# How do you find the asymptotes for g(x)= (x+2 )/( 2x^2)?

Nov 20, 2016

The vertical asymptote is $x = 0$
No slant asymptote
The horizontal asymptote is $y = 0$

#### Explanation:

As you cannot divide by $0$
Domain, ${D}_{g} \left(x\right) = \mathbb{R} - \left\{O\right\}$

$x \ne 0$
Therefore, $x = 0$ is a vertical asymptote

The degree of the numerator is $<$ the degree of the denominator.
So, we don't have a slant asymptote.

For the limits of $y$, we take the terms of highest coefficients

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

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

Therefore, $y = 0$ is a horizontal asymptote

graph{(x+2)/(2x^2) [-7.02, 7.03, -3.51, 3.51]}