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