Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^3}}}{{{x^2} - 1}} - \frac{{{x^3}}}{{{x^2} + 1}}} \right) \cr
& {\text{evaluating the limit, we get:}} \cr
& = \frac{{{\infty ^3}}}{{{\infty ^2} - 1}} - \frac{{{\infty ^3}}}{{{\infty ^2} + 1}} = \frac{\infty }{\infty } \cr
& {\text{simplify }}\frac{{{x^3}}}{{{x^2} - 1}} - \frac{{{x^3}}}{{{x^2} + 1}} \cr
& \frac{{{x^3}}}{{{x^2} - 1}} - \frac{{{x^3}}}{{{x^2} + 1}} = \frac{{{x^5} + {x^3} - {x^5} + {x^3}}}{{\left( {{x^2} - 1} \right)\left( {{x^2} - 1} \right)}} = \frac{{2{x^3}}}{{{x^4} - 1}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{2{x^3}}}{{{x^4} - 1}}} \right) \cr
& {\text{evaluating the limit, we get:}} \cr
& = \frac{{2{{\left( \infty \right)}^3}}}{{{{\left( \infty \right)}^4} - 1}} = \frac{\infty }{\infty } \cr
& {\text{apply l'Hopital's Rule}}{\text{}}{\text{,}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{d/dx\left( {2{x^3}} \right)}}{{d/dx\left( {{x^4} - 1} \right)}}} \right) \cr
& = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{6{x^2}}}{{4{x^3}}}} \right) \cr
& = \frac{3}{2}\mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x}} \right) \cr
& {\text{evaluating the limit, we get:}} \cr
& = \frac{3}{2}\left( {\frac{1}{\infty }} \right) \cr
& = \frac{3}{2}\left( 0 \right) \cr
& = 0 \cr} $$
