Answer
$$\infty $$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^4} - {x^3} - 3x}}{{7{x^2} + 9}} \cr
& {\text{evaluating the limit use the rules 4 and 2 }}\left( {page\,\,\,128} \right) \cr
& = \frac{{\mathop {\lim }\limits_{x \to \infty } {x^4} - \mathop {\lim }\limits_{x \to \infty } {x^3} - \mathop {\lim }\limits_{x \to \infty } 3x}}{{\mathop {\lim }\limits_{x \to \infty } 7{x^2} + \mathop {\lim }\limits_{x \to \infty } 9}} \cr
& = \frac{{{{\left( \infty \right)}^4} - {{\left( \infty \right)}^3} - 3\left( \infty \right)}}{{7{{\left( \infty \right)}^2} + 9}} \cr
& = \frac{\infty }{\infty } \cr
& {\text{divive the numerator and denominator by the highest power of }}x{\text{ in}} \cr
& {\text{the denominator}}{\text{. the highhest power is }}{x^2} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{{x^4}}}{{{x^2}}} - \frac{{{x^3}}}{{{x^2}}} - \frac{{3x}}{{{x^2}}}}}{{\frac{{7{x^2}}}{{{x^2}}} + \frac{9}{{{x^2}}}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} - x - \frac{3}{x}}}{{7 + \frac{9}{{{x^2}}}}} \cr
& {\text{use the rule }}\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^n}}} = 0 \cr
& {\text{the denominator approaches to 7}}{\text{, while the numerator becomes a }} \cr
& {\text{positive number that is larger and larger in magnitude}}{\text{, so}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^4} - {x^3} - 3x}}{{7{x^2} + 9}} = \infty \cr} $$
