How do you evaluate the limit #sqrt(2x+3)# as x approaches #3#? Calculus Limits Determining Limits Algebraically 1 Answer Alan P. Dec 19, 2016 #lim_(xrarr3)sqrt(2x+3)=3# Explanation: Since #sqrt(2x+3)# is defined when #x=3# the limit as #x# approaches #3# is simply the value of #sqrt(2x+3)# when #3# is substituted for #x# #sqrt(2x+3)# with #x=3# #color(white)("XXX")= sqrt(2 * 3 +3)=sqrt(9) =3# 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 4895 views around the world You can reuse this answer Creative Commons License