\varepsilon$. Final answer: $\boxed{\delta = 0.1}$