Question

How do you find the vertical, horizontal or slant asymptotes for `x/(1-x)^2`?

Answer 1

The vertical asymptote is `x=-1`
The vertical asymptote is `y=0`
No slant asymptote

Explanation:

Let `f(x)=x/(1-x)^2`

As you cannot divide by `0`, so a vertical asymptote is `x=1`

The degree of the numerator `<` the degree of the denominator, so there is no slant asymptote.

`lim_(x->-oo)f(x)=lim_(x->-oo)x/(x^2)=lim_(x->-oo)1/x=0^(-)`

`lim_(x->+oo)f(x)=lim_(x->+oo)x/(x^2)=lim_(x->+oo)1/x=0^(+)`

So, the horizontal asymptote is `y=0`

`lim_(x->1^(-))f(x)=lim_(x->1^(-))x/(1-x)^2=+oo`

`lim_(x->1^(+))f(x)=lim_(x->1^(+))x/(1-x)^2=+oo`

graph{x/(1-x)^2 [-8.89, 8.885, -4.444, 4.44]}