How do you find #lim root3(t+4)-root3t# as #t->oo#? Calculus Limits Limits at Infinity and Horizontal Asymptotes 1 Answer Cesareo R. Mar 1, 2017 #0# Explanation: #root3(t+4)-root3t = root3(t)(root3(1+4/t)-1)# but #root3(1+4/t) = 1+1/3(4/t)+(1/3(1/3-1))/(2!)(4/t)^2+cdots# and #root3(t)(root3(1+4/t)-1) = root3(t)(-1/3(4/t)-(1/3(1/3-1))/(2!)(4/t)^2-cdots) =# #=1/t^(2/3)(-4/3+f(1/t))# so #lim_(t->oo)root3(t+4)-root3t = lim_(t->oo)1/t^(2/3)(-4/3+f(1/t))=0# 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 1303 views around the world You can reuse this answer Creative Commons License