As the title said, I need to calculate these 3 limits-
$\lim_{x\rightarrow \frac{\pi k}{2}}\sin\left ( \frac{x}{2} \right )\left \lfloor \sin x \right \rfloor$ for $k=0,1,2$.
The thing is, we didn't even learn continuinity yet so I can't use the fact that $\sin$ is continuous. The only limits we are given as theorems are the $\lim$s of $\sin,\cos$ at $x\rightarrow 0$ (and we can use substitutions, that might lead us to these limits).
The solutions I came up with revolve showing for $k=2$ that the sided limits aren't equal (which I'm not fond of, as I need to use after inequalities for each side, the locality of limits (which means that if in some neighbourhood, 2 functions are identical, then their limits are equal at any point in the neighbourhood)). for $k=0,1$ I wanted to do the same concept, but two show the 2 sided limits are equal.
I really want to avoid the ways I came up with. I think they're very tedious considering we have to do the same thing for every point (in addition, for $k=1,2$ I need to use substitutions to get to limits at 0)
Is there any more simple way? Any help is appreciated in advance.
EDIT- I was able to do it gracefully for $k=0,1$, so now I need only $k=2$.