Show that $f(x)=\begin{cases}x&x\in \mathbb{Q}\\ 0&x\in \mathbb{R\setminus Q}\end{cases}$ doesn´t approach some $l$ as $x\rightarrow a\neq 0$

Question

Let $$f(x)=\begin{cases}x&x\in \mathbb{Q}\\ 0&x\in \mathbb{R\setminus Q}\end{cases}$$ and $0<\epsilon$

(1) $\;\;f(x)$ approaches some $l$ (namely $0$) as $x\rightarrow a=0$

for

$$|f(x)-0|<\epsilon\iff |0-0|<\epsilon\land |x-0|<\epsilon$$ We know that $|0-0|<\epsilon$ is true and we allso know that $\exists x\mid |x-0|<\epsilon$

Therfore (1) is true.

How can I prove, in a similar fashion, that $x\rightarrow a\neq 0 \Rightarrow\nexists l, f(x)\rightarrow l$?

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I´m new to limits in this fashion so I would also appreciate if you told me if (1) was successfully proven. Thanks in advance.

Answer 1

Hint: near $a$ there are both rational and irrational numbers.

Answer 2

Let $a\neq 0$ and suppose that $\lim\limits_{x\rightarrow a}f(x)=\ell$ exists, thus for all sequence $(x_n)$ such that $\lim\limits_{n\rightarrow +\infty}x_n=a$, we have $\lim\limits_{n\rightarrow +\infty}f(x_n)=\ell$. Let $u_n=\frac{\lfloor na\rfloor}{n}$ and $v_n=\frac{\lfloor na\sqrt{2}\rfloor}{n\sqrt 2}$, we have that $\lim\limits_{n\rightarrow +\infty}u_n=\lim\limits_{n\rightarrow +\infty}v_n=a$ but for all $n\in\mathbb{N}^*$, we have $f(u_n)=u_n$ because $u_n\in\mathbb{Q}$ and $f(v_n)=0$ because $v_n\notin\mathbb{Q}$ so $\lim\limits_{n\rightarrow +\infty}f(u_n)\neq\lim\limits_{n\rightarrow +\infty}f(v_n)$ because $a\neq 0$. Finally $\lim\limits_{x\rightarrow a}f(x)$ doesn't exist.