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

Jan 9, 2017

The vertical asymptote is $x = 0$
No oblique asymptote.
The horizontal asymptote is $y = 0$

#### Explanation:

The domain of $g \left(x\right)$ is ${D}_{g} \left(x\right) = \mathbb{R} - \left\{0\right\}$

As we cannot divide by $0$, $x \ne 0$

The vertical asymptote is $x = 0$

The degree of the numerator is $<$ than the degree of the denominator, so there is no oblique asymptote.

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

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

The horizontal asymptote is $y = 0$

graph{(y-(x+2)/(2x^2))(y)(y-100x)=0 [-7.02, 7.024, -3.51, 3.51]}