Answer
$${\text{The limit does not exist}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2} - 7{x^4}}}{{9{x^2} + 5x - 6}} \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} - 7{x^4}}}{{9{x^2} + 5x - 6}} = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {2{x^2} - 7{x^4}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {9{x^2} + 5x - 6} \right)}} \cr
& {\text{evaluate the limits}}{\text{, replacing }}\infty {\text{ for }}x \cr
& \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {2{x^2} - 7{x^4}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {9{x^2} + 5x - 6} \right)}} = \frac{{2{{\left( \infty \right)}^2} - 7{{\left( \infty \right)}^4}}}{{9{{\left( \infty \right)}^2} + 5\left( \infty \right) - 6}} = \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{{2{x^2}}}{{{x^4}}} - \frac{{7{x^4}}}{{{x^4}}}}}{{\frac{{9{x^2}}}{{{x^4}}} + \frac{{5x}}{{{x^4}}} - \frac{6}{{{x^4}}}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{2}{{{x^2}}} - 7}}{{\frac{9}{{{x^2}}} + \frac{5}{{{x^2}}} - \frac{6}{{{x^4}}}}} = \frac{{\mathop {\lim }\limits_{x \to \infty } \frac{2}{{{x^2}}} - \mathop {\lim }\limits_{x \to \infty } 7}}{{\mathop {\lim }\limits_{x \to \infty } \frac{9}{{{x^2}}} + \mathop {\lim }\limits_{x \to \infty } \frac{5}{{{x^2}}} - \mathop {\lim }\limits_{x \to \infty } \frac{6}{{{x^4}}}}} \cr
& = \frac{{2\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^2}}} - \mathop {\lim }\limits_{x \to \infty } 7}}{{\mathop {9\lim }\limits_{x \to \infty } \frac{1}{{{x^2}}} + 5\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^2}}} - 6\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{{2\left( 0 \right) - 7}}{{9\left( 0 \right) + 5\left( 0 \right) - 6\left( 0 \right)}} = \frac{{ - 7}}{0} = - \infty \cr
& {\text{The limit does not exist}} \cr} $$
