Understand the range of the sine function.

The sine function, $\sin \theta$, always oscillates between $-1$ and 1, regardless of the value of $\theta$.
So, we know that:

$$-1 \le \sin \theta \le 1$$

Since we are evaluating the limit as $\theta \to \infty$, we can assume $\theta$ is a positive number. Dividing all parts of the inequality by $\theta$ (which is positive) does not change the direction of the inequalities:

$$\frac{-1}{\theta} \le \frac{\sin \theta}{\theta} \le \frac{1}{\theta}$$

Now, we find the limit of the left and right sides of the inequality as $\theta \to \infty$:

$$\lim_{\theta \to \infty} \frac{-1}{\theta} = 0$$ $$\lim_{\theta \to \infty} \frac{1}{\theta} = 0$$

Since the limit of the lower bound ($\frac{-1}{\theta}$) is 0 and the limit of the upper bound ($\frac{1}{\theta}$) is 0 as $\theta \to \infty$, by the Squeeze Theorem, the limit of the function in between must also be 0.

$$\lim_{\theta \to \infty} \frac{\sin \theta}{\theta} = 0$$

The final answer is $\boxed{0}$.