例1求下列函数的极限 x+3 (1)lim(x2-3x+4)(2)lim x-1 x→1x2-3x+4 (3)1inx2+x-2 x→1x2+3x-4 AE (1) lim(x2-3x+4)=limx2-3limx+lim4 =(imx)2-3×1+4=12-3+4 x→1 =2 (2)由(1),分母的极限不为零,故由商的极限 运算法则,有 x+3 im(x+3) lim x+ lim3 1+3 x2-3x+4im(x2-37=2=22 lim x→>1 →1例1 求下列函数的极限 2 1 (1) lim( 3 4) x x x → − + 2 1 (1) lim( 3 4) x x x → 解 − + (2) 由(1)，分母的极限不为零，故由商的极限 运算法则，有 2 1 3 lim x 3 4 x → x x + − + 2 1 1 1 lim 3lim lim 4 x x x x x → → → = − + 2 1 3 (2) lim x 3 4 x → x x + − + 2 2 1 2 (3) lim x 3 4 x x → x x + − + − 2 2 1 (lim ) 3 1 4 1 3 4 x x → = −  + = − + = 2 1 2 1 lim( 3) lim( 3 4) x x x x x → → + = − + 1 1 lim lim3 2 x x x → → + = 1 3 2 + = = 2