How do you find the limit of #(x+5)/(25-x^2)# as x approaches #5^-#? Calculus Limits Determining Limits Algebraically 1 Answer Konstantinos Michailidis May 21, 2016 It is #lim_(x->5^-) (x+5)/(25-x^2)=lim_(x->5^-) (x+5)/((5-x)*(5+x))= lim_(x->5^-) cancel(x+5)/((5-x)*cancel(5+x))= lim_(x->5^-) 1/(x-5)=-oo# 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 1300 views around the world You can reuse this answer Creative Commons License