Continuity of function $\lim_{t \rightarrow x} f(t)$ Let a function $f:A \rightarrow \mathbb{R}$, defined on a set $A\subset \mathbb{R}$ without isolated points, has a finite limit $g(x)$ in each point $x\in A$.
I wish to show that $g: A \rightarrow \mathbb{R}$, $g(x)=\lim_{t \rightarrow x} f(t)$, is continuous.  

Let $x$ in $A$ and $\varepsilon\gt0$. By definition of $g(x)$ as a limit, there exists $\eta\gt0$ such that every $z$ in $A$ such that $|z-x|\lt2\eta$ and $z\ne x$ is such that $|f(z)-g(x)|\leqslant\varepsilon$.
Let $x'$ in $A$ such that $|x'-x|\lt\eta$ and $x'\ne x$.  Then every $z$ in $A$ such that $|z-x'|\lt\min\{|x'-x|,\eta\}$ and $z\ne x'$ is such that $z\ne x$ and $|z-x|\lt2\eta$, hence $|f(z)-g(x)|\leqslant\varepsilon$. In particular, the limit $g(x')$ of $f(z)$ when $z\to x'$ with $z$ in $A$ and $z\ne x'$, is such that $|g(x')-g(x)|\leqslant\varepsilon$.
This is valid for every $x'$ in $A$ such that $|x'-x|\lt\eta$ and $x'\ne x$, hence $g$ is continuous at $x$.
