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

##### 1 Answer

#### Answer:

Vertical:

Horizontal:

#### Explanation:

First, notice that the

#f(x)=1/(3(x-1)),color(white)(xxxx)x!=0#

However, canceling the **removable discontinuity**, at

**Vertical asymptotes:**

Vertical asymptotes will occur when the denominator equals

#3(x-1)=0#

Solved, this gives

#x=1#

Thus the vertical asymptote occurs at

Even though

**Horizontal asymptotes:**

Since the degree of the denominator is larger than the degree of the numerator, the horizontal asymptote is the line

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

Don't be fooled—there is a hole at