高数极限求解LIM X趋向于2 （X平方+AX+B)/(X-2)=5 求A B

lim(x->2)[(x²+Ax+B)/(x-2)]=5
=lim(x->2)(2x+A)=5,用洛必达法则
=2(2)+A=5
A=1
lim(x->2)[(x²+x+B)/(x-2)]=5
令x²+x+B=(x-2)(x+K)
x²+x+B=x²+(K-2)x-2K
K-2=1,K=3,所以B=-2K=-6
lim(x->2)[(x²+x-6)/(x-2)]
=lim(x->2)[(x-2)(x+3)/(x-2)]
=lim(x->2)(x+3)=2+3=5
验算正确,所以A=1,B=-6