# Limits at non-accumulation points

1. Jan 16, 2008

### breez

Why are limits of functions not defined at non-accumulation points?

For example, take the function f(x) = k, for x in Z

Then based on the epsilon delta definition of a limit, for any epsilon > 0, we can always find a delta, for which 0 < |x-x_0| < delta implies |f(x)-k| = 0 < epsilon. Thus, the limit of every non-accumulation point of f(x) has limit = k.

This example seems to contradict the fact that limits are undefined at non-accumulation points.

2. Jan 16, 2008

### mathwonk

i think they mean that if the function were defined on a given set, then we could extend it sometimes to non accumulation points using limits, but there is no unique way to do this at non accum. points.

3. Jan 16, 2008

### HallsofIvy

Staff Emeritus
Why should f(x)= k in your example?

If x0 is not an accumulation point of (the domain of) f, then there exist some delta such that any x such that 0< |x- x0|< delta is not in the domain of f. Then 0< |x- x0|< delta does not imply |f(x)- k|= 0, because f(x) is not defined and so |f(x)- f(x0| has no value.

4. Jan 17, 2008

### breez

Interesting, I asked my teacher this question and he told me to prove:

If c is not an accumulation point of the domain of
f, then for every number L we have

lim f(x) = L.
x-->c

How do you go about proving this?