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

1 Answer
Oct 21, 2016

Answer:

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]}