How do you find #lim (1-2t^-1+t^-2)/(3-4t^-1)# as #t->0#? Calculus Limits Limits at Infinity and Horizontal Asymptotes 1 Answer Guillaume L. May 10, 2018 #Lim_"t->0"(1-2t^-1+t^-2)/(3-4t^-1)=±oo# depending on if #t->+oo# or #t->-oo# Explanation: #Lim_"t->0"(1-2t^-1+t^-2)/(3-4t^-1)# #=Lim_"t->0"(t²-2t+1)/(3t²-4t)# #=Lim_"t->0"-1/(4t)# #=±oo# depending on if #t->+oo# or #t->-oo# \0/ here's our answer! 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 1361 views around the world You can reuse this answer Creative Commons License