Answer
(a)$ \frac{1}{8} $
(b)$ \frac{1}{8} $
Work Step by Step
(a)\begin{align*}
\lim _{x \rightarrow \infty} \frac{1}{8-\left(5 / x^{2}\right)}&=\frac{1}{8-\lim _{x \rightarrow \infty}\left(5 / x^{2}\right)}=\frac{1}{8-0}\\
&=\frac{\left[\frac{1}{8}\right]}{8-0}=\frac{\left[\frac{1}{8}\right]}{8}\\
&= \frac{1}{8}
\end{align*}
(b)\begin{align*}
\lim _{x \rightarrow-\infty} \frac{1}{8-\left(5 / x^{2}\right)}&=\frac{1}{8-\lim _{x \rightarrow-\infty}\left(5 / x^{2}\right)}\\
&=\frac{1}{8-0}\\
&= \frac{1}{8}
\end{align*}