# Trying to prove a limit of a simple multi case function

Full disclosure: This is my attempt at solving a homework assignment

Another full disclosure: This is my first time using LaTeX, so pardon any errors

We were asked to prove / disprove the existance of the limit as $x$ approaches some $x_0$ for the following function:

$$f(x) = \begin{cases} 0 &\mbox{if } x \in \mathbb{R}\backslash\mathbb{Q}\\ sin|x| & \mbox{if } n \in \mathbb{Q} \end{cases}$$

1. $x_0 \in \{ \pi * n, n \in \mathbb{Z}\}$
2. $x_0 \in \mathbb{R} \backslash \{ \pi * n, n \in \mathbb{Z} \}$

This is what I was thinking, would love to hear if it's missing something or if it's enough as a proof. Also other ideas would be greatly appreciated

For 1, I know that $\pi$ is irrational and an irrational number times a rational number (n) is still irrational, for any irrational $x_0$ the function returns the first case which is 0, giving us the constant function 0 for every $x_0$ in that range, whose limit is 0.

For 2, Let $x_0$ be a rational number (for every irrational one I get the constant function 0 as before), since it's rational I get $sin(x_0)$ the limit of which is just the value of $sin(x_0)$

I was wondering if I'm missing something here as this seems to simple to be true judging by the other questions we got for our assignment. Any help is appreciated!

-

Let's generalize. Suppose $g:\mathbb R\to \mathbb R$ is a continuous function (like $\sin |x|$ in your example) and we define $$f(x) = \begin{cases}0,\quad & \text{if } x\in\mathbb R\setminus \mathbb Q \\ g(x) \quad & \text{if } x\in \mathbb Q \end{cases}$$ For a given point $x_0$, there are two possibilities:
1. $g(x_0)=0$. Then $f(x_0)=0$ as well. Also, since $-|g(x)|\le f(x)\le |g(x)|$ for all $x\in\mathbb R$, the squeeze theorem implies $$\lim_{x\to x_0}|f(x)| = 0$$ So, $f$ is continuous at $x_0$.
2. $g(x_0)\ne 0$. Since rational numbers are dense, there is a sequence of rational numbers $x_n$ converging to $x_0$. We have $$\lim_{n\to\infty} f(x_n) = \lim_{n\to\infty} g(x_n) = g(x_0)\tag{1}$$ Since irrational numbers are also dense, there is a sequence of irrational numbers $y_n$ converging to $x_0$. We have $$\lim_{n\to\infty} f(y_n) = 0\tag{2}$$
From (1) and (2) it follows that $f$ is not continuous at $x_0$.