当x∈(0,π/2)时,证明：x/(1+x²)＜arctanx＜x,

Question

当x∈(0,π/2)时,证明：x/(1+x²)＜arctanx＜x,

Answer 1

arctanx导数=1/(1+x^2)
x/(1+x^2)导数=(1-x^2)/(1+x^2)^2.
令f(x)=arctanx-x/(1+x^2)
则f(x)‘=1/(1+x^2)-(1-x^2)/(1+x^2)^2=2x^2/(1+x^2)^2在x∈(0,π/2)时f(x)>0,单调增,
又因为f(0)=0,所以f(x)>0,即arctanx-x/(1+x^2)>0,得x/(1+x²)＜arctanx
同理,令g(x)=x-arctanx,g(x)'=1-1/(1+x^2)=x^2/(1+x^2),在x∈(0,π/2)时g(x)>0,单调增,
又因为g(0)=0,所以g(x)>0,即x-arctanx>0,得arctanx＜x
综上所诉,即证x/(1+x²)＜arctanx＜x成立 （注：x²=x^2）