Answer
$ - 6$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {h - 3} \right)}^2} - 9}}{h} \cr
& {\text{Evaluate the limit by using the Direct Substitution Property}} \cr
& \mathop {\lim }\limits_{u \to a} f\left( u \right) = f\left( a \right) \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {h - 3} \right)}^2} - 9}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {0 - 3} \right)}^2} - 9}}{0} = \frac{0}{0}{\text{ Indeterminate}} \cr
& \cr
& {\text{Expanding the binomial }} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2} - 6h + 9 - 9}}{h} \cr
& {\text{Simplifying}} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2} - 6h}}{h} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {h - 6} \right)}}{h} \cr
& = \mathop {\lim }\limits_{h \to 0} \left( {h - 6} \right) \cr
& \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{h \to 0} \left( {h - 6} \right) = 0 - 6 = - 6 \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {h - 3} \right)}^2} - 9}}{h} = - 6 \cr} $$
