1. The problem is to find the limit $$\lim_{x \to 0} \frac{\sin 3x}{x}$$. 2. We use the standard limit rule: $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$. 3. To apply this, rewrite the expression by introducing a factor inside the sine function: $$\lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot 3 = 3 \cdot \lim_{x \to 0} \frac{\sin 3x}{3x}$$ 4. Since $$\lim_{u \to 0} \frac{\sin u}{u} = 1$$, where $$u = 3x$$, we have: $$3 \cdot 1 = 3$$ 5. Therefore, the limit is 3. Answer: C. 3