1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{1}{x} \left( \frac{1}{2} - \frac{1}{3x + 2} \right)$$. 2. **Rewrite the expression inside the parentheses:** $$\frac{1}{2} - \frac{1}{3x + 2} = \frac{(3x + 2) - 2}{2(3x + 2)} = \frac{3x}{2(3x + 2)}$$. 3. **Substitute back into the limit:** $$\lim_{x \to 0} \frac{1}{x} \cdot \frac{3x}{2(3x + 2)} = \lim_{x \to 0} \frac{3x}{x \cdot 2(3x + 2)}$$. 4. **Cancel common factor $x$ in numerator and denominator:** $$\lim_{x \to 0} \frac{\cancel{3x}}{\cancel{x} \cdot 2(3x + 2)} = \lim_{x \to 0} \frac{3}{2(3x + 2)}$$. 5. **Evaluate the limit by substituting $x = 0$:** $$\frac{3}{2(3 \cdot 0 + 2)} = \frac{3}{2 \cdot 2} = \frac{3}{4}$$. **Final answer:** $$\boxed{\frac{3}{4}}$$