# How do you find vertical, horizontal and oblique asymptotes for ( x + 5) /( 6 - x^3)?

Feb 9, 2017

The vertical asymptote is $x = \sqrt[3]{6}$
The horizontal asymptote is $y = 0$
No oblique asymptote

#### Explanation:

Let $f \left(x\right) = \frac{x + 5}{6 - {x}^{3}}$

As you cannot divide by $0$, $x \ne \sqrt[3]{6}$

The vertical asymptote is $x = \sqrt[3]{6}$

As the degree of the numerator is $<$ than the degree of the denominator, there is no oblique asymptote

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{x}{- {x}^{3}} = {\lim}_{x \to - \infty} - \frac{1}{x} ^ 2 = {0}^{-}$

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} \frac{x}{- {x}^{3}} = {\lim}_{x \to + \infty} - \frac{1}{x} ^ 2 = {0}^{-}$

The horizontal asymptote is $y = 0$
graph{(y-(x+5)/(6-x^3))(y)=0 [-20.27, 20.27, -10.14, 10.14]}