Question

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

Answer 1

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]}