How do you find the limit of #(1+3/x)^(2x)# as x approaches negative infinity? Calculus Limits Determining Limits Algebraically 1 Answer Eddie Jun 22, 2016 #e^6# Explanation: #L = lim_{x \to - \infty} \ (1+3/x)^(2x)# let #1/p = 3/x, \qquad p = x/3# #\implies L = lim_{p \to - \infty} \ (1+1/p)^(6p) = lim_{p \to - \infty} \ ((1+1/p)^(p))^6# From well known limit: # \implies L = 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 7365 views around the world You can reuse this answer Creative Commons License