Answer
$ - \frac{1}{4}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( { - 2 + h} \right)}^{ - 1}} + {2^{ - 1}}}}{h} \cr
& {\text{Evaluate the limit by using the Direct Substitution Property}} \cr
& \mathop {\lim }\limits_{h \to a} f\left( h \right) = f\left( a \right) \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( { - 2 + h} \right)}^{ - 1}} + {2^{ - 1}}}}{h} = \frac{{{{\left( { - 2 + 0} \right)}^{ - 1}} + {2^{ - 1}}}}{0} = \frac{0}{0}{\text{ Indeterminate}} \cr
& \cr
& {\text{Simplifying}}{\text{, recall that }}{x^{ - 1}} = \frac{1}{x} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( { - 2 + h} \right)}^{ - 1}} + {2^{ - 1}}}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{h - 2}} + \frac{1}{2}}}{h} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{2 + h - 2}}{{2\left( {h - 2} \right)}}}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{h}{{2\left( {h - 2} \right)}}}}{h} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{h}{{2h\left( {h - 2} \right)}} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{1}{{2\left( {h - 2} \right)}} \cr
& \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{1}{{2\left( {h - 2} \right)}} = \frac{1}{{2\left( {0 - 2} \right)}} = - \frac{1}{4} \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( { - 2 + h} \right)}^{ - 1}} + {2^{ - 1}}}}{h} = - \frac{1}{4} \cr} $$
