#### Answer

$\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=1/2$

#### Work Step by Step

$$g(x)=\frac{1}{2+\frac{1}{x}}$$
(a) As $x\to\infty$, $1/x$ will approach $0$.
Therefore, $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{1}{2+\frac{1}{x}}=\frac{1}{2+0}=\frac{1}{2}$$
(b) As $x\to-\infty$, $1/x$ will approach $-1/\infty$ and this approaches $0$ as well.
Therefore, $$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\frac{1}{2+\frac{1}{x}}=\frac{1}{2+0}=\frac{1}{2}$$
A graph of the function $f(x)$ is enclosed below, which shows that $f(x)$ approaches $1/2$ as $x$ approaches either $\infty$ or $-\infty$.