1. **State the problem:** Find an open interval about $c=4$ such that for all $x$ in this interval, the inequality $|f(x) - L| < \varepsilon$ holds, where $f(x) = x + 1$, $L = 5$, and $\varepsilon = 0.01$. 2. **Formula and explanation:** We want to find $\delta > 0$ such that if $0 < |x - c| < \delta$, then $|f(x) - L| < \varepsilon$. 3. **Apply the function:** Since $f(x) = x + 1$, we have $$|f(x) - L| = |(x + 1) - 5| = |x - 4|.$$ 4. **Set the inequality:** We want $$|x - 4| < 0.01.$$ 5. **Interpretation:** This means $x$ must lie within $0.01$ units of $4$, so the open interval is $$(4 - 0.01, 4 + 0.01) = (3.99, 4.01).$$ 6. **Choose $\delta$:** We can take $\delta = 0.01$ so that whenever $0 < |x - 4| < 0.01$, it follows that $|f(x) - 5| < 0.01$. **Final answer:** $$\boxed{\delta = 0.01}$$ This means the open interval about $c=4$ is $(3.99, 4.01)$ where the inequality holds.