How to prove $\lim_{x\to 0} \lfloor\frac{\sin(x)}{x}\rfloor \cot(x)=0$? 
How to prove $\displaystyle \lim_{x\to 0} \left \lfloor\frac{\sin(x)}{x}\right\rfloor \cot(x)=0$?

We know that,
For $x>0$, it holds $\sin x<x$, so
$$
0<\frac{\sin x}{x}<1 \tag{*}
$$
For $x<0$
$$
\frac{\sin x}{x}=\frac{\sin(-x)}{-x}
$$
so the inequality (*) holds for every $x\ne0$. Hence
$$\displaystyle\lim_{x\to 0} \Big \lfloor\frac{\sin(x)}{x}\Big\rfloor \cot(x)=0.$$
On the other hand, $\displaystyle\lim_{x\to 0} \cot(x)=\infty$.
 A: @Angel your idea is ok. To be honest there is nothing to show here. Since $$\Big \lfloor\frac{\sin(x)}{x}\Big\rfloor=0 \quad \text{for } x\in [-\pi,\pi]\setminus\{0\} $$ what was spotted by you. So $$\displaystyle\lim_{x\to 0} \Big \lfloor\frac{\sin(x)}{x}\Big\rfloor \cot(x) =\displaystyle\lim_{x\to 0} 0\times \cot(x) = \displaystyle\lim_{x\to 0} 0=0.$$ The behavior of the $\cot$ function in the neighborhood of zero do not change anything since the values will be multiplied by zero so this is true zero function on $[-\pi,\pi]\setminus\{0\}$.
This problem is very similar (the same in the light of: Heine definition) to the following limit $$\lim_{n\to \infty} 0\times n$$ However this is the limit of zeros.
A: $\cot(x)$ only approaches infinity while the floor function is already at $0$, so the limit is $0$.
It is known that $\lim\limits_{x\to0}\frac{\sin x}{x} = 1$. Just use the Taylor Series for the sine function.
So now if you can prove that the derivative of the function inside the floor function is positive in the left neighbourhood of $x=0$ and negative in the right neighbourhood, you can deduce the value of your limit immediately.
Just calculate the derivative of $\frac{\sin x}{x}$ and show those properties.
