Answer
(a) $ \infty $
(b) $ \infty $
Work Step by Step
(a) \begin{align*}
\lim_{x\to \infty}g(x)&= \lim_{x\to \infty} \frac{3x^7+5x^2-1}{6x^3-7x+3}\\
&= \lim_{x\to \infty} \frac{3x^7/x^3+5x^2/x^3-1/x^3}{6x^3/x^3-7x/x^3+3/x^3}\\
&= \frac{ \lim_{x\to \infty}(3x^5)+ \lim_{x\to \infty}(5/x)- \lim_{x\to \infty}(1/x^3)}{ \lim_{x\to \infty}(6)- \lim_{x\to \infty}(7/x^2)+ \lim_{x\to \infty}(3/x^3)}\\
&= \frac{ \lim_{x\to \infty}(3x^5) }{ 6 }\\
&= \infty
\end{align*}
(b) \begin{align*}
\lim_{x\to -\infty}g(x)&= \lim_{x\to -\infty} \frac{3x^7+5x^2-1}{6x^3-7x+3}\\
&= \lim_{x\to -\infty} \frac{3x^7/x^3+5x^2/x^3-1/x^3}{6x^3/x^3-7x/x^3+3/x^3}\\
&= \frac{ \lim_{x\to- \infty}(3x^5)+ \lim_{x\to -\infty}(5/x)- \lim_{x\to -\infty}(1/x^3)}{ \lim_{x\to -\infty}(6)- \lim_{x\to -\infty}(7/x^2)+ \lim_{x\to- \infty}(3/x^3)}\\
&= \frac{ \lim_{x\to- \infty}(3x^5) }{ 6 }\\
&= \infty
\end{align*}
