根据函数单调性定义证明：f(x)=x/(x^2+1)在(-1,1)上为增函数

任取x1,x2∈（-1,1）Δx=x1-x2>0Δy=x2/(x2^2+1)-x1/(x1^2+1)=[x2(x1^2+1)-x1(x2^2+1)]/[(x2^2+1)(x1^2+1)]=(x2*x1^2-x1*x2^2+x2-x1)/[(x2^2+1)(x1^2+1)]=[x1*x2(x1-x2)+x2-x1]/[(x2^2+1)(x1^2+1)]=(1-x1*x2)(x2-x1)/[(x2^2+1)(x1^2+1)]>0所以为增函数