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

Mar 26, 2016

$\textcolor{b l u e}{y = 0 \text{ is an asymptote}}$

$\textcolor{b l u e}{x = 1 \text{ is an asymptote}}$

#### Explanation:

The basic rule is that you are 'not allowed' to divide by 0. Proper term for this is 'undefined'.

Given:$\text{ } \frac{x}{{\left(x - 1\right)}^{2}}$

Expanding the brackets

$\text{ } \frac{x}{{x}^{2} - 2 x + 1}$

$\textcolor{b l u e}{\text{Consider "x" becoming very large}}$

As $x$ becomes significantly large enough then the constants in the above equation become more and more insignificant. So there is a tendency for it to become:

$\text{ } \frac{\cancel{x}}{\cancel{x} \left(x - 2\right)}$ which again has another constant.

So as it becomes bigger still there is a tendency for it to approach $\frac{1}{x}$

${\lim}_{x \to - \infty} \frac{1}{x} = 0 {\textcolor{w h i t e}{}}^{-}$
${\lim}_{x \to + \infty} \frac{1}{x} = 0 {\textcolor{w h i t e}{}}^{+}$

Thus $\textcolor{b l u e}{y = 0 \text{ is an asymptote}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Consider the case of the denominator } = 0}$

${\left(x - 1\right)}^{2} = 0 \text{ }$ is undefined

So $\left(x - 1\right) = 0$ is undefined

Thus $\textcolor{b l u e}{x = 1 \text{ is an asymptote}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~