Analyzing the continuity of $\sin(x\left \lfloor{\frac{\pi }{x}} \right \rfloor)$ at 0

Question

There is a piecewise function $f$ defined as following:

$f:\left\{\begin{matrix} \sin(x\left \lfloor{\frac{\pi }{x}} \right \rfloor) \Leftarrow x\in \left ] 0,2\pi \right ]\\ 0 \Leftarrow x=0 \end{matrix}\right.$

I'm asked to study the continuity of the function at $x=0$.

As far as I know, in order to demonstrate that a function $f$ is continued at a certain point one has to prove that the limit of the function on both sides is equal that the function itself at that point. But I can't wrap my head around how I should deal with the division by 0 in the floor function.

Any help would be appreciated, thank you for your time.

Answer 1

For $x\in(0,2\pi],\;$ we have

$$\frac{\pi}{x}-1<\lfloor \frac{\pi}{x} \rfloor \leq \frac{\pi}{x}$$

and

$$\pi-x<x\lfloor \frac{\pi}{x} \rfloor\leq \pi$$

$$\implies \lim_{x\to 0^+} x\lfloor \frac{\pi}{x} \rfloor =\pi$$

thus

$$\lim_{x\to 0^+}f(x)=\sin(\pi)=0=f(0)$$

the function $f$ is right continuous at $x=0$.