The priority of limits How are the two expressions different?
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}}{x}\bigg\rfloor$$
and $$\bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor$$
If limit is inside the floor function, Do I have to apply the limits first?
If this is the case, then, $$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}}{x}\bigg\rfloor=0$$
$$\bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor=1$$
Am I solving this right?
Also how can I calculate,
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor$$ 
Thank you.

Yes you are right indeed since
$$\frac{\sin x}{x}<1 \implies \bigg \lfloor\frac{\sin{x}}{x}\bigg\rfloor=0 \implies \lim_{x\to0} \bigg \lfloor\frac{\sin{x}}{x}\bigg\rfloor=0$$
and since
$$\lim_{x\to0} \frac{\sin{x}}{x}=1 \implies \bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor=1
$$
For the latter for $x$ sufficiently small we have 
$$1<\frac{\sin{x}\cdot \tan{x}}{x^2}<2$$
indeed $\sin x>x-\frac{x^3} 6$ and $\tan x>x+\frac{x^3} 3$ and 
$$\frac{\sin{x}\cdot \tan{x}}{x^2}>1+\frac{x^2}6-\frac{x^6}{18}$$
therefore
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor=1$$
