Estimating a limit numerically near x = 0

First, combine the numerator into a single fraction: $\dfrac{1}{x+2} - \dfrac{1}{2} = \dfrac{2 - (x+2)}{2(x+2)} = \dfrac{-x}{2(x+2)}$. Dividing this by $x$ (for $x \neq 0$) gives $f(x) = \dfrac{-x}{2(x+2)x} = \dfrac{-1}{2(x+2)}$, which is much easier to evaluate near $x=0$.

Now compute values as $x \to 0^-$: for $x = -0.1, -0.01, -0.001$, $f(x) \approx -0.2632, -0.2506, -0.2501$, approaching $-0.25$. As $x \to 0^+$: for $x = 0.1, 0.01, 0.001$, $f(x) \approx -0.2381, -0.2494, -0.2499$, also approaching $-0.25$. Since both sides trend toward the same value, we conclude the limit exists and equals $-0.25$.