Show that: lim$_{n\rightarrow \infty}\,f(x_n)=f(\text{lim}_{n \rightarrow \infty} \, x_n)$ Let $f:X \rightarrow Y$ be a function. Prove that if $f$ is continuous, then for every convergent sequence $(x_n)$ lim$_{n\rightarrow \infty}\,f(x_n)=f(\text{lim}_{n \rightarrow \infty} \, x_n)$
My attempt: Assume $f$ is continuous. Then $\forall \epsilon >0, \exists \delta > 0$ such that $\forall p \in X$ and $q \in X: d(p,q) < \delta \implies d(f(q),f(q)) < \epsilon.$
And also assume that $(x_n)$ converges. Then $\exists L$ such that $\forall \epsilon >0, \exists N$ such that $n>N \implies d(x_n, L) < \epsilon.$
I need to use these definitions to show lim$_{n\rightarrow \infty}\,f(x_n)=f(\text{lim}_{n \rightarrow \infty} \, x_n)$
. Any hints?
Edit: lim$_{n \rightarrow \infty}(x_n) \in X$
 A: If you define $L:=\lim_{n \rightarrow \infty}x_n$, and you know that $L$ is in the domain of your function, then you have $|f(x_n) - f(L)| < \epsilon$ whenever $|x_n - L|$ is less than some appropriate $\delta$. But you know by hypothesis that you can find an $N$ where this is true for every $n>N$. 
That's a pretty big hint, but I'll let you polish it into a proof.
A: I'm replying with a possible answer I need confirmation on:
Let $L:=\lim_{n \rightarrow \infty}x_n$, and let $L$ be in the domain of $f$.
We want to prove the sequence ${f(x_n)}$ converges to $f(L)$, namely $\forall\epsilon >0 \ \exists  N$ such that $|f(x_n)-f(L)|<\epsilon$ for $n>N$.
$f$ is continuous in $x=L$ which means that $\forall\epsilon >0 \ \exists\delta$ such that $\forall x_n\in Dom(f), 0<|x_n-L|<\delta \to |f(x_n)-f(L)|<\epsilon $ 
We know $\lim_{n \rightarrow \infty}x_n=L$ which means that for any given $\delta$ I can find an $N$ for which $|x_m-L|<\delta$ for $n>N$
This means that given any $\epsilon$ I can find $N$ such that $|f(x_n)-f(L)|<\epsilon$, as I wanted to prove. 
In short, by giving a value for $\epsilon$, the continuity gives a value for $\delta$, and given a value for $\delta$ the fact that $\lim_{n \rightarrow \infty}x_n=L$ gives a value for N
