y=x^2-x,y'=2x-1,y'(1)=1,曲线y=f(x)和y=x^2-x在点(1,0)处有公共的切线,则f(1)=0,f'(1)=1,n→∞时nf[n/(n+2)]=f[n/(n+2)]/(1/n)=f'[n/(n+2)]*2/(n+2)^2/(-1/n^2)→-2f'（1）=-2.
