limx→1{(x2+ax+b)/[sin(x2-1)]}=3，求a，b的值．
【正确答案】：由于limx→1{(x2+ax+b)/[sin(x2-1)]} =limx→1[(x2+ax+b)/(x2-1)]=3, 因此一定有limx→1{(x2+ax+b)=0，故设x2ax+b=(x-1)(x+c)， 所以3=limx→1{[(x-1)(x+c)]/(x2-1)}=limx→1[(x+c)/(x+1)]=(1+c)/2, 因此c=5,x2+ax+b=(x-1)(x+5)=x2+4x-5,所以a=4,b=-5.
