A function is defined as follows: $$ f(x) = \begin{cases} x^2 & x \in \mathbb{R} - \mathbb{Q} \\ -x^2 & x \in \mathbb{Q} \end{cases} $$
Show that $\lim_{x \to a} f(x)$ does not exist for $a \ne 0$.
I'm really stuck on this problem. Is there a way to show this using epsilon-delta definitions or work with series? Thanks in advance!
Let $a\neq 0$. It suffices to choose two sequences $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ such that $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=a,$$ but $$ \lim_{n\to\infty}f(a_n)\neq\lim_{n\to\infty}f(b_n). $$ In your particular case, if $a\in\mathbb{Q}$ you can take $a_n=a+\frac{1}{n}\in\mathbb{Q}$ and $b_n=a+\frac{\pi}{n}\in\mathbb{R}\setminus\mathbb{Q}$. If $a\in\mathbb{R}\setminus\mathbb{Q}$ you can always suppose that there exists a sequence $A_k\in\mathbb{Q}$ such that $A_k\to a$ (since $\mathbb{Q}$ is dense in $\mathbb{R}$), and choose for instance $a_n=A_n+\frac{1}{n}$ and $b_n=a+\frac{1}{n}$. In both cases $$ \lim_{n\to\infty}f(a_n)=-\lim_{n\to\infty}f(b_n). $$