# What are the asymptote(s) and hole(s), if any, of  f(x) =(x-1)/(x^4-1) ?

Jan 18, 2018

$f \left(x\right)$ has a vertical asymptote at $x = - 1$, a hole at $x = 1$ and a horizontal asymptote $y = 0$. It has no oblique asymptotes.

#### Explanation:

$f \left(x\right) = \frac{x - 1}{{x}^{4} - 1}$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \left(x + 1\right) \left({x}^{2} + 1\right)}$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{1}{\left(x + 1\right) \left({x}^{2} + 1\right)}$

with exclusion $x \ne - 1$

Note that ${x}^{2} + 1 > 0$ for any real values of $x$

When $x = - 1$ the denominator is zero and the numerator is non-zero. So $f \left(x\right)$ has a vertical asymptote at $x = - 1$

When $x = 1$ both the numerator and denominator of the defining expression for $f \left(x\right)$ are zero, but the simplified expression is well defined and continuous at $x = 1$. So there is a hole at $x = 1$.

As $x \to \pm \infty$ the denominator of the simplified expression $\to \infty$, while the numerator is constant $1$. Hence the function tends to $0$ and has a horizontal asymptote $y = 0$

$f \left(x\right)$ has no oblique (a.k.a. slant) asymptotes. In order for a rational function to have an oblique asymptote, the numerator must have degree exactly one more than the denominator.

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