Answer
$$\eqalign{
& {\text{Vertical asymptote}}:{\text{ }}x = 0 \cr
& {\text{Horizontal asymptotes: }}y = - \frac{2}{\pi }{\text{ and }}y = \frac{2}{\pi } \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{{{\tan }^{ - 1}}x}} \cr
& {\text{*To calculate the vertical asymptotes we need to find the values }} \cr
& {\text{at which the function is undefined}}{\text{. Set the denominator to 0}} \cr
& {\tan ^{ - 1}}x = 0 \cr
& x = 0 \cr
& {\text{Then}}{\text{, we can conclude that }}x = 0{\text{ is a vertical asymptote}}{\text{.}} \cr
& \cr
& {\text{*Evaluate }}\mathop {\lim }\limits_{x \to - \infty } f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to \infty } f\left( x \right) \cr
& \mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{1}{{{{\tan }^{ - 1}}x}} \cr
& = \frac{1}{{\mathop {\lim }\limits_{x \to - \infty } \left[ {{{\tan }^{ - 1}}x} \right]}} \cr
& = \frac{1}{{ - \pi /2}} \cr
& = - \frac{2}{\pi } \cr
& and \cr
& \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{{{\tan }^{ - 1}}x}} \cr
& = \frac{1}{{\mathop {\lim }\limits_{x \to \infty } \left[ {{{\tan }^{ - 1}}x} \right]}} \cr
& = \frac{1}{{\pi /2}} \cr
& = \frac{2}{\pi } \cr
& {\text{Then}}{\text{, we can conclude that }}y = - \frac{2}{\pi }{\text{ and }}y = \frac{2}{\pi }{\text{ are}} \cr
& {\text{horizontal asymptotes}}{\text{.}} \cr
& \cr
& {\text{Summary:}} \cr
& {\text{Vertical asymptote}}:{\text{ }}x = 0 \cr
& {\text{Horizontal asymptotes: }}y = - \frac{2}{\pi }{\text{ and }}y = \frac{2}{\pi } \cr} $$
