How do you find #lim (1-5t^-1)/(4+6t^-1)# as #t->0#? Calculus Limits Limits at Infinity and Horizontal Asymptotes 1 Answer Gerardina C. Feb 11, 2017 #5/6# Explanation: Since #t^-1=1/t# you could rewrite: #lim_(t->0) (1-5/t)/(4-6/t)=lim_(t->0) ((t-5)/cancelt)/((4t-6)/cancelt)=lim_(t->0)(t-5)/(4t-6)=5/6# 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 1362 views around the world You can reuse this answer Creative Commons License