Answer
$$\frac{2}{5}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2} + 5}}{{5{x^2} - 1}} \cr
& {\text{use the rule }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,\,\left( {{\text{see page 128}}} \right).{\text{ Then}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2} + 5}}{{5{x^2} - 1}} = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {2{x^2} + 5} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {5{x^2} - 1} \right)}} \cr
& {\text{evaluate the limits}}{\text{, replacing }}\infty {\text{ for }}x \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2} + 5}}{{5{x^2} - 1}} = \frac{{2{{\left( \infty \right)}^2} + 5}}{{5{{\left( \infty \right)}^2} - 1}} = \frac{\infty }{\infty }{\text{ indeterminate form}} \cr
& {\text{divide each term in the numerator and denominator by }}{x^2} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{2{x^2}}}{{{x^2}}} + \frac{5}{{{x^2}}}}}{{\frac{{5{x^2}}}{{{x^2}}} - \frac{1}{{{x^2}}}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{2 + \frac{5}{{{x^2}}}}}{{5 - \frac{1}{{{x^2}}}}} = \frac{{\mathop {\lim }\limits_{x \to \infty } 2 + \mathop {\lim }\limits_{x \to \infty } \frac{5}{{{x^2}}}}}{{\mathop {\lim }\limits_{x \to \infty } 5 - \mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^2}}}}} \cr
& {\text{evaluate the limits}}{\text{, use the rules }}\mathop {\lim }\limits_{x \to \infty } k = k{\text{ and }}\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^n}}} = 0 \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{2 + \frac{5}{{{x^2}}}}}{{5 - \frac{1}{{{x^2}}}}} = \frac{{2 + 0}}{{5 - 0}} \cr
& = \frac{2}{5} \cr} $$
