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

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!

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!

Answer

Explanation:

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

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

Step2: Apply the limit

As (r\rightarrow\infty), (\lim_{r\rightarrow\infty}\frac{1}{r^{n}} = 0) for (n>0). So, (\lim_{r\rightarrow\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})