1. **Problem Statement:** Find the derivative of the function $$f(x) = \frac{1 - x}{2x}$$ using the limit definition of the derivative. 2. **Limit Definition of the Derivative:** The derivative of a function $$f(x)$$ at a point $$x$$ is given by: $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$ This means we find the slope of the secant line between $$x$$ and $$x+h$$ and then take the limit as $$h$$ approaches zero to get the slope of the tangent line. 3. **Calculate $$f(x+h)$$:** $$ f(x+h) = \frac{1 - (x+h)}{2(x+h)} = \frac{1 - x - h}{2x + 2h} $$ 4. **Set up the difference quotient:** $$ \frac{f(x+h) - f(x)}{h} = \frac{\frac{1 - x - h}{2x + 2h} - \frac{1 - x}{2x}}{h} $$ 5. **Find a common denominator for the numerator:** $$ \frac{(1 - x - h)(2x) - (1 - x)(2x + 2h)}{(2x + 2h)(2x)} $$ 6. **Expand the numerator:** $$ (1 - x - h)(2x) = 2x - 2x^2 - 2xh $$ $$ (1 - x)(2x + 2h) = 2x - 2x^2 + 2h - 2xh $$ 7. **Subtract the two expressions:** $$ [2x - 2x^2 - 2xh] - [2x - 2x^2 + 2h - 2xh] = 2x - 2x^2 - 2xh - 2x + 2x^2 - 2h + 2xh = -2h $$ 8. **So the difference quotient becomes:** $$ \frac{-2h}{(2x + 2h)(2x)} \cdot \frac{1}{h} = \frac{-2h}{(2x + 2h)(2x)h} = \frac{-2}{(2x + 2h)(2x)} $$ 9. **Take the limit as $$h \to 0$$:** $$ f'(x) = \lim_{h \to 0} \frac{-2}{(2x + 2h)(2x)} = \frac{-2}{(2x)(2x)} = \frac{-2}{4x^2} = -\frac{1}{2x^2} $$ **Final answer:** $$ f'(x) = -\frac{1}{2x^2} $$