# How do you find the slant asymptote of #(x^2)/(x-1)#?

##### 1 Answer

Dec 21, 2015

#### Answer:

Re-express as the sum of a polynomial and a term whose limit as

#x^2/(x-1) = x+1+1/(x-1)#

so the oblique asymptote is

#### Explanation:

#x^2/(x-1) = (x^2-x+x)/(x-1) = (x(x-1)+x)/(x-1) = x + x/(x-1)#

#= x + (x-1+1)/(x-1) = x+1+1/(x-1)#

Note that:

#1/(x-1)->0# as#x->oo# ,

So the oblique asymptote is