lim [√(x+1) -1 ] / (x/2)
= lim 2 [√(x+1) -1][√(x+1) +1] / x[√(x+1) +1]
= lim 2 / [√(x+1) +1]
= 1
故 √(x+1) - 1 和 x/2 是等价无穷小