1. The problem is to find the limit: $$\lim_{x \to 0} \frac{\ln(1 + 3x)}{3 \sin(x)}$$ 2. To solve this, we use the property that we can multiply and divide by the same expression to rewrite the limit in a more convenient form. Here, we multiply and divide by $x$: $$\lim_{x \to 0} \frac{\ln(1 + 3x)}{3 \sin(x)} = \lim_{x \to 0} \frac{\ln(1 + 3x)}{3x} \cdot \frac{x}{\sin(x)}$$ 3. Notice that multiplying by $\frac{x}{x} = 1$ does not change the value but allows us to separate the limit into two parts. 4. We rewrite the expression as: $$\lim_{x \to 0} \left( \frac{\ln(1 + 3x)}{3x} \right) \cdot \left( \frac{x}{\sin(x)} \right)$$ 5. This separation is useful because we know the standard limits: - $$\lim_{x \to 0} \frac{\ln(1 + 3x)}{3x} = 1$$ (from the derivative of $\ln(1+u)$ at $u=0$) - $$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$, so $$\lim_{x \to 0} \frac{x}{\sin(x)} = 1$$ 6. Therefore, the original limit is: $$1 \times 1 = 1$$ 7. In summary, the $\frac{x}{\sin(x)}$ term comes from multiplying and dividing by $x$ to separate the limit into known standard limits. Final answer: $$\boxed{1}$$