# How do you find the vertical, horizontal or slant asymptotes for x/(1-x)^2?

Nov 13, 2016

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

#### Explanation:

Let $f \left(x\right) = \frac{x}{1 - x} ^ 2$

As you cannot divide by $0$, so a vertical asymptote is $x = 1$

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

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

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

So, the horizontal asymptote is $y = 0$

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

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

graph{x/(1-x)^2 [-8.89, 8.885, -4.444, 4.44]}