Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} + 6x + 8}}{{{x^3} + 2x + 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{{{x^2} + 6x + 8}}{{{x^3} + 2x + 1}} = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {{x^2} + 6x + 8} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {{x^3} + 2x + 1} \right)}} \cr
& {\text{evaluate the limits}}{\text{, replacing }}\infty {\text{ for }}x \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} + 6x + 8}}{{{x^3} + 2x + 1}} = \frac{{{{\left( \infty \right)}^2} + 6\left( \infty \right) + 8}}{{{{\left( \infty \right)}^3} + 2\left( \infty \right) + 1}} = \frac{\infty }{\infty }{\text{ indeterminate form}} \cr
& {\text{divide each term in the numerator and denominator by }}{x^3} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{{x^2}}}{{{x^3}}} + \frac{{6x}}{{{x^3}}} + \frac{8}{{{x^3}}}}}{{\frac{{{x^3}}}{{{x^3}}} + \frac{{2x}}{{{x^3}}} + \frac{1}{{{x^3}}}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{x} + \frac{6}{{{x^2}}} + \frac{8}{{{x^3}}}}}{{1 + \frac{2}{{{x^2}}} + \frac{1}{{{x^3}}}}} = \frac{{\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} + \mathop {\lim }\limits_{x \to \infty } \frac{6}{{{x^2}}} + \mathop {\lim }\limits_{x \to \infty } \frac{8}{{{x^3}}}}}{{\mathop {\lim }\limits_{x \to \infty } 1 + \mathop {\lim }\limits_{x \to \infty } \frac{2}{{{x^2}}} + \mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^3}}}}} \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{{\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} + 6\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^2}}} + 8\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^3}}}}}{{\mathop {\lim }\limits_{x \to \infty } 1 + \mathop {\lim }\limits_{x \to \infty } \frac{2}{{{x^2}}} + \mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^3}}}}} = \frac{{0 + 0 + 0}}{{1 + 0 + 0}} = 0 \cr
& then \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} + 6x + 8}}{{{x^3} + 2x + 1}} = 0 \cr} $$
