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

Question

1083 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-02T14:27:36

vertical at #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#