How do you find #lim (x^2-x+4)/(3x^2+2x-3)# as #x->oo#? Calculus Limits Limits at Infinity and Horizontal Asymptotes 1 Answer bp Mar 19, 2017 #1/3# Explanation: Divide numerator and the denominator by #x^2#. Accordingly, it is #(1-1/x +4/x^2)/(3+2/x -3/x^2)#. Now applying the limit #x-> oo#, it becomes = #1/3# Answer link Related questions What kind of functions have horizontal asymptotes? How do you find horizontal asymptotes for #f(x) = arctan(x)# ? How do you find the horizontal asymptote of a curve? How do you find the horizontal asymptote of the graph of #y=(-2x^6+5x+8)/(8x^6+6x+5)# ? How do you find the horizontal asymptote of the graph of #y=(-4x^6+6x+3)/(8x^6+9x+3)# ? How do you find the horizontal asymptote of the graph of y=3x^6-7x+10/8x^5+9x+10? How do you find the horizontal asymptote of the graph of #y=6x^2# ? How can i find horizontal asymptote? How do you find horizontal asymptotes using limits? What are all horizontal asymptotes of the graph #y=(5+2^x)/(1-2^x)# ? See all questions in Limits at Infinity and Horizontal Asymptotes Impact of this question 1364 views around the world You can reuse this answer Creative Commons License