Question

what would be the limit of ((1-x^3)/x^2+9x))^5 as x goes to negative infinity?

Answer 1

Is this the correct problem?
\[
\lim_{x \rightarrow -\infty}\left( \frac{1-x^3}{x^2+9x} \right )^5
\]

Answer 2

yes

Answer 3

\[
\lim_{x \rightarrow -\infty}\left( \frac{1-x^3}{x^2+9x} \right )^5 =
\lim_{x \rightarrow -\infty}\left( = \frac{x^2(1/x^2-x)}{x^2(1+9/x)} \right )^5 \\
\lim_{x \rightarrow -\infty}\left( \frac{((1/x^2-x)}{(1+9/x)} \right )^5 = \infty\\
\]

Answer 4

ohh okay thank you (:

Answer 5

you are welcome.