11(0 )a12(x0)…a1n(x0) A(x)= 21(0 22(0 2n(0 (x0)an2(x0) X nn 0 容易验证下面的等式是成立的 设 lim A()=A, lim b(x)=B x->x0 x→x (1)lim(4(x)±B(x)=A±B x→)x 011 0 12 0 1 0 21 0 22 0 2 0 0 1 0 2 0 0 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) n n m m mn a x a x a x a x a x a x A x a x a x a x     =         容易验证下面的等式是成立的： 设 则 0 0 lim ( ) , lim ( ) x x x x A x A B x B → → = = 0 (1) lim( ( ) ( )) x x A x B x A B →  = 