Limits at non-accumulation points

[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.

[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.

[HallsofIvy]

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.

[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?