# How do you find the asymptotes for f(x) = x / (3x(x-1))?

Jan 4, 2016

Vertical: $x = 1$
Horizontal: $y = 0$

#### Explanation:

First, notice that the $x$ terms will cancel, leaving the function as

$f \left(x\right) = \frac{1}{3 \left(x - 1\right)} , \textcolor{w h i t e}{\times \times} x \ne 0$

However, canceling the $x$ terms leaves a hole, or a removable discontinuity, at $x = 0$.

Vertical asymptotes:

Vertical asymptotes will occur when the denominator equals $0$.

$3 \left(x - 1\right) = 0$

Solved, this gives

$x = 1$

Thus the vertical asymptote occurs at $x = 1$.

Even though $3 x$ is in the denominator in the original function, its cancellation makes it just a hole and not also a vertical asymptote.

Horizontal asymptotes:

Since the degree of the denominator is larger than the degree of the numerator, the horizontal asymptote is the line $y = 0$.

graph{x/(3x(x-1)) [-10, 10, -5, 5]}

Don't be fooled—there is a hole at $x = 0$, despite appearances.