Question

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

Answer 1

The vertical asymptotes are `x=1/2` and `x=-1/2`
And the horizontal asymptotes are `y=1/2` and `y=-1/2`

Explanation:

Let `f(x)=x/sqrt(4x^2-1)`
To look for vertical asymptotes, we look at the denominator `!=0`
`4x^2-1>=0` so `x^2>=1/4`
and `x>=+-1/2`
we remove the `=+-1/2` as we cannot divide by 0
so the vertical asymptotes are `x=1/2` and `x=-1/2`
For horizontal asymptotes we find the limit `x->+-oo`
Rearrange `f(x)=1/(sqrt(4x^2/x^2-1/x^2)`
`=1/(sqrt(4-1/x^2)`
And the limit as `x->oo` is `=1/sqrt4=+-1/2`
so horizontal asymptotes are `y=1/2` and `y=-1/2`
You can all this on the graph

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