Here's the question:
Let $f(x)=\sin \left( \frac{1}{x} \right)$ for $x\ne 0$ and let $f(0)=0$. Show that $f$ is discontinuous on $\mathbb{R}$ and still has the intermediate value property on $\mathbb{R}$.
It's easy to show that $f$ is discontinuous at $x=0$. Now, I go on to prove that $f$ has the intermediate value property. Let $a,b \in \mathbb{R}$ with $a<b$. Assume that $y$ be some real such that $f(a) < y < f(b)$. If $y=0$ we are done for $f(0)=y$. If $y\ne 0$, I was hoping to use the continuity of sine on $\mathbb{R}$ but nothing seems work out.
Hints would be appreciated. I do not need full solution.
If $0<a<b$ or $a<b<0$, then you just use the fact that $f$ is continuous in $\mathbb{R}\setminus\{0\}$.
If $a\leqslant0<b$, you use the fact that there's some natural $n$ such that$$\frac1{n\pi}<b$$and, since $y\in[-1,1]$, there's a $c\in\left[\frac1{(n+2)\pi},\frac1{n\pi}\right]$ such that $f(c)=y$.
Can you do the remaining case ($a<b\leqslant0$)?