# Thread: limit of function of a sequence equals the function of the limit of that sequence?

1. ## 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 ?

2. Just let $\alpha = \varepsilon~.$

3. Originally Posted by Plato
Just let $\alpha = \varepsilon~.$
so it is just like this?

alpha = e

so |xn - c| < alpha so for all e > 0 there exists n* in N such that for all n >= n*
|f(xn) - f(c) | < e

>> f(xn) converges to f(c)
>> lim (f(xn) = f(c) = f(lim (xn)) as n approaches to infinity.