Question

limx→00（x+1）（x2+1）（xn+1）/【（nx）n次方+1】n+1/2次方

Answer 1

lim (x->∞) (x+1)(x^2+1)...(x^n+1) /[(nx)^n + 1] ^( (n+1)/2)

分子 highest order of x = x^( n(n+1)/2)

分母 highest order of x = [n^(n(n+1)/2)] x^( n(n+1)/2)

lim (x->∞) (x+1)(x^2+1)...(x^n+1) /[(nx)^n + 1] ^( (n+1)/2)

= 1/[n^(n(n+1)/2)]