Prove: if $f$ and $g$ are continuous on $(a,b)$ and $f(x)=g(x)$ for every $x$ in a dense subset of $(a,b)$, then $f(x)=g(x)$ for all $x$ in $(a,b)$. 
Quoting:" Prove: if $f$ and $g$ are continuous on $(a,b)$ and $f(x)=g(x)$ for every $x$ in a dense subset of $(a,b)$, then $f(x)=g(x)$ for all $x$ in $(a,b)$."

Let $S \subset (a,b)$ be a dense subset such that every point $x \in (a,b)$ either belongs to S or is a limit point of S.
There exists $x_n \in S$ such that $\lim\limits_{n \rightarrow \infty} x_n =x$.
As $f$ and $g$ are continuous on (a,b), 
$$f(x)= \lim\limits_{n \rightarrow \infty} f(x_n)= \lim\limits_{n \rightarrow \infty} g(x_n)= g(x)$$


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*How can I conclude that "$f(x)=g(x)$ for all $x$ in $(a,b)$"?

*Is there a proof with $\epsilon$, $\delta$ definition of limits?


Much appreciated
 A: Let $S$ be the dense subset of $(a,b)$. Like you wrote, for any $x \in (a,b)$, there exists $x_n \in S$ such that $x_n \to x$. Then $$f(x)=f(\lim_n x_n)=\lim_nf(x_n)=\lim_ng(x_n)=g(x)$$
A: Assuming these are Real-valued functions of a Real variable. Consider the $0$ function on a dense subset. Since it is uniformly -continuous, we can extended it into the complement using sequential continuity  $( x_n \rightarrow x ) \rightarrow (f(x_n) \rightarrow f(x))$, which implies $f(x)=0$. Assume $f(x)= a>0 $ (Wolg) for some $x$ in the complement of the dense subset  . Then for any 'hood $(a- \delta, a+ \delta  )$ of a we have a $y$ with $|f(x)-f(y)|=|a-f(y)|\geq |a-0|> a/2 $. So using $\epsilon < a/2 $ you can show that the only continuous extension is the zero function, i.e., $f(x)=g(x)$ in the complement of the dense subset.
A: Let $\epsilon >0$ and $x \in (a,b) \cap S^c$.
$\forall t>0, \exists z \in S$ such that $|x-z|<t$  
We have that $f,g$ are continuous at $x$ thus $\exists \delta_1, \delta_2>0$ such that $$|f(y)-f(x)|< \epsilon /2, \forall y: |y-x|< \delta_1 $$ $$|g(y)-g(x)|< \epsilon /2, \forall y:|y-x|< \delta_2$$
Take $t= \min \{\delta_1,\delta_2\}$,thus $\exists y_0 \in S$ such that $|y_0-x|< t$.
We have that $|f(x)-g(x)| \leqslant |f(x)-f(y_0)|+|f(y_0)-g(y_0)|+|g(y_0)-g(x)|< \epsilon /2+0 +\epsilon /2 =\epsilon$
Thus we proved that $|f(x)-g(x)|< \epsilon ,\forall \epsilon>0$,
thus $f(x)=g(x)$
