How do you find the slant asymptote of (x^2)/(x-1)x2x−1?

1 Answer
George C.
Dec 21, 2015

Re-express as the sum of a polynomial and a term whose limit as x->oox→∞ is 00:

x^2/(x-1) = x+1+1/(x-1)x2x−1=x+1+1x−1

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

Explanation:

x^2/(x-1) = (x^2-x+x)/(x-1) = (x(x-1)+x)/(x-1) = x + x/(x-1)x2x−1=x2−x+xx−1=x(x−1)+xx−1=x+xx−1

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

Note that:

1/(x-1)->01x−1→0 as x->oox→∞,

So the oblique asymptote is y=x+1y=x+1

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