Đáp án:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\sqrt {{x^2} + 3} + 4x}}{{\sqrt {4{x^2} + 1} - x}}\\
= \mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\dfrac{{\sqrt {{x^2} + 3} }}{x} + \dfrac{{4x}}{x}}}{{\dfrac{{\sqrt {4{x^2} + 1} }}{x} - \dfrac{x}{x}}}\\
= \mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\sqrt {1 + \dfrac{3}{{{x^2}}}} + 4}}{{\sqrt {4 + \dfrac{1}{{{x^2}}}} - 1}}\\
= \dfrac{{\sqrt 1 + 4}}{{\sqrt 4 - 1}}\\
= 5
\end{array}$
