设函数f(x)=arctanx,g(x)=x,x＞0
f(0)=0,g(0)=0
f'(x)=1/(1+x²)＞0,g'(x)=1＞0
f'(x)-g'(x)=1/(1+x²)-1=-x²/(1+x²)≤0
即f'(x)≤g'(x)
因为[0,+∞)上f(x)与g(x)单调递增且f'(x)≤g'(x)
所以x>arctan(x)
则arctanx/x