Answer
$$\frac{1}{6}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x - 3}}{{x - 9}} \cr
& {\text{use the rule }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,\,\left( {{\text{see page 128}}} \right).{\text{ Then}} \cr
& \mathop {\lim }\limits_{x \to 9} \frac{{3{x^2} - 2x - 21}}{{x - 3}} = \frac{{\mathop {\lim }\limits_{x \to 9} \left( {\sqrt x - 3} \right)}}{{\mathop {\lim }\limits_{x \to 9} \left( {x - 9} \right)}} \cr
& {\text{evaluate the limits}}{\text{, replacing }}9{\text{ for }}x \cr
& \mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x - 3}}{{x - 9}} = \frac{{\sqrt 9 - 3}}{{9 - 9}} = \frac{0}{0}{\text{ indeterminate form}} \cr
& {\text{rationalizing}} \cr
& = \mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x - 3}}{{x - 9}} \times \frac{{\sqrt x + 3}}{{\sqrt x + 3}} \cr
& = \mathop {\lim }\limits_{x \to 9} \frac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}{{\left( {x - 9} \right)\left( {\sqrt x + 3} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 9} \frac{{{{\left( {\sqrt x } \right)}^2} - {{\left( 3 \right)}^2}}}{{\left( {x - 9} \right)\left( {\sqrt x + 3} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 9} \frac{{x - 9}}{{\left( {x - 9} \right)\left( {\sqrt x + 3} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 9} \frac{1}{{\sqrt x + 3}} \cr
& {\text{evaluating the limit when }}x \to 9 \cr
& = \frac{1}{{\sqrt 9 + 3}} \cr
& = \frac{1}{6} \cr} $$
