x趨向於1,lim(ax+b)/(x^2-3x+2)=2,則a,b分別是多少

x趨向於1,lim(ax+b)/(x^2-3x+2)=2,則a,b分別是多少


lim(ax+b)/(x^2-3x+2)
=lim(ax+b)/(x-1)(x-2)
=-lim(ax+b)/(x-1)
=-lim(ax-a+a+b)/(x-1)
=-lim(a+(a+b)/(x-1))=2
x->1所以x-1->∞
原式極限存在,所以a+b=0且-a+0=2
a=-2,b=2