What are the asymptote(s) and hole(s), if any, of # f(x) =(1-x)^2/(x^2-1)#?

1 Answer
Nov 21, 2015

Answer:

#f(x)# has a horizontal asymptote #y=1#, a vertical asymptote #x=-1# and a hole at #x=1#.

Explanation:

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

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

with exclusion #x != 1#

As #x->+-oo# the term #2/(x+1) -> 0#, so #f(x)# has a horizontal asymptote #y = 1#.

When #x = -1# the denominator of #f(x)# is zero, but the numerator is non-zero. So #f(x)# has a vertical asymptote #x = -1#.

When #x = 1# both the numerator and denominator of #f(x)# are zero, so #f(x)# is undefined and has a hole at #x=1#. Note that #lim_(x->1) f(x) = 0# is defined. So this is a removable singularity.