# How do you find the vertical, horizontal and slant asymptotes of: f(x)=x/(x-1)^2?

Dec 18, 2016

The vertical asymptote is $x = 1$
No slant asymptote.
The horizontal asymptote is $y = 0$

#### Explanation:

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{1\right\}$

As you cannot divide by $0$, $x \ne 1$

The vertical asymptote is $x = 1$

The degree of the numerator is $<$ than the degree of the denominator, there is no slant asymptote.

To calculate the limits as $x \to \pm \infty$, we take the termsof highest degree in the numerator and the denominator

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{x}{x} ^ 2 = {\lim}_{x \to - \infty} \frac{1}{x} = {0}^{-}$

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} \frac{x}{x} ^ 2 = {\lim}_{x \to + \infty} \frac{1}{x} = {0}^{+}$

The horizontal asymptote is $y = 0$

graph{x/(x-1)^2 [-10, 10, -5, 5]}