F(x) = x if x is rational, 0 if x is irrational.

[Mjjellen]

Homework Statement

F(x) = x if x is rational, 0 if x is irrational.
Use the δ, ε definition of the limit to prove that lim(x→0)f(x)=0.
Use the δ, ε definition of the limit to prove that lim(x→a)f(x) does not exist for any a≠0.

Homework Equations

lim(x→a)f(x)=L
0<|x-a|<δ, |f(x)-L|<ε

The Attempt at a Solution

I was mostly having troubles writing my initial equation, I was stumped very early on by filling in the values for the epsilon part of the equation, if that's how one is supposed to go by this problem. If not, any other advice that can be given to me?

 

[SammyS]

For the limit as x → 0 :

For rational numbers, f(x) = x.
What does δ need to be so that if 0 < |x - 0| < δ , then |f(x) - 0| < ε ?

For irrational numbers, any δ will work. Why?
So, use the same δ you pick for the rationals.
​

Do you know how to show that lim_{x → a}g(x) doesn't exist, in general? Of course using δ, ε .

 

[punto2003]

I have the exact same exercise and i cannot prove the second part
"Use the δ, ε definition of the limit to prove that lim(x→a)f(x) does not exist for any a≠0."

I was wondering if anyone could help me with that

Thank you in advance

 

[D H]

Assuming a limit did exist, would that lead to any contradictions?

 

[punto2003]

Yes, i think that i must assume that the limit exists,lets say that it is L.
The problem is that i don't know what i have to do next.
Probably i will have to work with to different cases,one if x is rational and
one if x is irrational.Another problem is that I don't know what ε Ι have to use
in order to get to the contradiction.

 

[D H]

Show that no matter what value L you pick as $\lim_{x \to c} f(x)$ means that you can find some $\epsilon>0$ but you cannot find a $\delta>0$ that satisfies the definition of the limit.