Chapter 2 - Limits - 2.7 Limits at Infinity - Exercises - Page 82: 13

$$-\infty $$

\begin{align*} \lim _{x \rightarrow -\infty}\frac{7x^2-9}{4x+3}&= \lim _{x \rightarrow -\infty}\frac{x^2}{x}\frac{7-9x^{-2}}{4+3x^{-1}}\\ &=\lim _{x \rightarrow -\infty}\frac{x^2}{x}\lim _{x \rightarrow -\infty}\frac{7-9x^{-2}}{4+3x^{-1}}\\ &=\frac{7}{4}\lim _{x \rightarrow -\infty} x \\ &=-\infty.\\ \end{align*}