Evaluating a limit by combining fractions over a common denominator

First combine the terms in the numerator over a common denominator: $\dfrac{1}{2+x} - \dfrac{1}{2} = \dfrac{2 - (2+x)}{2(2+x)} = \dfrac{-x}{2(2+x)}$. This turns the original expression into $\dfrac{-x}{2(2+x)} \cdot \dfrac{1}{x}$, since dividing by $x$ is the same as multiplying by $1/x$.

The factor of $x$ in the numerator cancels with the $x$ we are dividing by, leaving $\dfrac{-1}{2(2+x)}$ for $x \neq 0$. Now direct substitution is valid: plugging in $x = 0$ gives $\dfrac{-1}{2(2)} = -\dfrac{1}{4}$.