# lim$_{n\rightarrow \infty}\,f(x_n)=f(\text{lim}_{n \rightarrow \infty} \, x_n)$ $f$ continuous and $x_n$ convergent

I know this question has already been asked (For a continuous function $f$ and a convergent sequence $x_n$, lim$_{n\rightarrow \infty}\,f(x_n)=f(\text{lim}_{n \rightarrow \infty} \, x_n)$) but I need confirmation on my attemp and I'm not sure how to go about it without starting a new question

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

• Your answer looks fine to me! – Laars Helenius Mar 9 '15 at 17:33
• @Shomar How do you know that $L$ is in the domain of $f$? Thanks – GniruT Jan 28 '16 at 14:51