How do you find the asymptotes of #y=sqrt(x^2+x+1) - sqrt(x^2-x)#?
1 Answer
There are 2 horizontal asymptotes. On the right
Explanation:
# = (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
As
# = 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
# = 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#