# How do you find all horizontal and vertical asymptotes of f(x)=arctan x- 1/(x-1)?

Feb 12, 2017

Curvilinear asymptote : $y = a r c \tan x$.
Vertical asymptote: $\uparrow x = 1 \downarrow$.
Horizontal asymptotes: $y = \pm \frac{\pi}{2.}$

#### Explanation:

The form

$y = Q \left(x\right) + \frac{R \left(x\right)}{P \left(x\right)}$, reveals asymptotes

$y = Q \left(x\right) = a r c \tan x \mathmr{and} P \left(x\right) = x - 1 = 0$.

The first is a curvilinear asymptotes that has its outer asymptotes

$y = \pm \frac{\pi}{2}$.

See below the grandeur of the clustering, on either side of x = 1,

when general values are allowed to arc tan x. It is indeed marching

to $\infty$, in the opposite directions of the common-to-all asymptote

x = 1.

Here, the horizontal asymptotes are  y = (2 k + 1 ) pi/2, k = 0, +-1,

+-2 +-3, ....

These are the asymptotes of the curvilinear asymptotes

$y = k \pi + a r c \tan x , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$

graph{(x - tan (y + 1 / ( x - 1 )))(x - 1) = 0[-2 3 -5 5]}