Chapter 2 - Limits - 2.7 Limits at Infinity - Exercises - Page 83: 26

$-\frac{4}{5}.$

We have \begin{align*} \lim _{x \rightarrow- \infty}\frac{4x-3}{\sqrt{25x^2+4x}} &= \lim _{x \rightarrow- \infty}\frac{x(4-\frac{3}{x})}{x\sqrt{25+\frac{4}{x}}} \\ &=\frac{4-0}{\sqrt{25+0}} \\ &=\frac{4}{5}. \end{align*}