Answer
(a) $2$
(b) $2$
Work Step by Step
(a) Since
\begin{align*}
\lim_{x\to \infty } f(x)&=\lim_{x\to \infty } \frac{2x^3+7}{x^3-x^2+x+7}\\
&=\lim_{x\to \infty }\frac{2x^3/x^3+7/x^3}{x^3/x^3-x^2/x^3+x/x^3+7/x^3}\\
&=\lim_{x\to \infty }\frac{2 +7/x^3}{1-1/x +1/x +7/x^3}\\
&=\frac{\lim_{x\to \infty }(2)+\lim_{x\to \infty }(7/x^3)}{\lim_{x\to \infty }(1)-\lim_{x\to \infty }(1/x) +\lim_{x\to \infty }(1/x) +\lim_{x\to \infty }(7/x^3)}\\
&=\frac{2+0}{1-0+0+0}=2
\end{align*}
(b) Since
\begin{align*}
\lim_{x\to -\infty } f(x)&=\lim_{x\to -\infty } \frac{2x^3+7}{x^3-x^2+x+7}\\
&=\lim_{x\to- \infty }\frac{2x^3/x^3+7/x^3}{x^3/x^3-x^2/x^3+x/x^3+7/x^3}\\
&=\lim_{x\to- \infty }\frac{2 +7/x^3}{1-1/x +1/x +7/x^3}\\
&=\frac{\lim_{x\to- \infty }(2)+\lim_{x\to -\infty }(7/x^3)}{\lim_{x\to- \infty }(1)-\lim_{x\to -\infty }(1/x) +\lim_{x\to- \infty }(1/x) +\lim_{x\to -\infty }(7/x^3)}\\
&=\frac{2+0}{1-0+0+0}=2
\end{align*}
