高数无穷小的题：1.2.3.

1 lim [S(0,x^2)ln(1+t^2)dt/x^(2k/3)= lim 2xln(1+x^4)/x^(2k/3-1)2k/3=lim 2x^5/x^(2k/3-1)2k/3=A
2k/3-1=5,k=9
9阶无穷小
2 （1）lim ln（e^2x+2x^2）/kx^n=lim(e^2x+2x^2-1)/kx^n=lim(2e^2x+4x)/knx^(n-1)=1
n-1=0,n=1,2/kn=1,k=2
(2)limS ./kx^n=limln(1+arctanx)/knx^(n-1)=limarctanx/knx^(n-1)=limx/knx^(n-1)=1
n-1=1,n=2,1/kn=1,k=1/2
(3)lim{sin(x+x^3)-x}/kx^n=lim{(1+3x^2)cos(x+x^3)-1}/knx^(n-1)=lim{6xcos(x+x^3)-(1+3x^2)^2 sin(x+x^3)}/kn(n-1)x^(n-2)=
3 lima/x^n=limcosx^2/nx^(n-1)=A n=1
limb/x^n=lim2xtanx/nx^(n-1)=lim2x^2/nx^(n-1)=A ,n=3
limr/x^n=lim 0.5x^(-1/2)sinx^(3/2)/nx^(n-1)=lim 0.5x/nx^(n-1),n=2
阶数从高到低b,r,a

1. 当x→0时【0，x²】[∫ln(1+t²)dt]是x^(2/3)的几阶无穷小？

   x→0lim{【0，x²】[∫ln(1+t²)dt]/x^(2/3)}=x→0lim[2xln(1+x⁴)]/[(2/3)x^(-1/3)]

   =x→0lim[3x^(4/3)ln(1+x⁴)]=0，故当x→0时...