1. Limit Proof

Define f: R R by f(x) = x if x is rational and f(x) = 0 if x is irrational. Prove that f has a limit at c iff c = 0.
This has me totally stumped...

2. well if c is 0 it is pretty easy application of definitions because both of the piecewise functions converge to 0 at 0, so you can just choose the smaller of the two deltas to get it within any given epsilon.

For the other direction, I suggest a proof by contradiction. That is is say c is not 0. Anywhere else on the real line it is pretty clear that if you took epsilon to be say |c|/3 there is no way you can choose a delta small enough to ensure this function is continuous there. I think |c|/2 works actually, but better safe than sorry in my opinion.

Oh I should mention it is important that both the rationals and irrationals are dense, so any delta ball will contain points from both pieces.

3. Originally Posted by bearej50
Define f: R R by f(x) = x if x is rational and f(x) = 0 if x is irrational. Prove that f has a limit at c iff c = 0.

This has me totally stumped...
Assume $\lim_{x\rightarrow}{f(x)} = m$ and $c\neq 0$

Then we have two cases :

case I : m=0

In this case we have for all ε>0 ,there exists a δ>0 and such that :

for all x :if xεD(f) and 0<|x-c|<δ ,then |f(x)|<ε.......................................... ..........1

BUT in the interval (c-δ,c+δ) -{c} x can take both rational and irrational values and so we will have :

|f(x)| = |x| and |f(x)|=0 and if we choose ε<|x| we will have a contradiction

case II : $m\neq 0$.

In this case we have for all ε>0 ,there exists a δ>0 and such that :

for all x : if xεD(f) and 0<|x-c|<δ ,then |f(x)-m|<ε.............................................. ................................2

Αnd in this case again in the interval (c-δ,c+δ)-{c} x can take both rational and irrational values.

Hence |f(x)-m|=|x-m| and |f(x)-m|= |m| and if we put ε =|m| we will end with a contradiction

Conversely:

Let c=0 we must show : $\lim_{x\rightarrow 0}{f(x)} =m$

AGAIN we have two cases :

case I : m=0

In this case given ε>0 we must find a δ>0 and such that :

for all x:if 0<|x|<δ and xεD(f) ,then |f(x)|<ε.......................................... ...........................3

Again when x, rational |f(x)| =|x| and if we choose δ<ε ,then $\lim_{x\rightarrow 0}{f(x)} = 0$ and hence the limit exists ,

while in the case where ,x is irrational :

|f(x)|=0< ε and the limit again exists.

HENCE $\lim_{x\rightarrow 0}{f(x)} =0$ and the limit exists

Thus we do not have to consider the case where $m\neq 0$

Note the above problem is based on the following theorem:

LET z be a real No. Given ε>0 .there exists a rational No, a such that |a-c|<ε.

Τhe same theorem holds when ,a is irrational

4. ^ Thank you for being so helpful. I would +rep you but they wont let me

5. Originally Posted by bearej50
^ Thank you for being so helpful. I would +rep you but they wont let me
just clicking the thanks button is fine. chances are your reputation isn't high enough to give a lot of point through +rep anyway.