Calculate limit with floor function or prove it doesnt exist 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

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