Answer
$$\frac{1}{3}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} + 2x - 5}}{{3{x^2} + 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{{{x^2} + 2x - 5}}{{3{x^2} + 2}} = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {{x^2} + 2x - 5} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {3{x^2} + 2} \right)}} \cr
& {\text{evaluate the limits}}{\text{, replacing }}\infty {\text{ for }}x \cr
& \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {{x^2} + 2x - 5} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {3{x^2} + 2} \right)}} = \frac{{{{\left( \infty \right)}^2} + 2\left( \infty \right) - 5}}{{3{{\left( \infty \right)}^2} + 2}} = \frac{\infty }{\infty }{\text{ indeterminate form}} \cr
& {\text{divide each term in the numerator and denominator by }}{x^2} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{{x^2}}}{{{x^2}}} + \frac{{2x}}{{{x^2}}} - \frac{5}{{{x^2}}}}}{{\frac{{3{x^2}}}{{{x^2}}} + \frac{2}{{{x^2}}}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{1 + \frac{2}{x} - \frac{5}{{{x^2}}}}}{{3 + \frac{2}{{{x^2}}}}} = \frac{{\mathop {\lim }\limits_{x \to \infty } 1 + \mathop {\lim }\limits_{x \to \infty } \frac{2}{x} - \mathop {\lim }\limits_{x \to \infty } \frac{5}{{{x^2}}}}}{{\mathop {\lim }\limits_{x \to \infty } 3 + \mathop {\lim }\limits_{x \to \infty } \frac{2}{{{x^2}}}}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to \infty } 1 + 2\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} - 5\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^2}}}}}{{\mathop {\lim }\limits_{x \to \infty } 3 + 2\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{{1 + 2\left( 0 \right) - 5\left( 0 \right)}}{{3 + 2\left( 0 \right)}} \cr
& = \frac{1}{3} \cr} $$
