How do you find the horizontal asymptote #y = (1-x^2)/(x-1)#?

Question

2097 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-18T17:25:20

Because the numerator has degree greater than that of the denominator, there is no horizontal asymptote.

#y = (1-x^2)/(x-1)#

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

# = (x(-x+1/x))/(x(1-1/x))#

# = (-x+1/x)/(1-1/x)#

As #x# increases without bound, #y# decreases without bound
and
as #x# decreases without bound, #y# increases without bound.

The is no number #k#, and no line #y=k# that #y# is getting close to.

Answer 2 · 2015-09-18T22:24:28

#y = (1-x^2)/(x-1) = -x-1# with exclusion #x != 1#

This is a line of slope #-1#. It has no asymptotes.

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

#= -(x+1) = -x-1#

with exclusion #x != 1#

So this is a line of slope #-1# with an excluded point.

It has no asymptotes.