How do you find the asymptotes for #f(x)=(x^3+1)/(x^3+x)#?

1 Answer
Mar 5, 2016

Answer:

Reformulate #f(x)# and analyse to find that it has a horizontal asymptote #y=1# and vertical asymptote #x=0#

Explanation:

#f(x) = (x^3+1)/(x^3+x) = (x^3+x+1-x)/(x^3+x) = 1+(1-x)/(x^3+x) = 1+(1-x)/(x(x^2+1))#

As #x->+-oo# the term #(1-x)/(x(x^2+1)) -> 0#, so #f(x)->1#

So there is a horizontal asymptote #y=1#

When #x=0# then denominator of #f(x)# is zero but the numerator is non-zero. So #f(x)# has a vertical asymptote at #x=0#

Note that #x^2 >= 0# for any Real value of #x#, so #x^2+1# is always non-zero, resulting in no more vertical asymptotes.

graph{(x^3+1)/(x^3+x) [-10, 10, -5, 5]}