[proofplan] We prove continuity directly from the relative $\varepsilon$-$\delta$ definition on the domain $E$. Given an arbitrary tolerance $\varepsilon > 0$, we choose $\delta$ small enough both to remain inside the radius where the local Lipschitz estimate is valid and to make the Lipschitz bound smaller than $\varepsilon$. The assumed estimate then converts the smallness of $|x-a|$ into the smallness of $|f(x)-f(a)|$. [/proofplan] [step:Choose a radius that fits inside the Lipschitz neighbourhood and the target tolerance] Let $\varepsilon > 0$ be arbitrary. Define the positive real number $\delta$ by \begin{align*} \delta := \min\left\{r,\frac{\varepsilon}{C}\right\}. \end{align*} Since $r > 0$, $\varepsilon > 0$, and $C > 0$, both $r$ and $\varepsilon/C$ are positive, so $\delta > 0$. [guided] We must prove continuity at $a$ relative to $E$: for every $\varepsilon > 0$, we need a number $\delta > 0$ such that every $x \in E$ with $|x-a| < \delta$ satisfies $|f(x)-f(a)| < \varepsilon$. Let $\varepsilon > 0$ be fixed. The hypothesis gives a useful estimate only when $|x-a| < r$, so our $\delta$ must be no larger than $r$. The same estimate gives $|f(x)-f(a)| \leq C|x-a|$, so to force the right-hand side below $\varepsilon$, it is enough to make $|x-a| < \varepsilon/C$. We satisfy both requirements by defining \begin{align*} \delta := \min\left\{r,\frac{\varepsilon}{C}\right\}. \end{align*} This number is positive because $r > 0$, $\varepsilon > 0$, and $C > 0$. [/guided] [/step] [step:Use the local Lipschitz estimate to obtain the epsilon bound] Let $x \in E$ satisfy $|x-a| < \delta$. Since $\delta \leq r$, we have $|x-a| < r$, so the assumed local Lipschitz estimate applies and gives \begin{align*} |f(x)-f(a)| \leq C|x-a|. \end{align*} Using $|x-a| < \delta$ and $C > 0$, we obtain \begin{align*} C|x-a| < C\delta. \end{align*} Since $\delta \leq \varepsilon/C$, multiplication by $C > 0$ gives \begin{align*} C\delta \leq \varepsilon. \end{align*} Combining these inequalities yields \begin{align*} |f(x)-f(a)| < \varepsilon. \end{align*} [/step] [step:Conclude continuity at the point relative to $E$] For the arbitrary $\varepsilon > 0$, we have found $\delta > 0$ such that every $x \in E$ with $|x-a| < \delta$ satisfies $|f(x)-f(a)| < \varepsilon$. This is precisely the $\varepsilon$-$\delta$ definition of continuity of $f: E \to \mathbb R$ at $a$ relative to the domain $E$. Hence $f$ is continuous at $a$. [/step]