What is the limit of #(1+x)^(1/x) # as x approaches 0?

1 Answer
Dec 19, 2016

Answer:

#e#

Explanation:

From the binomial expansion

#(1+x)^n = 1 + nx + (n(n-1))/(2!)x^2+(n(n-1)(n-2))/(3!)x^3+ cdots +#

we have

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

#=1+1+(1(1-x))/(2!)+(1(1-x)(1-2x))/(3!)+cdots+#

so

#lim_(x->0)(1+x)^(1/x) =lim_(x->0)1+1+(1(1-x))/(2!)+(1(1-x)(1-2x))/(3!)+cdots+=sum_(k=0)^oo1/(k!) = e#