# How do you find the vertical, horizontal or slant asymptotes for  x/sqrt(4x^2-1)?

Jul 2, 2016

vertical at $x = \pm \frac{1}{2}$

horizontal asymptotes are ${\lim}_{x \to \pm \infty} f \left(x\right) = \pm \frac{1}{2}$

#### Explanation:

$\frac{x}{\sqrt{4 {x}^{2} - 1}}$

for vertical we look at when the numerator is zero

ie $\sqrt{4 {x}^{2} - 1} = 0$
$4 {x}^{2} = 1$
$x = \pm \frac{1}{2}$

because $\sqrt{4 {x}^{2} - 1} > 0 \forall x \notin \left(- \frac{1}{2} , \frac{1}{2}\right)$

the limit follows the sign of the numerator so

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

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

for horizontal asymptote we look at

${\lim}_{x \to \pm \infty} \frac{x}{\sqrt{4 {x}^{2} - 1}}$

${\lim}_{x \to \pm \infty} s g n \left(x\right) \frac{1}{\sqrt{4 - \frac{1}{x} ^ 2}}$

${\lim}_{x \to \pm \infty} \frac{1}{2} s g n \left(x\right) = \pm \frac{1}{2}$