When is the function Continuous? 
In my assignment I have to determine when is the function continuous.

This is the function:
\begin{equation}
g(x) = \begin{cases} \left\lfloor  {\sin\frac{1}{x}}\right\rfloor&\text{if} \space x \ne 0 \\
0 & \text{ if }x=0
\end{cases}
\end{equation}
I have to find when it is continuous on the interval $(\frac{1}{2\pi},\infty)$
I have proved that the function is not continuous when $x \to \frac {1}{\pi}$ and when $x\to\frac{2}{\pi}$
However, I suspect there are more that these values.
How can I find where is the function continuous?
Thanks.
 A: The function $f: \>x\mapsto y:={1\over x}$ maps the interval $\ \left]{1\over2\pi},\infty\right[\ $ homeomorphically onto the open interval $J:=\ ]0,2\pi[\ $ of the $y$-axis. We therefore have to check at which points $y\in J$  the function
$$h(y):=\lfloor\sin y\rfloor\qquad (y\in J)$$
is discontinuous. Now $\sin$ is continuous everywhere, but $\lfloor\cdot\rfloor$ has jump discontinuities at the integers. Therefore we have to analyze the points $y\in J$ where $\sin y$ assumes an integer value. These are the points ${\pi\over2}$, $\pi$, and ${3\pi\over2}$. One has $h\bigl({\pi\over2}\bigr)=1$, but $h(y)=0$ immediately to the left of ${\pi\over2}$. Then $h(\pi)=0$, but $h(y)=-1$ immediately to the right of $\pi$. Finally $h(y)=h\bigl({3\pi\over2}\bigr)=-1$ in a full neighborhood of ${3\pi\over2}$.  
Therefore $h$ is discontinuous only at the points $y={\pi\over2}$ and $y=\pi$, and your function $g:=h\circ f$ is discontinuous only at the two points $x={1\over\pi}$ and $x={2\over\pi}$ which you have spotted yourself.
A: The function $g$ is not continuous at $x=0$, since there are points $x$ arbitrarily close to $0$ such that $\sin(1/x)=1$.
The function $g$ is continuous at $a$ such that $\sin(1/a)$ is defined and not an integer. This is because the floor function is continuous at non-integers.
At points $a$ such that $\sin(1/a)=-1$, the function $g$ is continuous. For if $\sin(1/a)=-1$, we have $\lfloor\sin(1/t)\rfloor=-1$ for all $t$ close enough to $a$.
The function $g$ is not continuous at $x=a$ if $\sin(1/a)=1$. For if $t$ is near $a$, we have $g(t)=0$.
Finally, if $\sin(1/a)=0$, the function $g$ is not continuous at $a$.  For $\sin(1/x)$ changes sign at $a$, so there are $t$ arbitrarily close to $a$ such that $g(t)=-1$.
