Cho \(a\) là hằng số dương, \(\mathop {\lim }\limits_{x \to a} \frac{{a\left( {\sqrt {3{x^2} + {a^2}} - 2a} \right)}}{{x - a}}\) bằng 

Ta có \(\mathop {\lim }\limits_{x \to a} \frac{{a\left( {\sqrt {3{x^2} + {a^2}} - 2a} \right)}}{{x - a}} = \mathop {\lim }\limits_{x \to a} \frac{{a\left( {\sqrt {3{x^2} + {a^2}} - 2a} \right)\left( {\sqrt {3{x^2} + {a^2}} + 2a} \right)}}{{(x - a)\left( {\sqrt {3{x^2} + {a^2}} + 2a} \right)}}\)

\( = \mathop {\lim }\limits_{x \to a} \frac{{a\left( {3{x^2} + {a^2} - 4{a^2}} \right)}}{{\left( {x - a} \right)\left( {\sqrt {3{x^2} + {a^2}} + 2a} \right)}} = \mathop {\lim }\limits_{x \to a} \frac{{a\left( {3{x^2} - 3{a^2}} \right)}}{{(x - a)\left( {\sqrt {3{x^2} + {a^2}} + 2a} \right)}}\)

\( = \mathop {\lim }\limits_{x \to a} \frac{{3a\left( {x - a} \right)\left( {x + a} \right)}}{{(x - a)\left( {\sqrt {3{x^2} + {a^2}} + 2a} \right)}} = = \mathop {\lim }\limits_{x \to a} \frac{{3a\left( {x + a} \right)}}{{\left( {\sqrt {3{x^2} + {a^2}} + 2a} \right)}} = \frac{{3a\left( {a + a} \right)}}{{\sqrt {3{a^2} + {a^2}} + 2a}} = \frac{{6{a^2}}}{{4a}} = \frac{{3a}}{2}\). Chọn A.