Answer
$f(x)=2\tan^{-1} x+\pi-2$
$g(x)=2\tan^{-1} (-x)+\pi-2$
Work Step by Step
Let's consider the functions $f(x)$ and $g(x)$:
$f(x)=2\tan^{-1} x+\pi-2$
$g(x)=2\tan^{-1} (-x)+\pi-2$
We have:
$\displaystyle\lim_{x\rightarrow-\infty} f(x)=-2$
$\displaystyle\lim_{x\rightarrow\infty} f(x)=4$
$\displaystyle\lim_{x\rightarrow -\infty} g(x)=4$
$\displaystyle\lim_{x\rightarrow\infty} g(x)=-2$
Therefore:
$\displaystyle\lim_{x\rightarrow \infty} f(x)\not=\displaystyle\lim_{x\rightarrow \infty} g(x)$
