How do you find the asymptotes of #y=sqrt(x^2+x+1) - sqrt(x^2-x)#?

1 Answer
Jim H
Jul 8, 2017

There are 2 horizontal asymptotes. On the right #y=1# is an asymptote and on the left #y=-1# is an asymptote.

Explanation:

#y# never becomes infinite, so there is no vertical asymptote.

#((sqrt(x^2+x+1) - sqrt(x^2-x)))/1 * ((sqrt(x^2+x+1) + sqrt(x^2-x)))/((sqrt(x^2+x+1) + sqrt(x^2-x))) = ((x^2+x+1)-(x^2-x))/(sqrt(x^2+x+1) + sqrt(x^2-x))#

# = (2x+1)/(sqrt(x^2+x+1) + sqrt(x^2-x))#

# = (x(2+1/x))/(sqrt(x^2)(sqrt(1+1/x+1/x^2)+sqrt(1-1/x))#

Recall that #sqrt(x^2) = absx#.

As #xrarroo#, #x# is positive so we have

#lim_(xrarroo)(x(2+1/x))/(sqrt(x^2)(sqrt(1+1/x+1/x^2)+sqrt(1-1/x))) = lim_(xrarroo)(x(2+1/x))/(x(sqrt(1+1/x+1/x^2)+sqrt(1-1/x))#

# = lim_(xrarroo)(2+1/x)/(sqrt(1+1/x+1/x^2)+sqrt(1-1/x))#

# = (2+0)/(sqrt(1+0+0)+sqrt(x-0)) = 2/2 = 1#

As #xrarr-oo#, #x# is negative so we have

#lim_(xrarr-oo)(x(2+1/x))/(sqrt(x^2)(sqrt(1+1/x+1/x^2)+sqrt(1-1/x))) = lim_(xrarr-oo)(x(2+1/x))/((-x)(sqrt(1+1/x+1/x^2)+sqrt(1-1/x))#

# = lim_(xrarroo)-(2+1/x)/(sqrt(1+1/x+1/x^2)+sqrt(1-1/x))#

# = -(2+0)/(sqrt(1+0+0)+sqrt(x-0)) = -2/2 = -1#

Related questions
See all questions in Infinite Limits and Vertical Asymptotes
Impact of this question
7497 views around the world
You can reuse this answer
Creative Commons License