How do you find #lim sqrtx+sqrt(x+1)# as #x->oo#? Calculus Limits Limits at Infinity and Horizontal Asymptotes 1 Answer Andrea S. Mar 17, 2017 #lim_(x->oo) sqrtx+ sqrt(x+1) = +oo# Explanation: We have: #lim_(x->oo) sqrtx = +oo# and #lim_(x->oo) sqrt(x+1) = +oo# So: #lim_(x->oo) sqrtx+ sqrt(x+1) = +oo# 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 2626 views around the world You can reuse this answer Creative Commons License