Convergent sequences and functions on them Let $X$ and $Y$ be two Hausdorff spaces, and $f,g : X \to Y$ be two continuous functions. Let $S$ be a convergent sequence, say $S=(x_1,x_2,x_3,\ldots)$ with $x_k \in X$ for all $k \ge 1$, with limit $\ell \in X$.
Assume that $f(x_k) = g(x_k)$ for all $k \ge 1$. Clearly, since $f$ and $g$ are continuous, we know that $f(x_k) \to f(\ell)$ and $g(x_k) \to g(\ell)$ as  $k \to \infty$. I have two questions:


*

*Does it follow that $f(\ell) = g(\ell)$? (My gut feeling is "yes", but how does one prove this?)

*What can be said about $f|_U$ and $g|_U$, where $U$ is an open neighbourhood with $\ell \in U \subseteq X$?


EDIT I've added the condition that both $X$ and $Y$ be Hausdorff. 
 A: Answer to updated version:
Yes, and the assumption that $X$ is Hausdorff isn't needed.  As you mentioned, the sequence $(f(x_k))=(g(x_k))$ in $Y$ converges to $f(\ell)$ and to $g(\ell)$.  Because $Y$ is Hausdorff, limits of sequences are unique, so this implies that $f(\ell)=g(\ell)$.
I do not know what is being asked about $f|_U$ and $g|_U$, but for example, they need not be equal on any such neighborhood.  Perhaps the example $X=Y=\mathbb R$, $f=0$, $g(x)=x\sin(1/x)$, $g(0)=0$, $x_n=\dfrac{1}{\pi n}$ would be relevant?  (Or $f=0$, $g(x)=\max\{-x,0\}$, $x_n=1/n$.)  
For holomorphic functions on connected open subsets of $\mathbb C$, or real analytic functions on intervals in $\mathbb R$, we would have $f=g$ if $x_n\neq \ell$ for all $n$.
Added: To see that limits in Hausdorff spaces are unique, suppose that $(a_n)$ is a sequence in a Hausdorff space $Y$ converging to $a\in Y$ and let $b\neq a$ be another element of $Y$.  It will be shown that $(a_n)$ does not converge to $b$.
Let $U$ and $V$ be open neighborhoods of $a$ and $b$ respectively such that $U\cap V=\varnothing$.  Because $(a_n)\to a$, eventually $a_n\in U$.  This implies that eventually $a_n\not\in V$, and therefore $(a_n)\not\to b$.

Answer to original version:
If $Y$ is Hausdorff, then limits of sequences in $Y$ are unique when they exist, so yes.
Without this assumption there are counterexamples.  For example, if $Y=\{0,1\}$ with indiscrete topology, $X=Y$, $S=(0,0,0,\ldots)$, $\ell=1$, $f(0)=f(1)=0$, $g(0)=0$, $g(1)=1$.  ($X$ could be any space that has a sequence $(x_n)$ converging to a limit $\ell$ such that $x_n\neq \ell$ for all $n$, so e.g. $X=\mathbb R$ would also work, but I took $X=Y$ for economy.)
