1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{x - \sin 2x}{x + \sin 3x}$$. 2. **Recall the formula and rules:** For small angles, $$\sin x \approx x$$. This approximation helps simplify expressions involving sine near zero. 3. **Apply the approximation:** Replace $$\sin 2x$$ with $$2x$$ and $$\sin 3x$$ with $$3x$$ near zero: $$\lim_{x \to 0} \frac{x - \sin 2x}{x + \sin 3x} \approx \lim_{x \to 0} \frac{x - 2x}{x + 3x} = \lim_{x \to 0} \frac{x - 2x}{x + 3x}$$ 4. **Simplify numerator and denominator:** $$\frac{x - 2x}{x + 3x} = \frac{\cancel{x} - 2\cancel{x}}{\cancel{x} + 3\cancel{x}} = \frac{-x}{4x}$$ 5. **Cancel common factor $$x$$:** $$\frac{-\cancel{x}}{4\cancel{x}} = \frac{-1}{4}$$ 6. **Evaluate the limit:** Since the expression simplifies to a constant, the limit is: $$\boxed{-\frac{1}{4}}$$