for $$\lim_{x\to k}\lfloor x\rfloor \sin\frac{π x}2$$
find the limit for $k=0,1,2,3$
i started with
$$x-1<\lfloor x\rfloor\le x$$ $$\Downarrow$$ $$x\sin\frac{π x}2-\sin\frac{π x}2<\lfloor x\rfloor \sin\frac{π x}2\le x\sin\frac{π x}2$$
for $k= 0$ and for $k=2$
$$\lim\limits_{x\to k}x\sin\frac{π x}2-\sin\frac{π x}2=\lim\limits_{x\to k} x\sin\frac{π x}2= 0$$
and because of squeeze theorem $$\lim\limits_{x\to 0}\lfloor x\rfloor \sin\frac{π x}2=0$$
for $k=1$ and $k=3$ $$\lim\limits_{x\to k}x\sin\frac{π x}2-\sin\frac{π x}2=0$$ and
$$\lim\limits_{x\to k} x\sin\frac{π x}2= 1 , -1$$ so it means there is no limit at all? can i use sqeeze therom on the opposit and say that it represent the limit on the right and the limit on the left?
You can sometimes use the Squeeze Theorem to show that a limit does exist, but never to show that a limit does not exist. So your solutions are fine for $k=0$ and $k=2$ but you need to find an alternative for $k=1$ and $k=3$.
Hint. Try taking the limit from the right and left separately. If $3<x<4$ then $\lfloor x\rfloor=3$, so $$\lfloor x\rfloor\sin\frac{\pi x}{2}=3\sin\frac{\pi x}{2}\ .$$ If $2<x<3$ then $\lfloor x\rfloor=2$ and so $$\lfloor x\rfloor\sin\frac{\pi x}{2}=2\sin\frac{\pi x}{2}\ .$$ You can now find the limit for both of these as $x\to3$. If the two limits have the same value then it is the value of $$\lim_{x\to3}\lfloor x\rfloor\sin\frac{\pi x}{2}\ ,$$ if they are different (or if either does not exist) then this "two-sided" limit does not exist.