# Limits at non-accumulation points

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.

## Answers and Replies

mathwonk
Homework Helper
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.

HallsofIvy
Homework Helper
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.
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.

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?