Question #89992 Calculus Limits Determining Limits Algebraically 1 Answer Jim H Oct 7, 2016 The limit is #0# Explanation: For #h > 0#, we have #1+h > 1#. So, for #h > 0#, we have #f(1+h) = ((1+h)-1)^2#. This simplifies to #f(1+h) = h^2# #" "# (for #h > 0#). Therefore, #lim_(hrarr0^+)f(1+h) - lim_(hrarr0^+)h^2 = 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 1237 views around the world You can reuse this answer Creative Commons License