lim x趋向正无穷 [(1+x)^3/2]/[x^1/2]-x 答案是3/2 求详细过程与思路

Question

如题所述

Answer 1

解：lim x->+∞ [(1+x)^3/2]/[x^1/2]-x
=lim x->+∞ [(1+x)^(3/2)-x^(3/2)]/[x^(1/2)] (∞/∞，用罗比达法则)
=lim x->+∞ [3/2*(1+x)^(1/2)-3/2*x^(1/2)]/[1/2*x^(-1/2)]
=lim x->+∞ 3*[(1+x)^(1/2)-x^(1/2)]*x^(1/2)]
=lim x->+∞ 3*[√(1+x)-√x]*√x 分母分子同乘以[√(1+x)+√x]
=lim x->+∞ 3*[(1+x)-x]*√x /[√(1+x)+√x]
=lim x->+∞ 3√x /[√(1+x)+√x]
=lim x->+∞ 3 /[√(1/x+1)+1]
=3/(1+1)=3/2

Answer 2

lim x->+∞ [(1+x)^3/2]/[x^1/2]-x 分子分母同乘以 （√((1+x)³/x)+x）整理得
=lim x->+∞ (3x²+3x+1)/(√(x(1+x)³)+x²)
=lim x->+∞ 3/(√(1+1/x³)+1) + 3/(√((1+x)³/x)+x)+1/(√(x(1+x)³)+x²)
=3/(1+1)+0+0
=3/2