# What are the asymptotes for G(x) = (x-1)/(x-x^3)?

Jun 21, 2017

You may start by factorizing the denominator:

#### Explanation:

$x - {x}^{3} = x \left(1 - {x}^{2}\right) = x \left(1 - x\right) \left(1 + x\right)$

$x = 1$ leaves us with the undefined $\frac{0}{0}$

So $x = 0 \mathmr{and} x = - 1$ are the vertical asympotes.

If we realize that $x - 1 = - \left(1 - x\right)$ we can simplify, as long as $x \ne 1$

$\frac{\cancel{x - 1}}{- x \left(\cancel{x - 1}\right) \left(x + 1\right)} = - \frac{1}{x \left(x + 1\right)} = - \frac{1}{{x}^{2} + 1}$

And this goes to zero if $x$ gets large enough (either pos or neg)

So $y = 0$ is the horizontal asymptote.
graph{(x-1)/(x-x^3) [-16.6, 15.44, -7.7, 8.33]}

Jun 21, 2017

$\text{vertical asymptotes at " x=0" and } x = - 1$
$\text{horizontal asymptote at } y = 0$

#### Explanation:

$\text{simplifying } g \left(x\right)$

$g \left(x\right) = \frac{x - 1}{x \left(1 - {x}^{2}\right)} = - \frac{\cancel{\left(x - 1\right)}}{x \cancel{\left(x - 1\right)} \left(x + 1\right)}$

$\Rightarrow g \left(x\right) = - \frac{1}{x \left(x + 1\right)}$

$\text{the removal of " (x-1)" from the numerator/denominator}$
$\text{means there is a hole at x = 1}$

$\text{the graph of " -1/(x(x+1))" is the same as } \frac{x - 1}{x - {x}^{3}}$
$\text{but without the hole}$

The denominator of g(x) cannot be zero as this would make g(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{solve } x \left(x + 1\right) = 0 \Rightarrow x = 0 , x = - 1$

$\Rightarrow x = 0 \text{ and " x=-1" are the asymptotes}$

$\text{Horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , g \left(x\right) \to c \text{ ( a constant)}$

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

$g \left(x\right) = - \frac{\frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{x}{x} ^ 2} = - \frac{\frac{1}{x} ^ 2}{1 + \frac{1}{x}}$

as $x \to \pm \infty , g \left(x\right) \to \frac{0}{1 + 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{(x-1)/(x-x^3) [-10, 10, -5, 5]}