Is there any function in R whose limit exists only at all rational points and does not exist at all irrational points? Is there any function in R whose limit exists only at all rational points and does not exist at all irrational points?
Please give me answer. Probably no, I think but not confirmed.
 A: No. 
Let $f:\Bbb R\to \Bbb R.$
For $x\in \Bbb R$ let $V^+(x)=\lim_{x<y\to x}\sup \{|f(z)-f(z')|:z,z'\in (x,y)\}$ and let $V^-(x)=\lim_{x>y\to x}\sup \{|f(z)-f(z')|:z,z'\in (y,x)\}...$ We allow $V^+(x)$ and $V^-(x)$ to take values in $[0,\infty].$
For $q\in \Bbb Q^+$ let $D(q)=\{x: \max (V^+(x),V^-(x)\,)\ge q\}.$
$(1).$ Lemma: $D(q)$ is closed.
Proof: Let $x$ be a limit point of $D(q).$ Then $x$ is in the closure of $D(q)\cap (x, \infty)$ or $x$ is in the closure of $D(q)\cap (-\infty,x).$ Suppose $x$ is in the closure of $(x,\infty)$, because in what follows, the other case  can be handled similarly. 
For each $y>x$ and each $r\in (0,q)$ there exists $y'\in D(q)\cap (x,y).$ Let $s=\min (y-y',y'-x).$ Since $y'\in D(q)$ and $r<q$ there exists $\{z,z'\}\subset (y'-s,y)\subset (x,y)$, or $\{z,z'\}\subset \{y,y+s)\subset (x,y),$ such that $|f(z)-f(z')|>r.$ So $V^+(x)\ge r.$
So $\forall r\in (0,q)\,(V^+(x)\ge q),$ so $V^+(x)\ge q,$ so $x\in D(q).$
$(2).$ Let $S$ be the set of $x\in \Bbb R$ such that $\lim_{x\ne y\to x}f(y)$ exists. Then $\Bbb R\setminus S=\cup \{D(q):q\in \Bbb Q^+\}$  because $x\in \Bbb R \setminus S$ $\iff$ $\exists r>0 \,(V^+(x)>0\lor V^-(x)>0)$ $\iff$ $\exists r>0\, \forall q\in \Bbb Q \cap (0,r)\, (x\in D(q)\,)$ $\iff$ $\exists q\in \Bbb Q^+\, (x\in D(q)\,).$
Now $E(q)=\Bbb R \setminus D(q)$ is open because $D(q)$ is closed. 
So $S=\cap_{q\in \Bbb Q}E(q)$ is a $G_{\delta}$ subset of $\Bbb R.$
$(3).$ $\Bbb Q$ is NOT a $G_{\delta}$ subset of $\Bbb R.$
