How do you find #lim sqrtt(sqrt(t+2)-sqrt(t+1))# as #t->oo#? Calculus Limits Limits at Infinity and Horizontal Asymptotes 1 Answer Cesareo R. Feb 13, 2017 #1/2# Explanation: #sqrtt(sqrt(t+2)-sqrt(t+1))=sqrtt((t+2-(t+1)))/(sqrt(t+2)+sqrt(t+1))=# #(sqrtt/sqrtt) 1/((sqrt(1+2/t)+sqrt(1-1/t)))=1/(sqrt(1+2/t)+sqrt(1+1/t))# so #lim_(t->oo)sqrtt(sqrt(t+2)-sqrt(t+1))=lim_(t->oo)1/(sqrt(1+2/t)+sqrt(1+1/t))=1/2# 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