What is the limit of #(sqrt x) / (x + 4)# as x approaches infinity? Calculus Limits Determining Limits Algebraically 1 Answer A. S. Adikesavan Mar 12, 2016 0 Explanation: #lim to# infinity of the ratio of #x^m/x^n# is 0 or infinity, according as, when m < or < n. . Divide by #sqrtx# both numerator and denominator. The function becomes #1/(sqrtx + 4/sqrtx)# As #xto# infinity, the limit is 1#/#(infimity + 0) = 0. Answer link Related questions How do you find the limit #lim_(x->5)(x^2-6x+5)/(x^2-25)# ? How do you find the limit #lim_(x->3^+)|3-x|/(x^2-2x-3)# ? How do you find the limit #lim_(x->4)(x^3-64)/(x^2-8x+16)# ? How do you find the limit #lim_(x->2)(x^2+x-6)/(x-2)# ? How do you find the limit #lim_(x->-4)(x^2+5x+4)/(x^2+3x-4)# ? How do you find the limit #lim_(t->-3)(t^2-9)/(2t^2+7t+3)# ? How do you find the limit #lim_(h->0)((4+h)^2-16)/h# ? How do you find the limit #lim_(h->0)((2+h)^3-8)/h# ? How do you find the limit #lim_(x->9)(9-x)/(3-sqrt(x))# ? How do you find the limit #lim_(h->0)(sqrt(1+h)-1)/h# ? See all questions in Determining Limits Algebraically Impact of this question 1705 views around the world You can reuse this answer Creative Commons License