Đáp án đúng: B Ta có: $ B=\underset{x\to +\infty }{\mathop{\lim }}\,\dfrac{\left| x \right|\sqrt{4+\dfrac{1}{x} }+x.\sqrt[3]{8+\dfrac{1}{{}{ x ^ 2 }}-\dfrac{1}{{}{ x ^ 3 }}}}{\left| x \right|\sqrt[4]{1+\dfrac{3}{{}{ x ^ 4 }}}}$
$=\underset{x\to +\infty }{\mathop{\lim }}\,\dfrac{\sqrt{4+\dfrac{1}{x} }+\sqrt[3]{8+\dfrac{1}{{}{ x ^ 2 }}-\dfrac{1}{{}{ x ^ 3 }}}}{\sqrt[4]{1+\dfrac{3}{{}{ x ^ 4 }}}}=4 $