## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to\infty}\frac{\sin x}{[x]}=0$$
*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 $$A=\lim_{x\to\infty}\frac{\sin x}{[x]}=0$$