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