Answer
$\frac{3}{4}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{3{x^3} - 8x + 2}}{{4{x^3} - 5{x^2} - 2}} \cr
& {\text{Use properties of limits}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {3{x^3} - 8x + 2} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {4{x^3} - 5{x^2} - 2} \right)}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {3{x^3}} \right) - \mathop {\lim }\limits_{x \to \infty } \left( {8x} \right) + \mathop {\lim }\limits_{x \to \infty } \left( 2 \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {4{x^3}} \right) - \mathop {\lim }\limits_{x \to \infty } \left( {5{x^2}} \right) - \mathop {\lim }\limits_{x \to \infty } \left( 2 \right)}} = \frac{\infty }{\infty }{\text{ Ind}} \cr
& {\text{Divide the numerator and denominator by }}{x^3} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{3{x^3} - 8x + 2}}{{4{x^3} - 5{x^2} - 2}} = \mathop {\lim }\limits_{x \to \infty } \frac{{3{x^3} - 8x + 2}}{{4{x^3} - 5{x^2} - 2}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{3{x^3}}}{{{x^3}}} - \frac{{8x}}{{{x^3}}} + \frac{2}{{{x^3}}}}}{{\frac{{4{x^3}}}{{{x^3}}} - \frac{{5{x^2}}}{{{x^3}}} - \frac{2}{{{x^3}}}}} \cr
& {\text{Use properties of limits}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( 3 \right) - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{8}{{{x^2}}}} \right) + \mathop {\lim }\limits_{x \to \infty } \left( {\frac{2}{{{x^3}}}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( 4 \right) - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{5}{{{x^2}}}} \right) - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{2}{{{x^3}}}} \right)}} \cr
& {\text{Evaluate the limits}} \cr
& {\text{ = }}\frac{{3 - 0 + 0}}{{4 - 0 - 0}} \cr
& = \frac{3}{4} \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{3{x^3} - 8x + 2}}{{4{x^3} - 5{x^2} - 2}} = \frac{3}{4} \cr} $$
