How do you find the vertical, horizontal and slant asymptotes of: #f(x)=x/(x-1)^2#?

1 Answer
Dec 18, 2016

Answer:

The vertical asymptote is #x=1#
No slant asymptote.
The horizontal asymptote is #y=0#

Explanation:

The domain of #f(x)# is #D_f(x)=RR-{1}#

As you cannot divide by #0#, #x!=1#

The vertical asymptote is #x=1#

The degree of the numerator is #<# than the degree of the denominator, there is no slant asymptote.

To calculate the limits as #x->+-oo#, we take the termsof highest degree in the numerator and the denominator

#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^(+)#

The horizontal asymptote is #y=0#

graph{x/(x-1)^2 [-10, 10, -5, 5]}