Answer
$$ \frac{\sqrt{3 }}{4}$$
Work Step by Step
Given $$ \lim _{x\to \infty }\frac{\sqrt{x+3x^2}}{4x-1}$$
Then
\begin{aligned}
\lim _{x\to \infty }\frac{\sqrt{x+3x^2}}{4x-1}&= \lim _{x\to \infty }\frac{\sqrt{\frac{x}{x^2}+3\frac{x^2}{x^2}}}{4\frac{x}{x}-\frac{1}{x}} \\
&=\lim _{x\to \infty }\frac{\sqrt{\frac{1}{x }+3 }}{4 -\frac{1}{x}}\\
&=\lim _{x\to \infty }\frac{\sqrt{0+3 }}{4 -0}\\
&= \frac{\sqrt{3 }}{4}
\end{aligned}
