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