Calculate limit with floor function or prove it doesnt exist

Question

Please help calculating the following limit:

$$ \lim_{x\to\frac{\pi}{2}} \frac{\lfloor{}\sin(x)\rfloor}{\lfloor x\rfloor} $$

I used $$ t = x - \frac{\pi}{2} $$ and got:

$$ \lim_{t\to 0} \frac{\lfloor{}\sin(t+\frac{\pi}{2})\rfloor}{\lfloor t+\frac{\pi}{2}\rfloor} $$

for t close to 0 we get from arithmetic of limits that the denominator is 1 but not sure how to go from here..

Thanks

Answer 1

In a punctured neighborhood of $\pi/2$ we have $\lfloor\sin x\rfloor=0$ and $\lfloor x\rfloor=\color{red}1$, so the limit is $0$.

Answer 2

The floor function is continuous on $\Bbb R\setminus \Bbb Z$ so $$\lim_{x\to\frac\pi2}\lfloor x\rfloor=\left\lfloor\frac\pi2\right\rfloor=1$$ and since

$$0\le \sin x<1\,\forall x\in I:=(0,\pi)\setminus\{\frac\pi2\}$$ hence $\lfloor \sin x\rfloor=0\;\forall x\in I$ so the desired limit is $0$.

Answer 3

In the range $\left(1,\pi\right)$ but at $\pi/2$, as the numerator is zero and the denominator nonzero, the function is zero.

Answer 4

We have $\forall x\in(0,\pi)\;\; 0<\sin(x)<1 $ and $\lfloor \sin(x) \rfloor =0$.

the limit is zero.