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

1 Answer
Tom
Jan 6, 2016

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^-#

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