$\sin(2\pi x[x])$ is continuous on $\mathbb R$

Question

How to show that $f(x)=\sin(2\pi x[x])$ is continuous on $\mathbb R$?

Proof:I used the result

$\vert \sin(x)-\sin(y)\vert <\vert x - y\vert $

Replacing $x\rightarrow x[x]$ and $y\rightarrow y[y]$,we get

$\vert \sin(2\pi x[x])-\sin(2\pi y[y])\vert <2\pi\vert x[x] - y[y]\vert $

I wanted to some how show that $\vert x[x] - y[y]\vert <\vert x-y\vert$

So,that for every $\epsilon >0$ there exists $\delta=\epsilon/2 \pi>0$ such that

$\vert x-y\vert <\delta\implies \vert \sin(2\pi x[x])-\sin(2\pi y[y])\vert <\epsilon$

Need amendments in above proof...

Thank you...

Answer 1

In $(k,k+1)$, $\sin(2\pi x\lfloor x\rfloor)=\sin(2k\pi x)$ and this function is notoriously continuous.

Then

$$\lim_{x\to k^-}\sin(2\pi x\lfloor x\rfloor)=\lim_{x\to k^-}\sin(2\pi x(k-1))=0,$$

$$\lim_{x\to k^+}\sin(2\pi x\lfloor x\rfloor)=\lim_{x\to k^+}\sin(2\pi xk)=0,$$ and $$\sin(2\pi k\lfloor k\rfloor)=0.$$

Answer 2

HINT

Note the discontinuities of $x[x]$ only happen at $\left\{\left. a + \frac12\right| a \in \mathbb{Z}\right\}$, which has period $1$. The idea is, when you take the sine, you can transform the period to make it exactly $2\pi$, the natural period of the sine function.