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...
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.
Assume $\displaystyle \lim_{x\rightarrow}{f(x)} = m$ and $\displaystyle c\neq 0$
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 : $\displaystyle m\neq 0$.
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 :$\displaystyle \lim_{x\rightarrow 0}{f(x)} =m$
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 $\displaystyle \lim_{x\rightarrow 0}{f(x)} = 0$ and hence the limit exists ,
while in the case where ,x is irrational :
|f(x)|=0< ε and the limit again exists.
HENCE $\displaystyle \lim_{x\rightarrow 0}{f(x)} =0$ and the limit exists
Thus we do not have to consider the case where $\displaystyle m\neq 0$
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