Answer
$$\infty $$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{2{x^3} - x - 3}}{{6{x^2} - x - 1}} \cr
& {\text{evaluating the limit use the rules 4 and 2 }}\left( {page\,\,\,128} \right) \cr
& = \frac{{\mathop {\lim }\limits_{x \to \infty } 2{x^3} - \mathop {\lim }\limits_{x \to \infty } x - \mathop {\lim }\limits_{x \to \infty } 3}}{{\mathop {\lim }\limits_{x \to \infty } 6{x^2} - \mathop {\lim }\limits_{x \to \infty } x - \mathop {\lim }\limits_{x \to \infty } 1}} \cr
& = \frac{{2{{\left( \infty \right)}^3} - \infty - 3\left( \infty \right)}}{{6{{\left( \infty \right)}^2} - \infty - 1}} \cr
& = \frac{\infty }{\infty } \cr
& {\text{divive the numerator and denominator by the highest power of }}x{\text{ in}} \cr
& {\text{the denominator}}{\text{. the highhest power is }}{x^2} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{2{x^3}}}{{{x^2}}} - \frac{x}{{{x^2}}} - \frac{3}{{{x^2}}}}}{{\frac{{6{x^2}}}{{{x^2}}} - \frac{x}{{{x^2}}} - \frac{1}{{{x^2}}}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{2x - \frac{1}{x} - \frac{3}{{{x^2}}}}}{{6 - \frac{1}{x} - \frac{1}{{{x^2}}}}} \cr
& {\text{use the rule }}\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^n}}} = 0 \cr
& {\text{the denominator approaches to 6}}{\text{, while the numerator becomes a }} \cr
& {\text{positive number that is larger and larger in magnitude}}{\text{, so}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{2{x^3} - x - 3}}{{6{x^2} - x - 1}} = \infty \cr} $$