Answer
$$ - \frac{1}{8}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{1}{{{x^2} - 4}} - \frac{{\sqrt {x - 1} }}{{{x^2} - 4}}} \right] \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{8}{{{x^2} - 4}} - \frac{x}{{x - 2}}} \right] = \frac{1}{{{2^2} - 4}} - \frac{{\sqrt {2 - 1} }}{{{2^2} - 4}} = \infty - \infty \cr
& {\text{Simplify}} \cr
& = \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{{1 - \sqrt {x - 1} }}{{{x^2} - 4}}} \right] \cr
& {\text{Rationalizing}} \cr
& = \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{{1 - \sqrt {x - 1} }}{{{x^2} - 4}} \times \frac{{1 + \sqrt {x - 1} }}{{1 + \sqrt {x - 1} }}} \right] \cr
& = \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{{1 - \left( {x - 1} \right)}}{{\left( {{x^2} - 4} \right)\left( {1 + \sqrt {x - 1} } \right)}}} \right] \cr
& = \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{{1 - x + 1}}{{\left( {{x^2} - 4} \right)\left( {1 + \sqrt {x - 1} } \right)}}} \right] \cr
& = \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{{2 - x}}{{\left( {{x^2} - 4} \right)\left( {1 + \sqrt {x - 1} } \right)}}} \right] \cr
& {\text{Factor}} \cr
& = \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{{2 - x}}{{\left( {x + 2} \right)\left( {x - 2} \right)\left( {1 + \sqrt {x - 1} } \right)}}} \right] \cr
& = - \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{{x - 2}}{{\left( {x + 2} \right)\left( {x - 2} \right)\left( {1 + \sqrt {x - 1} } \right)}}} \right] \cr
& = - \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{1}{{\left( {x + 2} \right)\left( {1 + \sqrt {x - 1} } \right)}}} \right] \cr
& {\text{Evaluate the limit}} \cr
& {\text{ = }} - \frac{1}{{\left( {2 + 2} \right)\left( {1 + \sqrt {2 - 1} } \right)}} \cr
& {\text{ = }} - \frac{1}{{\left( 4 \right)\left( 2 \right)}} \cr
& = - \frac{1}{8} \cr} $$
