#lim_(x->oo)((x+1)/(x-2))^(2x-1)=# ? Calculus Limits Determining Limits Algebraically 1 Answer Cesareo R. Nov 10, 2016 #e^6# Explanation: #((x+1)/(x-2))=((x-2)+3)/(x-2)=1+3/(x-2)# now calling #y = (x-2)/3# we have #((x+1)/(x-2))^(2x-1)=(1+1/y)^(6y+3)# so #lim_(x->oo)((x+1)/(x-2))^(2x-1)=lim_(y->oo)(1+1/y)^(6y)lim_(y->oo)(1+1/y)^3 = (lim_(y->oo)(1+1/y)^y)^6 (1+lim_(y->oo)1/y)^3=e^6# 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 5023 views around the world You can reuse this answer Creative Commons License