How do you find the limit of #sqrt(x+1)/(x-4)# as x approaches 3? Calculus Limits Determining Limits Algebraically 1 Answer Ernest Z. · Jim H Jul 15, 2015 #lim_{xto3}sqrt(x+1)/(x-4) = -2# Explanation: Let's try direct substitution. #sqrt(x+1)/(x-4) = sqrt(3+1)/(3-4) = sqrt4/(-1)= 2/(-1) = -2# So the function is continuous at #x = 3#. #lim_{xto3}sqrt(x+1)/(x-4) = -2# 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 1571 views around the world You can reuse this answer Creative Commons License