Question

Find the limit or show that it does not exist.

19. \(\mathop {\lim }\limits_{x \to \infty } \frac{{r - {r^3}}}{{2 - {r^2} + 3{r^3}}}\)

Short answer

The value of the limit is\( - \frac{1}{3}\).

Step-by-step solution

Use the property of limit

A function is continuous at a point “a” if \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\).

Evaluate the given limit

Evaluate the given limit as follows:

\(\begin{aligned}\mathop {\lim }\limits_{x \to \infty } \frac{{r - {r^3}}}{{2 - {r^2} + 3{r^3}}}{\rm{ }} &= \mathop {\lim }\limits_{r \to \infty } \frac{{{r^3}\left( {\frac{1}{{{r^2}}} - 1} \right)}}{{\left( {\frac{2}{{{r^3}}} - \frac{1}{r} + 3} \right){r^3}}}\\ &= \mathop {\lim }\limits_{r \to \infty } \frac{{\left( {\frac{1}{{{r^2}}} - 1} \right)}}{{\left( {\frac{2}{{{r^3}}} - \frac{1}{r} + 3} \right)}}\\ &= \frac{{\left( {\frac{1}{{{\infty ^2}}} - 1} \right)}}{{\left( {\frac{2}{{{\infty ^3}}} - \frac{1}{\infty } + 3} \right)}}\\ &= \frac{{0 - 1}}{{0 - 0 + 3}}\\ &= \frac{{ - 1}}{3}\end{aligned}\)

Thus, the value of the limit is\( - \frac{1}{3}\).