Answer
$$\lim_{x\to\infty}g(x)=\lim_{x\to-\infty}g(x)=\frac{1}{8}$$
Work Step by Step
$$g(x)=\frac{1}{8-\frac{5}{x^2}}$$
(a) As $x\to\infty$, $x^2$ also approaches $\infty$ and $5/x^2$ will approach $0$.
Therefore, $$\lim_{x\to\infty}g(x)=\lim_{x\to\infty}\frac{1}{8-\frac{5}{x^2}}=\frac{1}{8-0}=\frac{1}{8}$$
(b) As $x\to-\infty$, $x^2$ will approach $\infty$ and $5/x^2$ approaches $0$ as a result.
Therefore, $$\lim_{x\to-\infty}g(x)=\lim_{x\to-\infty}\frac{1}{8-\frac{5}{x^2}}=\frac{1}{8-0}=\frac{1}{8}$$
A graph of the function $g(x)$ is enclosed below, which shows that $g(x)$ approaches $1/8$ as $x$ approaches either $\infty$ or $-\infty$.
