Answer
$\frac{\sqrt{3}}{4}$
Work Step by Step
$\lim\limits_{x \to \infty}\frac{\sqrt{x+3x^2}}{4x-1}=\lim\limits_{x \to \infty}\frac{\sqrt{x+3x^2}}{4x-1}\times \frac{1/x}{1/x}$
$=\lim\limits_{x \to \infty}\frac{\sqrt{1/x+3}}{4-1/x}$ (Use the properties of limits)
$=\frac{\sqrt{\lim\limits_{x \to \infty}1/x+3}}{\lim\limits_{x \to \infty}(4-1/x)}$ (Use the limit $\lim\limits_{x \to \infty} 1/x =0$)
$=\frac{\sqrt{0+3}}{4-0}$
$=\frac{\sqrt{3}}{4}$
Thus, $\lim\limits_{x \to \infty}\frac{\sqrt{x+3x^2}}{4x-1}=\frac{\sqrt{3}}{4}$
