1-x,x<0 例5设f(x)= 求lmf(x) x2+1,x≥0x→0 解x=0是函数的分段点两个单侧极限为 f(x)=lim(1-x)=1, y →0 →0 lim f(x)=lim(x+1)=1 x→0 x→0 J=x+1 左右极限存在且相等, 0 故limf(x)=1.例5 , lim ( ). 1, 0 1 , 0 ( ) 0 2 f x x x x x f x x→    +  −  设 = 求 y o x 1 y = 1 − x 1 2 y = x + 解 x = 0是函数的分段点,两个单侧极限为 lim ( ) lim (1 ) 0 0 f x x x x = − → − → − = 1, lim ( ) lim ( 1) 2 0 0 = + → + → + f x x x x = 1, 左右极限存在且相等, lim ( ) 1. 0 = → f x x 故