Finding points of Continuity of functions There are three functions where I have to show the points of continuity of. While I think it makes a lot of sense, I'm struggling with the notation. I apologize for the lengthy post.
$(1)$ $f: \mathbb{R} \rightarrow \mathbb{R}$ with $f(x) = 1 \space for\space \lvert x\rvert \lt \sqrt{3}$ and $f(x) = -1 \space for\space \lvert x\rvert \geq \sqrt{3}$
$(2)$ $f: [0,\infty) \rightarrow \mathbb{R}$ with $f(x) = x^2\lfloor\frac{1}{x}\rfloor \space$ for $x \ne 0$ and $f(0) = 0$
$(3)$ $f:\mathbb{Q} \rightarrow \mathbb{R}$ with $f(x) = 1 \space for\space \lvert x\rvert \lt \sqrt{3}$ and $f(x) = -1 \space for\space \lvert x\rvert \geq \sqrt{3}$
$(1)$ I've tried solving the following: The possible points of discontinuity could be $\sqrt{3}$ and $-\sqrt{3}$. I'm really not sure how to put this formally. If we evaluate for $x = \sqrt{3}$:$\space$for values less than $\sqrt{3}$ we get 1. For values $\geq$$\sqrt{3}$ we get $-1$.
Comparing the left and right limits we get: $\lim_{x\to \sqrt{3},\space x \lt \sqrt{3}}=1 \ne -1 =\lim_{x\to -\sqrt{3},\space x \geq \sqrt{3}}$
So there is a point of discontinuity there. The same thing can be done for $-\sqrt{3}$. Therefore, since I'm asked for the points of continuity, we get $\mathbb{R}\setminus\{\sqrt{3},-\sqrt{3}\}$
For $(2)$ I'm not really sure. I've tried plotting it and realised, that after $x=1, \space \lfloor\frac{1}{x}\rfloor$ will always be $0$, therefore $f(x) = 0$ for $x>1$. However, this obviously is not a proof and for values between $0$ and $1$, this function seems to have many points of discontinuity.
How can I try to show this or even calculate all the points of discontinuity?
For $(3)$ I have to admit I can't really make much sense of. Since the inputs are rationals the outcome will be different from $(1)$. Is there an easy way to calculate or prove this? Does $\sqrt{3}$ being irrational have something to do with it?
Thanks in advance for your patience and apologies for my lack of knowledge and the mistakes I probably made while calculating $(1)$
 A: 1)) You already showed that at points $\pm \sqrt{3}$ the function $f$ is discontinuous. It remains to show
that $f$ is continuous at $x\ne \pm \sqrt{3}$. This holds because for each such $x$ there exists $\varepsilon>0$ such that
$f$ is constant on an interval $(x-\varepsilon, x+\varepsilon)$. Remark that since $\sqrt{3}$ is irrational, such $\varepsilon$
exists for each rational $x$, so the function $f$ in (3) is continuous too.
2)) If $x\ne 0$ and $1/x\not\in\Bbb N$ then there exists $\varepsilon>0$ such that
$\lfloor 1/y\rfloor=\lfloor 1/x\rfloor=\operatorname{const}$ when $y\in (x-\varepsilon, x+\varepsilon)$. Then
$f(y)= y^2\lfloor 1/x\rfloor $ for such $y$, so $f$ is continuous at $x$.
If $1/x\in\Bbb N$ then it is easy to check that $\lim_{y\to x+} f(y)=x^2\lfloor 1/x\rfloor=f(x)$,
but $\lim_{y\to x-} f(y)= x^2(\lfloor 1/x\rfloor+1) \ne f(x)$, so $f$ is discontinuos at $x$.
Finally, if $x=0$ then $$0=\lim_{y\to x+} f(y)\le \lim_{y\to x+} y^2\frac 1{y}=\lim_{y\to x+} y=0,$$
so $f$ is continuous at $x$.
