How do you find the limit of #(sqrt(x+4) -2) / x# as x approaches 0? Calculus Limits Determining Limits Algebraically 1 Answer Euan S. Aug 13, 2016 #1/4# Explanation: We have limit of indeterminate form, ie #0/0# so can use L'Hopital's rule: #lim_(xrarr0) (sqrt(x+4) - 2)/x = lim_(xrarr0)(d/(dx)(sqrt(x+4)-2))/(d/(dx)(x))# #=lim_(xrarr0)(1/(2sqrt(x+4)))/1 = 1/(2sqrt(0+4)) = 1/4# 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 5043 views around the world You can reuse this answer Creative Commons License