Answer
$\frac{7}{4}$
Work Step by Step
Divide the numerator and denominator by $x$, which is the highest power of $x$ occurring in the denominator. Then, we get
$\lim\limits_{x \to \infty}\frac{7x-9}{4x+3}=\lim\limits_{x \to \infty}\frac{\frac{7x}{x}-\frac{9}{x}}{\frac{4x}{x}+\frac{3}{x}}=\lim\limits_{x \to \infty}\frac{7-\frac{9}{x}}{4+\frac{3}{x}}=\frac{7-0}{4+0}=\frac{7}{4}$
