What are the asymptotes and removable discontinuities, if any, of #f(x)= (x^2-1)/(x-2)^2#?

1 Answer
Dec 25, 2017

Horizontal asymptote: #y=1#
Vertical asymptote: #x=2#

Explanation:

Domain:
#x-2!=0#
#D(f)=RR-{2}#
That means there may be vertical asymptote. We have to find one-sided limit in the point 2

#Lim_(xrarr2^+)(x^2-1)/(x-2)^2=3/0^+=+oo#

#Lim_(xrarr2^-)(x^2-1)/(x-2)^2=3/0^(-)=-oo#

Definition says: If the limit in the point which is not included in the domain has a finite value then it's not an asymptote
That means: We didnt't find finite value so there is an asymptote.

Vertical asymptote: #x=2#

Horizontal asymptote: in #+oo#
You have to find a limit of f(x). If you find a finite value then it's an asymptote. Some look like this: #y=k#, this is asymptote that is parallel to x axis. There is also asymptote in the form: #y=kx+q# (it's line)

#Lim_(xrarroo)(x^2-1)/(x-2)^2=Lim_(xrarroo)(x^2-1)/(x^2-4x+2)=#

#Lim_(xrarroo)(cancel(x^2)(1-1/x^2))/(cancel(x^2)(1-4/x+2/x^2))=1/1=1#
#y=1#

Horizontal asymptote: in #-oo#
It's the same: #y=1#