Answer
$${\text{The limit does not exist}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{ - 5{x^3} - 4{x^2} + 8}}{{6{x^2} + 3x + 2}} \cr
& {\text{use the rule }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,\,\left( {{\text{see page 128}}} \right).{\text{ Then}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{ - 5{x^3} - 4{x^2} + 8}}{{6{x^2} + 3x + 2}} = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( { - 5{x^3} - 4{x^2} + 8} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {6{x^2} + 3x + 2} \right)}} \cr
& {\text{evaluate the limits}}{\text{, replacing }}\infty {\text{ for }}x \cr
& \frac{{\mathop {\lim }\limits_{x \to \infty } \left( { - 5{x^3} - 4{x^2} + 8} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {6{x^2} + 3x + 2} \right)}} = \frac{{ - 5{{\left( \infty \right)}^3} - 4{{\left( \infty \right)}^2} + 8}}{{6{{\left( \infty \right)}^2} + 3\left( \infty \right) + 2}} = \frac{\infty }{\infty }{\text{ indeterminate form}} \cr
& {\text{divide each term in the numerator and denominator by }}{x^3} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{ - 5{x^3}}}{{{x^3}}} - \frac{{4{x^2}}}{{{x^3}}} + \frac{8}{{{x^3}}}}}{{\frac{{6{x^2}}}{{{x^3}}} + \frac{{3x}}{{{x^3}}} + \frac{2}{{{x^3}}}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{ - 5 - \frac{4}{x} + \frac{8}{{{x^3}}}}}{{\frac{6}{x} + \frac{3}{{{x^2}}} + \frac{2}{{{x^3}}}}} = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( { - 5} \right) - \mathop {\lim }\limits_{x \to \infty } \frac{4}{x} + \mathop {\lim }\limits_{x \to \infty } \frac{8}{{{x^3}}}}}{{\mathop {\lim }\limits_{x \to \infty } \frac{6}{x} + \mathop {\lim }\limits_{x \to \infty } \frac{3}{{{x^2}}} + \mathop {\lim }\limits_{x \to \infty } \frac{2}{{{x^3}}}}} \cr
& \frac{{\mathop {\lim }\limits_{x \to \infty } \left( { - 5} \right) - 4\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} + 8\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^3}}}}}{{6\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} + 3\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^2}}} + 2\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^3}}}}} \cr
& {\text{evaluate the limits}}{\text{, use the rules }}\mathop {\lim }\limits_{x \to \infty } k = k{\text{ and }}\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^n}}} = 0 \cr
& = \frac{{ - 5 - 4\left( 0 \right) + 8\left( 0 \right)}}{{6\left( 0 \right) + 3\left( 0 \right) + 2\left( 0 \right)}} = \frac{{ - 5}}{0} = - \infty \cr
& {\text{The limit does not exist}} \cr} $$