Answer
$$\lim_{x\to\infty}\frac{\sin x}{[x]}=0$$
Work Step by Step
*Remember that $\lim_{x\to\pm\infty}\frac{a}{x^n}=a\lim_{x\to\pm\infty}\frac{1}{x^n}=a\times0=0$
$$A=\lim_{x\to\infty}\frac{\sin x}{[x]}$$
We know that $-1\le\sin x\le1$
So, $$\frac{-1}{[x]}\le\frac{\sin x}{[x]}\le\frac{1}{[x]}$$
As $x\to\infty$, the greatest integer $[x]$ also approaches $\infty$.
Therefore, both $-1/[x]$ and $1/[x]$ will approach $0$. In other words, $$\lim_{x\to\infty}\frac{-1}{[x]}=\lim_{x\to\infty}\frac{1}{[x]}=0$$
So, according to the Sandwich Theorem, we can conclude that $$A=\lim_{x\to\infty}\frac{\sin x}{[x]}=0$$
