求极限,例题x趋于0 lim∫下限为0上限为x[∫下限为0上限为u^2arctan(1+t)dt]du/x(1-cosx)=2lim x趋于0∫下限为0上限为x[∫下限为0上限u^2 arctan(1+t)dt]du/x^3这个前面那个2是怎么来的!= x趋于0 2lim∫下限0上限为x^2 arctan(1+t)du/3x^2= x趋于0 2lim 2xarctan(1+x^2)/6x=2/3 乘 π/4=π/6
问题描述：
求极限,例题
x趋于0 lim∫下限为0上限为x[∫下限为0上限为u^2arctan(1+t)dt]du/x(1-cosx)
=2lim x趋于0∫下限为0上限为x[∫下限为0上限u^2 arctan(1+t)dt]du/x^3
这个前面那个2是怎么来的!
= x趋于0 2lim∫下限0上限为x^2 arctan(1+t)du/3x^2
= x趋于0 2lim 2xarctan(1+x^2)/6x
=2/3 乘 π/4
=π/6
答
1/(1-cosx)=1/[1-(1-2sin(x/2)·sin(x/2)·)]
=1/[2sin(x/2)·sin(x/2)]
等价于1/(2·x/2·x/2)=2/x^2(当x趋向于0时).
那个2就是这里的分子2.