How do you find the vertical, horizontal or slant asymptotes for $\frac{x}{\sqrt{4 x^{2} - 1}}$ $x/sqrt(4x^2-1)$ ?

Question

1225 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-02T14:27:36

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

horizontal asymptotes are $lim_{x to pm oo} f(x) = pm 1/2$

$x/sqrt(4x^2-1)$

for vertical we look at when the numerator is zero

ie $sqrt(4x^2-1) = 0$
$4x^2=1$
$x = pm 1/2$

because $sqrt(4x^2-1) >0 forall x notin (- 1/2,1/2)$

the limit follows the sign of the numerator so

$lim_{x to -(1/2) ^ -} = - oo$

$lim_{x to (1/2) ^+} = + oo$

for horizontal asymptote we look at

$lim_{x to pm oo} x/sqrt(4x^2-1)$

$lim_{x to pm oo} sgn(x) 1/sqrt(4-1/x^2)$

$lim_{x to pm oo} 1/2 sgn(x) = pm 1/2$

How do you find the vertical, horizontal or slant asymptotes for x/sqrt(4x^2-1)x√4x2−1 x/sqrt(4x^2-1)?