切点(1,0)在f(x)上f(1)=0y=x²-xy'=2x-1y'(1)=1=f'(1)lim(n→∞)nf[n/(n+2)]令t=1/n原极限=lim(t→0)f[(1/t)/(1/t+2)]·1/t=lim(t→0)f[1/(1+2t)]/t=lim(t→0)f(1)/t为0/0型，采用洛必达法则=lim(t→0)f'(1)=1
