I was solving limits and it said $\lim_{x→0}[x^2/\tan(x)\sin(x)]=0$

Question

I tried using $[x][y]=[xy]$ but clearly, that is wrong

I know that $\lim_{x→0}[\sin(x)x]=0,\lim_{x→0}[\sin⁡(x)x]=0$ and $\lim_{x→0}[\tan(x)x]=1$.

Answer 1

One can see that

$$\frac{x^2}{\tan(x)\sin(x)}=\frac{\cos(x)}{\left(\frac{\sin(x)}x\right)^2}$$

thus, the limit is $1$ without the floor function $[\cdot]$. With the floor function, we can deduce the limit will be $1$ or $0$, depending on which side $\frac{x^2}{\tan(x)\sin(x)}$ approaches $1$. We can deduce which side the limit falls on by observing the following:

$$\tan(x)\sin(x)-x^2>0\text{ for }|x|<r,\ r>0$$

which follows from taking the Maclaurin series expansion.

Thus, the limit is $0$.