\begin{align} f(x) = \begin{cases} x^2, & \text{if $n$ is rational} \\ -x^2, & \text{if $n$ is irrational} \end{cases} \end{align}
\begin{equation} a) \ \text{There is no $a$ where} \lim_{x \to a} f(x) \text{ exists} \\ b) \ \lim_{x \to a} f(x) \text{ exists only when $a = 0$} \\ c) \ \lim_{x \to a} f(x) \text{ exists for infinitely many $a$} \\ d) \ \text{Impossible to answer without more information} \end{equation}
I originally thought that the answer would be (a) here because there would be no continuity in this function whatsoever as x constantly switches between rational and irrational values, but I am unsure of how I could mathematically explain it. (I have not been taught the epsilon-delta definition, but I do know the squeeze theorem if that could somehow apply here)
Since $-x^2\le f(x) \le x^2$, and $\underset{x\rightarrow 0}{\lim}x^2 = \underset{x\rightarrow 0}{\lim}-x^2 = 0$, by the squeeze theorem, you have $\underset{x\rightarrow 0}{\lim}f(x) = 0$. For any other $a$, you can always find a sequence of rational numbers $a_n$ and a sequence of irrational numbers $b_n$, such that $\underset{n\rightarrow \infty}{\lim}a_n = \underset{n\rightarrow \infty}{\lim}b_n = a$, but then $\underset{n\rightarrow \infty}{\lim}f(a_n) = \underset{n\rightarrow \infty}{\lim} a_n^2 = a$, $\underset{n\rightarrow \infty}{\lim}f(b_n) = \underset{n\rightarrow \infty}{\lim} -b_n^2 = -a$. They are not equal since $a\not=0$, so the limit $\underset{x\rightarrow a}{\lim}f(x)$ does not exist. Thus the answer is b).