Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{3{x^3} + 2x - 1}}{{2{x^4} - 3{x^3} - 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{{3{x^3} + 2x - 1}}{{2{x^4} - 3{x^3} - 2}} = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {3{x^3} + 2x - 1} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {2{x^4} - 3{x^3} - 2} \right)}} \cr
& {\text{evaluate the limits}}{\text{, replacing }}\infty {\text{ for }}x \cr
& \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {3{x^3} + 2x - 1} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {2{x^4} - 3{x^3} - 2} \right)}} = \frac{{3{{\left( \infty \right)}^3} + 2\left( \infty \right) - 1}}{{2{{\left( \infty \right)}^4} - 3{{\left( \infty \right)}^3} - 2}} = \frac{\infty }{\infty }{\text{ indeterminate form}} \cr
& {\text{divide each term in the numerator and denominator by }}{x^4} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{3{x^3}}}{{{x^4}}} + \frac{{2x}}{{{x^4}}} - \frac{1}{{{x^4}}}}}{{\frac{{2{x^4}}}{{{x^4}}} - \frac{{3{x^3}}}{{{x^4}}} - \frac{2}{{{x^4}}}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{3}{x} + \frac{2}{{{x^3}}} - \frac{1}{{{x^4}}}}}{{2 - \frac{3}{x} - \frac{2}{{{x^4}}}}} = \frac{{\mathop {\lim }\limits_{x \to \infty } \frac{3}{x} + \mathop {\lim }\limits_{x \to \infty } \frac{2}{{{x^3}}} - \mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^4}}}}}{{\mathop {\lim }\limits_{x \to \infty } 2 - \mathop {\lim }\limits_{x \to \infty } \frac{3}{x} - \mathop {\lim }\limits_{x \to \infty } \frac{2}{{{x^4}}}}} \cr
& = \frac{{3\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} + 2\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^3}}} - \mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^4}}}}}{{\mathop {\lim }\limits_{x \to \infty } 2 - 3\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} - 2\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^4}}}}} \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{{3\left( 0 \right) + 2\left( 0 \right) - 0}}{{2 - 3\left( 0 \right) - 2\left( 0 \right)}} \cr
& = 0 \cr} $$
