1. **Stating the problem:** We want to find the limit $$\lim_{x \to 0} \frac{\ln(1 + 3x)}{3 \sin(x)}$$ 2. **Why split the expression?** Splitting the limit into parts helps us use known standard limits and simplifies evaluation. 3. **Rewrite the expression:** $$\lim_{x \to 0} \frac{\ln(1 + 3x)}{3 \sin(x)} = \lim_{x \to 0} \left( \frac{\ln(1 + 3x)}{3x} \cdot \frac{x}{\sin(x)} \right)$$ We split the original fraction into two parts multiplied together. 4. **Why multiply and divide by $x$?** We introduced $x$ in numerator and denominator to create expressions matching known limits: - $\lim_{x \to 0} \frac{\ln(1 + 3x)}{3x} = 1$ (because $\ln(1 + u) \approx u$ for small $u$) - $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ (a standard trigonometric limit) 5. **Rewrite the second part:** $$\frac{x}{\sin(x)} = \frac{1}{\frac{\sin(x)}{x}}$$ This is why the denominator $3 \sin(x)$ becomes $3x$ in the first fraction and the second fraction is inverted. 6. **Evaluate each limit separately:** - $\lim_{x \to 0} \frac{\ln(1 + 3x)}{3x} = 1$ - $\lim_{x \to 0} \frac{\sin(x)}{x} = 1 \implies \lim_{x \to 0} \frac{x}{\sin(x)} = 1$ 7. **Multiply the results:** $$1 \times 1 = 1$$ **Final answer:** $$\lim_{x \to 0} \frac{\ln(1 + 3x)}{3 \sin(x)} = 1$$ This splitting helps because each part matches a well-known limit, making the problem easier to solve.