Answer
$$ - \frac{3}{2}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{8}{{{x^2} - 4}} - \frac{x}{{x - 2}}} \right] \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{8}{{{x^2} - 4}} - \frac{x}{{x - 2}}} \right] = \frac{8}{{{2^2} - 4}} - \frac{2}{{2 - 2}} = \infty - \infty \cr
& {\text{Simplify}} \cr
& = \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{8}{{\left( {x + 2} \right)\left( {x - 2} \right)}} - \frac{x}{{x - 2}}} \right] \cr
& = \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{{8 - x\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}} \right] \cr
& = \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{{8 - {x^2} - 2x}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}} \right] \cr
& = - \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{{{x^2} + 2x - 8}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}} \right] \cr
& {\text{Factor}} \cr
& = - \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{{\left( {x + 4} \right)\left( {x - 2} \right)}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}} \right] \cr
& = - \mathop {\lim }\limits_{x \to {2^ + }} \left[ {\frac{{x + 4}}{{x + 2}}} \right] \cr
& {\text{Evaluate the limit}} \cr
& {\text{ = }} - \frac{{2 + 4}}{{2 + 2}} \cr
& {\text{ = }} - \frac{3}{2} \cr} $$
