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

1 Answer
George C.
Dec 21, 2015

Re-express as the sum of a polynomial and a term whose limit as #x->oo# is #0#:

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

so the oblique asymptote is #y = x+1#

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 #y=x+1#

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