# How do you find horizontal asymptotes for f(x) = arctan(x) ?

Jul 24, 2014

By definition, $\arctan x$ is the inverse function of the restriction of the tangent function $\tan$ to the interval $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ (see inverse cosine and inverse tangent ).

The tangent function has vertical asymptotes $x = - \frac{\pi}{2}$ and $x = \frac{\pi}{2}$, for $\tan x = \sin \frac{x}{\cos} x$ and $\cos \setminus \pm \frac{\pi}{2} = 0$.

Moreover, the graph of the inverse function ${f}^{- 1}$ of a one-to-one function $f$ is obtained from the graph of $f$ by reflection about the line $y = x$ (see finding inverse functions ), which transforms vertical lines into horizontal lines.

Thus, the vertical asymptotes $x = \setminus \pm \frac{\pi}{2}$ for $y = \tan x$ correspond in this reflection to the horizontal asymptotes $y = \setminus \pm \frac{\pi}{2}$ for $y = \arctan x$.

Here's a graph of arctan(x):