@N3buchadnezzar I'm trying now ...

@N3buchadnezzar but the name is not actually important

No, but I want some more information =)

@N3buchadnezzar like what?

Who discovered it, similar functions etc.

maybe do a serch for non-riemann integrable?

I guess it is a proof that a limit can still exist, even if the function is not continuous anywhere! I find that really neat, quite a cool function.

=)

no

@N3buchadnezzar yep

what limit?

@N3buchadnezzar I once read a book by Riesz and Nagy which had some excellent discussions of discontinuous/continous functions along those lines

@cassandra0 $$ \lim_{x \to 0} f(x) = 0 $$

@OldJohn Thomae's function?

it's continuous at 0

@N3buchadnezzar not sure

How can a function be continous at a single point?

Lookie look en.wikipedia.org/wiki/Nowhere_continuous_function

what is the definition of continuity?

@N3buchadnezzar easy - $\lim f(x) = a$ at only one point

Yeah, it just seems counter-intuitive.

@cassandra0 depends - probably sequential continuity in this context

f is cont at a if f(a) = lim {x->a} f(x)

@N3buchadnezzar I find continuous non-differentiable functions even more counter-intuitive :)

^^ Indeed

a limit can not exist if a function is not continuous there

@cassandra0 what??

it's a tautology from the definition

> The function f is continuous at some point c of its domain if the limit of f(x) as x approaches c through the domain of f exists and is equal to f(c).

@cassandra0 the function which is zero everywhere except for taking the value 1 when $x$ is zero is discontinuous and has a limit at 0

you are now talking about a different function

Functions, functions everywhere

it doesn't have a limit at 0

@cassandra0 Oh?

$$g(x) = \left\{ \begin{array}{cccl} 1 & \text{if} & x & \text{is irrational} \\ 0 & \text{if} & x & \text{is rational} \end{array} \right.$$

$$f(x) = \left\{ \begin{array}{cccl} x & \text{if} & x & \text{is irrational} \\ 0 & \text{if} & x & \text{is rational} \end{array} \right.$$

@cassandra0 my function has a perfectly good limit at zero - it is just not equal to the value of the function there

f is cont. at 0. g is discont. everywhere.

And my function is $f$ =)

@OldJohn, lim_{x -> 0} g(x) is not defined since there exist two different sequences (one rationals, one irrationals) that converge to different values

@cassandra0 yes - I am talking about the function I gave earlier - as a counterexample to your statement "a limit can not exist if a function is not continuous there"

which one?

@WillHunting where?

"the function which is zero everywhere except for taking the value 1 when x is zero is discontinuous and has a limit at 0" <- this is g no?

oh I see

@WillHunting Ah! But I have two

I understand, thanks

darn - gone quiet here - was it something I said?

g'night all

@OldJohn Nighty mate

That almost sounded like your age too.

Good night guys!

I spend too much time answering questions that no one reads.