Question

find the limit. (if the limit is infinite, enter ∞ or -∞, as approplim_{r→∞} \frac{r - r^{3}}{4 - r^{2} + 8r^{3}}-\frac{1}{8}well done!

Explanation:

Step1: Divide numerator and denominator by \(r^{3}\)

$$ \lim_{r ightarrow\infty}\frac{\frac{r}{r^{3}}-\frac{r^{3}}{r^{3}}}{\frac{4}{r^{3}}-\frac{r^{2}}{r^{3}}+\frac{8r^{3}}{r^{3}}}=\lim_{r ightarrow\infty}\frac{\frac{1}{r^{2}} - 1}{\frac{4}{r^{3}}-\frac{1}{r}+8} $$

Step2: Apply the limit

As \(r
ightarrow\infty\), \(\lim_{r
ightarrow\infty}\frac{1}{r^{n}} = 0\) for \(n>0\).
So, \(\lim_{r
ightarrow\infty}\frac{\frac{1}{r^{2}} - 1}{\frac{4}{r^{3}}-\frac{1}{r}+8}=\frac{0 - 1}{0-0 + 8}\)

Answer:

\(-\frac{1}{8}\)