Prove that $f(x)\to L$ as $x\to x_{0}\in E$ iff $f(x)\to L$ as $x\to x_{0}\in E\cap(x_{0}-\delta,x_{0}+\delta)$ for every $\delta > 0$.

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Let $X$ be a subset of $\textbf{R}$, let $E$ be a subset of $X$, let $x_{0}$ be an adherent point of $E$, let $f:X\rightarrow\textbf{R}$ be a function, and let $L$ be a real number. Let $\delta > 0$. Then we have that \begin{align*} \lim_{x\rightarrow x_{0};x\in E}f(x) = L \end{align*} if and only if \begin{align*} \lim_{x\rightarrow x_{0};x\in E\cap(x_{0}-\delta,x_{0}+\delta)}f(x) = L \end{align*}

My solution

Let us prove the implication $(1)\Rightarrow(2)$ first.

According to the definition of limit, for every $\varepsilon > 0$, there is a $\eta > 0$ such that for every $x\in E$ we have that \begin{align*} |x - x_{0}| \leq \eta \Rightarrow |f(x) - L| \leq \varepsilon \end{align*} Thus, if we take $\zeta = \min\{\delta,\eta\}$, the result follows.

Conversely, how do we prove the implication $(2)\Rightarrow(1)$?

Any contribution is appreciated.