# How do you find the Vertical, Horizontal, and Oblique Asymptote given g(x)=(2+x)/(x^2(5-x))?

Nov 5, 2016

The vertical asymptotes are $x = 0$ and $x = 5$
The horizontal asymptote is $y = 0$

#### Explanation:

As we cannot divide by $0$, so $x \ne 0$ and $x \ne 5$
$\therefore x = 0$ and $x = 5$ are vertical asymptotes
Rewrite the expression
$g \left(x\right) = \frac{\frac{2}{x} + 1}{5 x - {x}^{2}} = \frac{\frac{2}{x} + 1}{{x}^{2} \left(\frac{5}{x} - 1\right)}$

limit $g \left(x\right) = 0$
$x \to - \infty$

limit $g \left(x\right) = {0}^{-}$
$x \to + \infty$