Answer
$ - \frac{1}{3}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{r \to \infty } \frac{{r - {r^3}}}{{2 - {r^2} + 3{r^3}}} \cr
& {\text{Using properties of limits}} \cr
& \mathop {\lim }\limits_{r \to \infty } \frac{{r - {r^3}}}{{2 - {r^2} + 3{r^3}}} = \frac{{\mathop {\lim }\limits_{r \to \infty } \left( {r - {r^3}} \right)}}{{\mathop {\lim }\limits_{r \to \infty } \left( {2 - {r^2} + 3{r^3}} \right)}} \cr
& {\text{ }} = \frac{{\mathop {\lim }\limits_{r \to \infty } \left( r \right) - \mathop {\lim }\limits_{r \to \infty } \left( {{r^3}} \right)}}{{\mathop {\lim }\limits_{r \to \infty } \left( 2 \right) - \mathop {\lim }\limits_{r \to \infty } \left( {{r^2}} \right) + \mathop {\lim }\limits_{r \to \infty } \left( {3{r^3}} \right)}} \cr
& {\text{Evaluate the limits}} \cr
& {\text{ }} = \frac{\infty }{\infty } \cr
& {\text{Divide the numerator and denominator by }}{r^3} \cr
& \mathop {\lim }\limits_{r \to \infty } \frac{{r - {r^3}}}{{2 - {r^2} + 3{r^3}}} = \mathop {\lim }\limits_{r \to \infty } \frac{{\frac{r}{{{r^3}}} - \frac{{{r^3}}}{{{r^3}}}}}{{\frac{2}{{{r^3}}} - \frac{{{r^2}}}{{{r^3}}} + \frac{{3{r^3}}}{{{r^3}}}}} \cr
& = \mathop {\lim }\limits_{r \to \infty } \frac{{\frac{1}{{{r^2}}} - 1}}{{\frac{2}{{{r^3}}} - \frac{1}{r} + 3}} \cr
& {\text{Use properties of limits}} \cr
& = \frac{{\mathop {\lim }\limits_{r \to \infty } \left( {\frac{1}{{{r^2}}}} \right) - \mathop {\lim }\limits_{r \to \infty } \left( 1 \right)}}{{\mathop {\lim }\limits_{r \to \infty } \left( {\frac{2}{{{r^3}}}} \right) - \mathop {\lim }\limits_{r \to \infty } \left( {\frac{1}{r}} \right) + \mathop {\lim }\limits_{r \to \infty } \left( 3 \right)}} \cr
& {\text{Evaluate the limits}} \cr
& = \frac{{0 - 1}}{{0 - 0 + 3}} \cr
& = - \frac{1}{3} \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{r \to \infty } \frac{{r - {r^3}}}{{2 - {r^2} + 3{r^3}}} = - \frac{1}{3} \cr} $$
