Answer
$$3{x^2}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^3} - {x^3}}}{h} \cr
& {\text{use the rule }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,\,\left( {{\text{see page 128}}} \right).{\text{ Then}} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^3} - {x^3}}}{h} = \frac{{\mathop {\lim }\limits_{h \to 0} \left[ {{{\left( {x + h} \right)}^3} - {x^3}} \right]}}{{\mathop {\lim }\limits_{h \to 0} h}} = \frac{{{{\left( {x + 0} \right)}^3} - {x^3}}}{0} = \frac{0}{0}{\text{ indeterminate form}} \cr
& {\text{expand the binomial }}{\left( {x + h} \right)^3}{\text{ use }}{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} \cr
& then \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^3} - {x^3}}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{x^3} + 3{x^2}h + 3x{h^2} + {h^3} - {x^3}}}{h} \cr
& {\text{simplifying the numerator}} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{3{x^2}h + 3x{h^2} + {h^3}}}{h} \cr
& {\text{factor}} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {3{x^2} + 3xh + {h^2}} \right)}}{h} \cr
& = \mathop {\lim }\limits_{h \to 0} \left( {3{x^2} + 3xh + {h^2}} \right) \cr
& {\text{evaluating the limit when }}h \to 0 \cr
& = 3{x^2} + 3x\left( 0 \right) + {\left( 0 \right)^2} \cr
& = 3{x^2} \cr} $$
