Question

How do you find the asymptotes for `f(x) = x / sqrt(x^2 + 1)`?

Answer 1

We can rewrite this as follows

`f(x)=(x)/[absxsqrt(1+1/x^2)]`

Hence for `x->+oo` we have that `f(x)->1`

and

for `x->-oo` we have that `f(x)->-1`

Hence the horizontal asympotes are `y=1` and `y=-1`