limit of function of a sequence equals the function of the limit of that sequence?

ok I'm doing my homework when i face this problem:

suppose f is continuous at x = c in [a,b]. this implies that for any sequence {xn} in [a,b] and {xn} converges to c, we have the sequence {f(xn)} converges to f(c). In other words: lim f(xn) = f(lim (xn)) with n approaches to infinity. True or False?

I think it is true so maybe I can use definition to prove it? But I'm not sure.

f continuous at c so by definition, for any e>0 there exists alpha > 0 such that |x-c| < alpha >> |f(x)-f(c)|< e

and {xn} converges to c so also by definition for every e > 0, there is a n* such that for all n >= n* >> | xn - c| < e

how can i go from here to get to:

|f(xn) - f(c)| < e ?

Thanks very much in advance!