Answer
$$\frac{1}{4}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{ - 1/\left( {x + 2} \right) + 1/2}}{x} \cr
& {\text{use the rule }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,\,\left( {{\text{see page 128}}} \right).{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{ - 1/\left( {x + 2} \right) + 1/2}}{x} = \frac{{\mathop {\lim }\limits_{x \to 0} \left( { - 1/\left( {x + 2} \right) + 1/2} \right)}}{{\mathop {\lim }\limits_{x \to 0} x}} \cr
& {\text{evaluate the limits}}{\text{, replacing }}0{\text{ for }}x \cr
& = \frac{{\left( { - 1/\left( {0 + 2} \right) + 1/2} \right)}}{0} = \frac{{ - 1/2 + 1/2}}{0} = \frac{0}{0}{\text{ indeterminate form}} \cr
& {\text{Simplifying the numerator as}} \cr
& - \frac{1}{{x + 2}} + \frac{1}{2} = \frac{{ - 2 + x + 2}}{{2\left( {x + 2} \right)}} = \frac{x}{{2\left( {x + 2} \right)}} \cr
& then \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{ - 1/\left( {x + 2} \right) + 1/2}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{x}{{2\left( {x + 2} \right)}}}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{x}{{2x\left( {x + 2} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{1}{{2\left( {x + 2} \right)}} \cr
& {\text{evaluating the limit when }}x \to 0 \cr
& = \frac{1}{{2\left( {0 + 2} \right)}} \cr
& = \frac{1}{4} \cr} $$