Biết rằng Lim 2{x^3} + 6 căn 3/ 3 - {x^2= a căn 3  + b Tính \({a^2} + {b^2}.

Chọn A

Ta có

\[\mathop {\lim }\limits_{x \to - \sqrt 3 } \frac{{2{x^3} + 6\sqrt 3 }}{{3 - {x^2}}} = \mathop {\lim }\limits_{x \to - \sqrt 3 } \frac{{2(x + \sqrt 3 )({x^2} - x\sqrt 3 + {{\sqrt 3 }^2})}}{{(\sqrt 3 - x).(\sqrt 3 + x)}} = \mathop {\lim }\limits_{x \to - \sqrt 3 } 2\frac{{{x^2} - x\sqrt 3 + {{\sqrt 3 }^2}}}{{\sqrt 3 - x}} = 3\sqrt 3 = a\sqrt 3 + b.\]

\( \Rightarrow {a^2} + {b^2} = {3^2} = 9\)