Because $-1 \le \sin{x} \le 1$, we always have

$$-\mid{\frac{1}{x}}\mid < \frac{\sin{x}}{x} \le \mid{\frac{1}{x}}\mid$$

Meanwhile, noting that $\displaystyle\lim_{x\to\infty}\mid{\frac{1}{x}}\mid=0$ and applying the Sandwich theorem lead to

$$\lim_{x\to\infty}\frac{\sin{x}}{x}=\boxed{0}$$