How do you find the horizontal asymptote for $b(x)= (x^3-x+1)/(2x^4+x^3-x^2-1)$ ?

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Answer 1 · 2016-01-06T14:13:49

The secret is factorization :

$b(x)= (x^3-x+1)/(2x^4+x^3-x^2-1) =(x^3(1-1/x^2+1/x^3))/(x^4(2+1/x-1/x^2-1/x^4))$

$b(x)=(1-1/x^2+1/x^3)/(x(2+1/x-1/x^2-1/x^4)$

$lim_(x->+oo) (1-1/x^2+1/x^3)/(x(2+1/x-1/x^2-1/x^4)) = 1/(oo) = 0^+$

$lim_(x->-oo) (1-1/x^2+1/x^3)/(x(2+1/x-1/x^2-1/x^4)) = -1/(oo) = 0^-$

How do you find the horizontal asymptote for b(x)= (x^3-x+1)/(2x^4+x^3-x^2-1)b(x)= (x^3-x+1)/(2x^4+x^3-x^2-1)?