# Math Help - covergence of a function and sequences

1. ## covergence of a function and sequences

Hi can anyone help me with this question please:

If f:E->R is continous at c and (xn) is a sequence of points in E converging to c, prove that f(xn)->f(c) as n tends to infinity.

I think that c is an accumlation point and i tried to use its properties but i got nowhere

Thank you

2. Here is a very intuitive approach to this problem.
If $f$ is continuous at $c\in E$ then if $x \approx c$, x is near c, then $f(x)\approx f(c)$.

Now to say that $\left( {c_n } \right) \to c$ means that almost all the terms of $\left( {c_n } \right)$ are near $c$.
So does it not follow that almost all the terms of $\left( f({c_n }) \right)$ are near $f(c)$?

That is the outline of a proof.