Evaluating a limit by combining fractions over a common denominator

Direct substitution at $x=2$ gives $\frac{1/2 - 1/2}{0} = \frac{0}{0}$, an indeterminate form, so we need to simplify algebraically first. Combine the numerator $\frac{1}{x} - \frac{1}{2}$ over a common denominator: $\frac{1}{x} - \frac{1}{2} = \frac{2 - x}{2x}$.

Substituting this back gives $\dfrac{\frac{2-x}{2x}}{x-2} = \frac{2-x}{2x(x-2)}$. Since $2 - x = -(x-2)$, this simplifies to $\frac{-(x-2)}{2x(x-2)} = \frac{-1}{2x}$ for $x \neq 2$. Now direct substitution works: $\lim_{x\to 2} \frac{-1}{2x} = \frac{-1}{4}$.

The technique used was combining fractions over a common denominator. The numerator, one over x minus one half, was rewritten as a single fraction with denominator two x. This allowed a common factor to be canceled with the denominator, resolving the zero over zero form.