# How do you find the asymptotes for x/sqrt(4x^2-1)?

Oct 21, 2016

The vertical asymptotes are $x = \frac{1}{2}$ and $x = - \frac{1}{2}$
And the horizontal asymptotes are $y = \frac{1}{2}$ and $y = - \frac{1}{2}$

#### Explanation:

Let $f \left(x\right) = \frac{x}{\sqrt{4 {x}^{2} - 1}}$
To look for vertical asymptotes, we look at the denominator $\ne 0$
$4 {x}^{2} - 1 \ge 0$ so ${x}^{2} \ge \frac{1}{4}$
and $x \ge \pm \frac{1}{2}$
we remove the $= \pm \frac{1}{2}$ as we cannot divide by 0
so the vertical asymptotes are $x = \frac{1}{2}$ and $x = - \frac{1}{2}$
For horizontal asymptotes we find the limit $x \to \pm \infty$
Rearrange f(x)=1/(sqrt(4x^2/x^2-1/x^2)
=1/(sqrt(4-1/x^2)
And the limit as $x \to \infty$ is $= \frac{1}{\sqrt{4}} = \pm \frac{1}{2}$
so horizontal asymptotes are $y = \frac{1}{2}$ and $y = - \frac{1}{2}$
You can all this on the graph

graph{x/sqrt(4*x^2-1) [-2.5, 2.5, -1.25, 1.25]}