## (dis)prove this function does not have a limit at zero.

1. The problem statement, all variables and given/known data
The function g:ℝ→ℝ defined by "g(x) = 0 for x being rational and g(x) = 1 for x being irrational" does not have a limit at zero.

2. Relevant equations
I have to use the definition of the limit of a function at a point--I can't use the sequential criterion this time.

So the definition of the limit of a function at a point is: (assuming function and 0 being the cluster point)
A real number L is said to be a limit of f at c if, given any ε>0, there exists a δ>0 s.t. if x is in the domain, and 0<|x-c|<δ, then |f(x) - L|<ε.

Thus the negation of the definition is:
A real number L is NOT a limit of f at c if there exist ε>0 s.t. for all δ>0, x is in the domain, 0<|x-c|<δ, AND |f(x)-L|≥ε.

3. The attempt at a solution
I am going to prove this true by using the negation of the definition.

Proof:
Assume the limit of L ≤ 0. Choose ε=1/2.
By denseness of the irrationals, for all δ>0, there exists x in the irrationals s.t. 0<|x|<δ, but |f(x)-0|=|1-L|≥1 ≥ 1/2=ε. Thus L≤0 is not a limit of g.

Assume the limit of L ≥ 1. Choose ε=1/2.
By denseness of the rationals, for all δ>0, there exists x in the rationals s.t. 0<|x|<δ, but |f(x)-L|=|0-L|≥1 ≥ 1/2=ε. Thus L≥1 is not a limit of g.

Assume the limit of 0≤ L ≤ 1/2. Choose ε=1/2.
By denseness of the irrationals, for all δ>0, there exists x in the irrationals s.t. 0<|x|<δ, but |f(x)-L|=|1-L|≥1/2=ε. Thus 0≤ L ≤ 1/2 is not a limit of g.

Assume the limit of 1/2< L ≤ 1. Choose ε=1/2.
By denseness of the rationals, for all δ>0, there exists x in the rationals s.t. 0<|x|<δ, but |f(x)-L|=|0-L|>1/2=ε. Thus 1/2< L ≤ 1 is not a limit of g.

Thus, we have proved for ever L in the reals that L cannot be a limit of this function.
Q.E.D
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 Recognitions: Homework Help Science Advisor Way too many cases, don't you think? Use the triangle inequality. If r is rational and i is irrational then |f(r)-f(i)|=|f(r)-L+L-f(i)|<=|f(r)-L|+|f(i)-L|.
 Hmm, I don't quite see where this is going though. I understand the triangle inequality, but I don't understand its usefulness here. Do I need to show the epsilon as well? I'm assuming there is a way to "compact" what I did in to this inequality, but I'm having trouble seeing that. Thank you very much :)

## (dis)prove this function does not have a limit at zero.

Oh ok! I think I see it now! We have
1=|f(r)-f(i)|=|f(r)-L+L-f(i)|<=|f(r)-L|+|f(i)-L|<=ε+ε=2ε by our definition for all ε>0
Taking ε=1/4, we are given:
1<=1/2, which is our contradiction. Thus, g does not have a limit at 0.
Q.E.D.

How is that?

Recognitions:
Homework Help
 Quote by Hodgey8806 Oh ok! I think I see it now! We have 1=|f(r)-f(i)|=|f(r)-L+L-f(i)|<=|f(r)-L|+|f(i)-L|<=ε+ε=2ε by our definition for all ε>0 Taking ε=1/4, we are given: 1<=1/2, which is our contradiction. Thus, g does not have a limit at 0. Q.E.D. How is that?
Right. So the value of L doesn't really matter. No need to break into cases.
 Oh ok, great! Thank you so much! This is genius and, for me at least, it is elegant. Haha, my teacher will be very impressed I think.

Recognitions:
Homework Help
 Quote by Hodgey8806 Oh ok, great! Thank you so much! This is genius and, for me at least, it is elegant. Haha, my teacher will be very impressed I think.
The thinking, of course, is just that 0 and 1 can't possibly both be contained in an interval whose length is less than 1. The triangle inequality is just the formal way of saying that.
 I see that now, and this application I think will be very useful for future problems. I have seen similar proofs even in our text that use the triangle inequality in such a way. But it was always just a little "in the dark" for me. However, I think this will very much help me see how to do a problem like this much more elegantly!

 Tags limit definition, limit of a sequence, proof, real analysis, sequence