证明 记 t =ⁿ√(x+1)-1,则 x = (t+1)ⁿ-1,有 x→0 <==> t→0,于是, lim(x→0)[ⁿ√(x+1)-1]/(x/n) = lim(t→0)(nt)/[(t+1)ⁿ-1] = lim(t→0){n/∑(0≤k≤n-1)[(t+1)^k]} = 1,得证。