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

1 Answer
Nov 13, 2016

Answer:

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]}