حد چپ و راست تابع f در $x = 2$ برابرند پس: $\left( {a - 2} \right)\left( 2 \right) + 2x\left( { - 3} \right) = \left( {a - 2} \right)\left( 1 \right) + 2x\left( { - 2} \right)$
$ \Rightarrow a = - 2 \Rightarrow \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{x\left[ {4x} \right] - {x^4}}}{{{x^2} - \left[ {x + 2} \right] - 4}} = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{ - 8x - {x^4}}}{{{x^2} - x - 6}} = $
$ = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{ - x\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}{{\left( {x + 2} \right)\left( {x - 3} \right)}} = \frac{{2 \times 12}}{{ - 5}} = \frac{{ - 48}}{{10}} = - 4/8$