# How do you find the asymptotes for f(x)=(x^3+1)/(x^3+x)?

Mar 5, 2016

Reformulate $f \left(x\right)$ and analyse to find that it has a horizontal asymptote $y = 1$ and vertical asymptote $x = 0$

#### Explanation:

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

As $x \to \pm \infty$ the term $\frac{1 - x}{x \left({x}^{2} + 1\right)} \to 0$, so $f \left(x\right) \to 1$

So there is a horizontal asymptote $y = 1$

When $x = 0$ then denominator of $f \left(x\right)$ is zero but the numerator is non-zero. So $f \left(x\right)$ has a vertical asymptote at $x = 0$

Note that ${x}^{2} \ge 0$ for any Real value of $x$, so ${x}^{2} + 1$ is always non-zero, resulting in no more vertical asymptotes.

graph{(x^3+1)/(x^3+x) [-10, 10, -5, 5]}