Answer
$\frac{4}{5}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{4x + 3}}{{5x - 1}} \cr
& {\text{Use properties of limits}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {4x} \right) + \mathop {\lim }\limits_{x \to \infty } \left( 3 \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {5x} \right) - \mathop {\lim }\limits_{x \to \infty } \left( 1 \right)}} = \frac{\infty }{\infty }{\text{ Ind}} \cr
& {\text{Divide the numerator and denominator by }}x \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{4x + 3}}{{5x - 1}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{4x}}{x} + \frac{3}{x}}}{{\frac{{5x}}{x} - \frac{1}{x}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{4 + \frac{3}{x}}}{{5 - \frac{1}{x}}} \cr
& {\text{Use properties of limits}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( 4 \right) + \mathop {\lim }\limits_{x \to \infty } \left( {\frac{3}{x}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( 5 \right) - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x}} \right)}} \cr
& {\text{Evaluate the limits}} \cr
& {\text{ = }}\frac{{4 - 0}}{{5 - 0}} \cr
& = \frac{4}{5} \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{4x + 3}}{{5x - 1}} = \frac{4}{5} \cr} $$
