 10:00 PM
@N3buchadnezzar but the name is not actually important   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.
2
=) 10:04 PM
no  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 10:07 PM How can a function be continous at a single point? 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 10:09 PM
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). 10:18 PM
@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? $$g(x) = \left\{ \begin{array}{cccl} 1 & \text{if} & x & \text{is irrational} \\ 0 & \text{if} & x & \text{is rational} \end{array} \right.$$
$$f(x) = \left\{ \begin{array}{cccl} x & \text{if} & x & \text{is irrational} \\ 0 & \text{if} & x & \text{is rational} \end{array} \right.$$ 10:20 PM
@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

user19161
I finally got a star after many days with no stars. @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" 10:22 PM
which one? @WillHunting where?

user19161 "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 10:42 PM
darn - gone quiet here - was it something I said?
g'night all @OldJohn Nighty mate
That almost sounded like your age too.
2 11:13 PM
Good night guys! 11:54 PM