x20≤x<1 例25.讨论函数f(x12-x1xs2在x=1 处的连续性 解 f(1=1 H limf(x)=limx =1 x-1 lim f(x)=lim(2-x)=1 x→1+ lim f(x=f()=l 故f(x)在x=1处连续8 例25. 讨论函数 在x = 1 2 0 1 ( ) 2 1 2 x x f x x x    =   −   解 (1) 1 f = 故 f x x ( ) 1 . 在 = 处连续 2 1 1 lim ( ) lim 1 x x f x x → → − − 且 = = 1 1 lim ( ) lim(2 ) 1 x x f x x → → + + = − = 1 lim ( ) (1) 1 x f x f →  = = 处的连续性