How do you find vertical, horizontal and oblique asymptotes for (2x^3+4x-8)/(x^3-6x)?

Dec 1, 2016

The vertical asymptotes are $x = 0$, $x = \sqrt{6}$ and $x = - \sqrt{6}$
No oblique asymptote
the horizontal asymptote is $y = 2$

Explanation:

Let $f \left(x\right) = \frac{2 {x}^{3} + 4 x - 8}{{x}^{3} - 6 x}$

The denominator is

$= {x}^{3} - 6 x = x \left(x - \sqrt{6}\right) \left(x + \sqrt{6}\right)$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{0 , \pm \sqrt{6}\right\}$

As you cannot divide by $0$, $x \ne 0$, $x \ne \sqrt{6}$ and $x \ne - \sqrt{6}$

So the vertical asymptotes are $x = 0$, $x = \sqrt{6}$ and $x = - \sqrt{6}$

As the degree of the numerator $=$ the degree of the denominator, there is no oblique asymptote.

To calculate the limits as $x \to \pm \infty$, we take the terms of highest degree in the numerator and the denominator.

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} \frac{2 {x}^{3}}{x} ^ 3 = 2$

So, the horizontal asymptote is $y = 2$

graph{(2x^3+48-8)/(x^3-6x) [-28.86, 28.87, -14.43, 14.45]}