Answer
$\infty $
Work Step by Step
\begin{align*}
\lim _{x \rightarrow-\infty}\left(\frac{ 1-x^3}{ x^{2}+7x}\right)^{5}&=\lim _{x \rightarrow-\infty}\left(\frac{ 1/ x^{2}-x^3/ x^{2}}{ x^{2}/ x^{2}+7x/ x^{2}}\right)^{5}\\
&=\lim _{x \rightarrow-\infty}\left(\frac{ 1/ x^{2}-x }{ 1+7 / x }\right)^{5}\\
&=\left(\frac{ \lim _{x \rightarrow-\infty}(1/ x^{2})-\lim _{x \rightarrow-\infty}(x) }{\lim _{x \rightarrow-\infty}(1)+\lim _{x \rightarrow-\infty}(7 / x) }\right)^{5}\\
&=\left(\frac{0-\lim _{x \rightarrow-\infty}(x) }{1+0 }\right)^{5}\\
&=\left(\frac{ - (-\infty)}{1 }\right)^{5}\\
&=\infty
\end{align*}
