# What are the asymptote(s) and hole(s), if any, of  f(x) =x/(x^3-x)?

Mar 30, 2018

Holes 0
Vertical Asymptotes $\pm 1$
Horizontal Asymptotes 0

#### Explanation:

An vertical asymptote or a hole is created by a point in which the domain is equal to zero i.e. ${x}^{3} - x = 0$

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

So either $x = 0$ or ${x}^{2} - 1 = 0$

${x}^{2} - 1 = 0$ therefore $x = \pm 1$

An horizontal asymptote is created where the top and the bottom of the fraction don't cancel out. Whilst a hole is when you can cancel out.

So $\frac{\textcolor{red}{x}}{\textcolor{red}{x} \left({x}^{2} - 1\right)} = \frac{1}{{x}^{2} - 1}$

So as the $x$ crosses out 0 is merely a hole. Whilst as the ${x}^{2} - 1$ remains $\pm 1$ are asymptotes

For horizontal asymptotes one is trying to find what happens as x approaches infinity or negative infinity and whether it tends to a particular y value.

To do this divide both the numerator and denominator of the fraction by the highest power of $x$ in the denominator

$\lim x \to \infty \frac{\frac{x}{{x}^{3}}}{{x}^{3} / {x}^{3} - \frac{x}{x} ^ 3} = \lim x \to \infty \frac{\frac{1}{{x}^{2}}}{1 - \frac{1}{x} ^ 2} = \frac{\frac{1}{{\infty}^{2}}}{1 - \frac{1}{\infty} ^ 2} = \frac{0}{1 - 0} = \frac{0}{1} = 0$

To do this we have to know two rules

$\lim x \to \infty {x}^{2} = \infty$
and
$\lim x \to \infty \frac{1}{x} ^ n = \frac{1}{\infty} = 0 \mathmr{if} n > 0$

For limits to negative infinty we have to make all the $x$ into $- x$

$\lim x \to \infty = - \frac{x}{- {x}^{3} + x} = \frac{- \frac{x}{{x}^{3}}}{- {x}^{3} / {x}^{3} + \frac{x}{x} ^ 3} = \lim x \to \infty \frac{- \frac{1}{{x}^{2}}}{- 1 + \frac{1}{x} ^ 2} = \frac{- \frac{1}{{\infty}^{2}}}{- 1 + \frac{1}{\infty} ^ 2} = \frac{0}{- 1 + 0} = \frac{0}{-} 1 = 0$

So the horizontal asymptote as x approaches $\pm \infty$ is 0