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

Question

2112 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-04T21:05:38

Re-express as the sum of a linear polynomial and a rational expression that tends to zero as $x->+-oo$ to find slant asymptote:

$y = x-1$

$x^2/(x+1)$

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

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

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

Then $1/(x+1) -> 0$ as $x->+-oo$

So the slant asymptote (a.k.a. oblique asymptote) is $y = x-1$

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