Proving that the limit doesn't exist for $x$ other than $0$

Question

I have a problem/proof that I can't quite solve.

Let

$$f(x)=\begin{cases}x^2\quad &x \in \mathbb{R} \setminus \mathbb{Q}\\-x^2 \quad &x \in \mathbb{Q}.\\ \end{cases}$$

Prove that

$$\lim_{x\to n} f(x)$$

doesn't exist for $n$ other than $0$.

Maybe I can use the regular epsilon-delta proof? I tried that but wasn't sure of what assumptions I could make with the absolute values. I would appreciate any help.

Answer 1

Using well known definition of limit In terms of sequences by Eduard Heine, let's take $q \in \mathbb{Q}$. Then $x_n=\frac{1}{n}+q \in \mathbb{Q} $ and $x_n \to q, n \to \infty.$ So $f(x_n)=-x_n^2 \to -q^2$.

From another hand we know, that between each pair of rational numbers exist irrationals, so we can choose sequence $q< y_n<x_n $ and $y_n \in \mathbb{R} \setminus \mathbb{Q}$. As $y_n \to q$ and $f(y_n)=y_n^2 \to q^2$, then we obtain that for limit does not exist if not holds $q^2=-q^2$. But this is only for $q=0.$