Cho biết (mathop {lim }limits_{x to – 2} dfrac{{4{x^3} – 1}}{{3{x^2} + x + 2}}) bằng

(mathop {lim }limits_{x to  – 2} dfrac{{4{x^3} – 1}}{{3{x^2} + x + 2}} ) (= dfrac{{4.{{( – 2)}^2} – 1}}{{3.{{( – 2)}^2} + ( – 2) + 2}} = dfrac{{ – 33}}{{12}} = dfrac{{ – 11}}{4})