Answer
$\frac{1}{4}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x - 2}}{{x - 4}} \cr
& {\text{Use the quotient property of limits}} \cr
& \mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x - 2}}{{x - 4}} = \frac{{\mathop {\lim }\limits_{x \to 4} \left( {\sqrt x - 2} \right)}}{{\mathop {\lim }\limits_{x \to 4} \left( {x - 4} \right)}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to 4} \left( {\sqrt x } \right) - \mathop {\lim }\limits_{x \to 4} \left( 2 \right)}}{{\mathop {\lim }\limits_{x \to 4} \left( x \right) - \mathop {\lim }\limits_{x \to 4} \left( 4 \right)}} \cr
& {\text{Evaluate the limit}} \cr
& \frac{{\mathop {\lim }\limits_{x \to 4} \left( {\sqrt x } \right) - \mathop {\lim }\limits_{x \to 4} \left( 2 \right)}}{{\mathop {\lim }\limits_{x \to 4} \left( x \right) - \mathop {\lim }\limits_{x \to 4} \left( 4 \right)}} = \frac{{\sqrt 4 - 2}}{{4 - 4}}{\text{ = }}\frac{0}{0} \cr
& {\text{The limit is an indeterminate form type }}\frac{0}{0},{\text{ so we can apply}} \cr
& {\text{the L'Hospital's Rule}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x - 2}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \frac{{\frac{d}{{dx}}\left[ {\sqrt x - 2} \right]}}{{\frac{d}{{dx}}\left[ {x - 4} \right]}} \cr
& = \mathop {\lim }\limits_{x \to 4} \frac{{\frac{1}{{2\sqrt x }}}}{1} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{x \to 4} \frac{1}{{\sqrt x }} \cr
& {\text{Evaluate the limit when }}x \to 4 \cr
& = \frac{1}{2}\left( {\frac{1}{{\sqrt 4 }}} \right) \cr
& = \frac{1}{4} \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x - 2}}{{x - 4}} = \frac{1}{4} \cr} $$
Another method
Factor the denominator as $(\sqrt x-2)(\sqrt x+2)$, then simplify.
