Proposition: Let $(X,d), (X', d')$ be metric spaces. Let $E$ be a subset of $X$. Let $f: X \to X'$ be a function. Let $x_0 \in \bar E$ and $L \in X'$. Define a function $g: E \cup \{x_0\} \to X'$ by $g(x)=f(x)$ if $x$ is not equal to $x_0$ and $g(x)=L$ if $x=x0$. Then $\lim f(x) =L$ as $x \to x_0$ and $x \in E$ if and only if $g$ is continuous and the (*) condition holds: if $x_0 \in E$,then $f(x_0)=L$.
I am confused about proving the forward direction.
Here is my attempt: Since $\lim f(x) =L$ as $x \to x_0$ and $x \in E$, for all $ε > 0$, there exists $δ > 0 $ such that if $x \in E$, in particular $x \in E \setminus \{x_0\}$, and $d(x,x_0)<δ$,we have $d'(f(x),L)<ε$.
1.I think I have shown that $g$ is continuous on $E\setminus \{x_0\}$. But how to show $g$ is continous on $E \cup \{x_0\}$? 2. How to show the (*) condition?
Funcftion $g$ is continuous at $x_0$, if $\lim_{x\rightarrow x_0}g(x)=g(x_0)$. Now
\begin{equation}\lim_{x\rightarrow x_0}g(x)=\lim_{x\rightarrow x_0}f(x)=L=g(x_0).\end{equation} The last equality follows by definition of $g$ at $x_0$.
Since $f(x)=g(x)$ for $x\neq x_0$, your working shows that the limit point of $g(x)$ as $x$ approaches $x_0$ is $L$. The continuity follows since also by definition $g(x_0)=L$.