How do you find the horizontal asymptote for #b(x)= (x^3-x+1)/(2x^4+x^3-x^2-1)#? Precalculus Functions Defined and Notation Asymptotes 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^-# Answer link Related questions What is an asymptote? What are some examples of functions with asymptotes? How do asymptotes relate to boundedness? Why do some functions have asymptotes? What are the vertical asymptotes of #f(x) = (2)/(x^2 - 1)#? Is the x-axis an asymptote of #f(x) = x^2#? Where are the vertical asymptotes of #f(x) = tan x#? Where are the vertical asymptotes of #f(x) = cot x#? How do I find the vertical asymptotes of #f(x)=tan2x#? How do I find the vertical asymptotes of #f(x) = tanπx#? See all questions in Asymptotes Impact of this question 1394 views around the world You can reuse this answer Creative Commons License