What are the asymptote(s) and hole(s), if any, of # f(x) =(x-1)/(x^4-1) #?

1 Answer
Jan 18, 2018

Answer:

#f(x)# has a vertical asymptote at #x=-1#, a hole at #x=1# and a horizontal asymptote #y=0#. It has no oblique asymptotes.

Explanation:

#f(x) = (x-1)/(x^4-1)#

#color(white)(f(x)) = color(red)(cancel(color(black)((x-1))))/(color(red)(cancel(color(black)((x-1))))(x+1)(x^2+1))#

#color(white)(f(x)) = 1/((x+1)(x^2+1))#

with exclusion #x!=-1#

Note that #x^2+1 > 0# for any real values of #x#

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

When #x=1# both the numerator and denominator of the defining expression for #f(x)# are zero, but the simplified expression is well defined and continuous at #x=1#. So there is a hole at #x=1#.

As #x->+-oo# the denominator of the simplified expression #->oo#, while the numerator is constant #1#. Hence the function tends to #0# and has a horizontal asymptote #y=0#

#f(x)# has no oblique (a.k.a. slant) asymptotes. In order for a rational function to have an oblique asymptote, the numerator must have degree exactly one more than the denominator.

graph{1/((x+1)(x^2+1)) [-10, 10, -5, 5]}