[proofplan] We compare the two relevant $\varepsilon$-$\delta$ conditions. Continuity at $a$ controls $|f(x)-f(a)|$ for every $x \in E$ sufficiently close to $a$, including $x=a$. The punctured limit controls the same quantity only for $x \in E$ with $x \ne a$. Thus continuity immediately implies the limit, while the converse follows by treating the case $x=a$ separately, where the difference is exactly zero. [/proofplan] [step:Unpack continuity and the punctured limit through $E$] The assertion that $f$ is continuous at $a$ means that for every $\varepsilon > 0$ there exists $\delta > 0$ such that, for every $x \in E$, \begin{align*} |x-a| < \delta \implies |f(x)-f(a)| < \varepsilon. \end{align*} The assertion \begin{align*} \lim_{\substack{x \to a, x \in E \setminus \{a\}}} f(x) = f(a) \end{align*} means that for every $\varepsilon > 0$ there exists $\delta > 0$ such that, for every $x \in E$, \begin{align*} 0 < |x-a| < \delta \implies |f(x)-f(a)| < \varepsilon. \end{align*} The hypothesis that $a$ is not isolated in $E$ ensures that this punctured approach to $a$ through $E$ is a genuine local limiting condition rather than a condition over an eventually empty punctured neighbourhood. [/step] [step:Restrict the continuity estimate to punctured points] Assume that $f$ is continuous at $a$. Let $\varepsilon > 0$ be given. By continuity at $a$, choose $\delta > 0$ such that, for every $x \in E$, \begin{align*} |x-a| < \delta \implies |f(x)-f(a)| < \varepsilon. \end{align*} If $x \in E$ satisfies $0 < |x-a| < \delta$, then in particular $|x-a| < \delta$, so the same implication gives \begin{align*} |f(x)-f(a)| < \varepsilon. \end{align*} Therefore \begin{align*} \lim_{\substack{x \to a, x \in E \setminus \{a\}}} f(x) = f(a). \end{align*} [guided] Assume that $f$ is continuous at $a$. To prove the punctured limit statement, we must verify its $\varepsilon$-$\delta$ definition. Let $\varepsilon > 0$ be arbitrary. Continuity at $a$ gives a number $\delta > 0$ such that every point $x \in E$ with $|x-a| < \delta$ satisfies \begin{align*} |f(x)-f(a)| < \varepsilon. \end{align*} The punctured limit asks for the same inequality only under the stronger hypothesis $0 < |x-a| < \delta$. Since $0 < |x-a| < \delta$ implies $|x-a| < \delta$, the continuity estimate applies directly. Hence, for every $x \in E$ satisfying $0 < |x-a| < \delta$, we have \begin{align*} |f(x)-f(a)| < \varepsilon. \end{align*} This is exactly the definition of \begin{align*} \lim_{\substack{x \to a, x \in E \setminus \{a\}}} f(x) = f(a). \end{align*} [/guided] [/step] [step:Recover continuity by handling the point $x=a$ separately] Assume that \begin{align*} \lim_{\substack{x \to a, x \in E \setminus \{a\}}} f(x) = f(a). \end{align*} Let $\varepsilon > 0$ be given. By the punctured limit condition, choose $\delta > 0$ such that, for every $x \in E$, \begin{align*} 0 < |x-a| < \delta \implies |f(x)-f(a)| < \varepsilon. \end{align*} Now let $x \in E$ satisfy $|x-a| < \delta$. If $x \ne a$, then $0 < |x-a| < \delta$, so the punctured limit estimate gives \begin{align*} |f(x)-f(a)| < \varepsilon. \end{align*} If $x=a$, then \begin{align*} |f(x)-f(a)| = |f(a)-f(a)| = 0 < \varepsilon. \end{align*} Thus, for every $x \in E$, \begin{align*} |x-a| < \delta \implies |f(x)-f(a)| < \varepsilon. \end{align*} Since $\varepsilon > 0$ was arbitrary, $f$ is continuous at $a$. [/step] [step:Conclude the equivalence] The first implication proves that continuity of $f$ at $a$ implies agreement with the punctured limit through $E$. The second implication proves that agreement with the punctured limit through $E$ implies continuity of $f$ at $a$. Therefore $f$ is continuous at $a$ if and only if \begin{align*} \lim_{\substack{x \to a, x \in E \setminus \{a\}}} f(x) = f(a). \end{align*} [/step]