By # 4519, we have

$$\sin{x} \le x \le \tan{x},\qquad \left(0 \le x < \frac{\pi}{2}\right)$$

Dividing this relation by $\sin{x}$ leads to

$$1\le \frac{x}{\sin{x}} \le \frac{1}{\cos{x}}$$

When $x$ is negative and $-\frac{\pi}{2} < x < 0$, we have the following relation (note that in this case, $\sin{x}<0$, therefore dividing $\sin{x}$ will change the direction of inequality)

$$\sin{x} \ge x \ge \tan{x} \implies 1 \le \frac{x}{\sin{x}} \le \frac{1}{\cos{x}}$$

Therefore, by setting $\delta = \frac{\pi}{2}$, we have

$$\mid{x-0}\mid < \delta = \frac{\pi}{2} \implies 1 \le \frac{x}{\sin{x}} < \frac{1}{\cos{x}}$$

The left side of the above inequality is a constant independent of $x$. The right side approaches $1$ when $x\to 0$. Therefore, we conclude

$$\lim_{x\to 0}\ \frac{x}{\sin{x}} =1$$