How to calculate the limit of #((1+x)^{x}-1)^x# when #x->+0# ?

Question

644 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-04T21:40:50

#1#

For small #abs x#

#(1+x)^x = 1+x^2+(x^3(x-1))/(2!)+O(x^4)#

then

#((1+x)^x-1)^x = (x^2+(x^3(x-1))/(2!)+O(x^4))^x = #

#=x^(2x)(1+(x(x-1))/(2!) + O(x^2))^x#

so

#lim_(x->0)((1+x)^x-1)^x = lim_(x->0)x^(2x) lim_(x->0)(1+(x(x-1))/(2!) + O(x^2))^x = 1 xx 1 = 1#

NOTE

We assumed as known the textbook result

#lim_(x->0)x^x = 1#